How can I evaluate $$\sum_{n=1}^\infty\frac{2n}{3^{n+1}}$$? I know the answer thanks to Wolfram Alpha, but I'm more concerned with how I can derive that answer. It cites tests to prove that it is convergent, but my class has never learned these before. So I feel that there must be a simpler method.

In general, how can I evaluate $$\sum_{n=0}^\infty (n+1)x^n?$$

  • 6
    $\begingroup$ I believe this is an arithmo-geometric series. You can find information here: artofproblemsolving.com/wiki/… $\endgroup$
    – Jason Kim
    Commented Jul 7, 2018 at 19:11
  • $\begingroup$ Express 2 as 3-1 then decompose it into 2 terms. Then add 1 and subtract 1 to the numerator of the term with 3^(n+1) as denominator. Then apply telescoping series for first two terms and u will get an infinite g.p with 3rd term. $\endgroup$
    – Infinite
    Commented Mar 13, 2022 at 17:16

24 Answers 24


No need to use Taylor series, this can be derived in a similar way to the formula for geometric series. Let's find a general formula for the following sum: $$S_{m}=\sum_{n=1}^{m}nr^{n}.$$

Notice that \begin{align*} S_{m}-rS_{m} & = -mr^{m+1}+\sum_{n=1}^{m}r^{n}\\ & = -mr^{m+1}+\frac{r-r^{m+1}}{1-r} \\ & =\frac{mr^{m+2}-(m+1)r^{m+1}+r}{1-r}. \end{align*}
Hence $$S_m = \frac{mr^{m+2}-(m+1)r^{m+1}+r}{(1-r)^2}.$$
This equality holds for any $r$, but in your case we have $r=\frac{1}{3}$ and a factor of $\frac{2}{3}$ in front of the sum. That is \begin{align*} \sum_{n=1}^{\infty}\frac{2n}{3^{n+1}} & = \frac{2}{3}\lim_{m\rightarrow\infty}\frac{m\left(\frac{1}{3}\right)^{m+2}-(m+1)\left(\frac{1}{3}\right)^{m+1}+\left(\frac{1}{3}\right)}{\left(1-\left(\frac{1}{3}\right)\right)^{2}} \\ & =\frac{2}{3}\frac{\left(\frac{1}{3}\right)}{\left(\frac{2}{3}\right)^{2}} \\ & =\frac{1}{2}. \end{align*}

Added note:

We can define $$S_m^k(r) = \sum_{n=1}^m n^k r^n.$$ Then the sum above considered is $S_m^1(r)$, and the geometric series is $S_m^0(r)$. We can evaluate $S_m^2(r)$ by using a similar trick, and considering $S_m^2(r) - rS_m^2(r)$. This will then equal a combination of $S_m^1(r)$ and $S_m^0(r)$ which already have formulas for.

This means that given a $k$, we could work out a formula for $S_m^k(r)$, but can we find $S_m^k(r)$ in general for any $k$? It turns out we can, and the formula is similar to the formula for $\sum_{n=1}^m n^k$, and involves the Bernoulli numbers. In particular, the denominator is $(1-r)^{k+1}$.

  • 6
    $\begingroup$ @Eric How do you make the transformation $\sum_{n=1}^m={r-r^{m+1} \over 1-r}$? Secondly at this step you can substitute the series with this explicit formula as the series converges (obviously because it is finite). If the series was infinite you couldn't have done that (unless $|r| \lt 1$) as it would diverge. However later you apply this formula to an infinite series $\sum_{n=1}^{\infty}{2n \over 3^n+1}$. Could you explain why you consider it to be suitable for an infinite series, albeit it was initially brought out for finite series? $\endgroup$ Commented Jun 5, 2014 at 16:38
  • 12
    $\begingroup$ Small nitpick: "equality holds for any $r$" should be "equality holds for any $r\neq1$" $\endgroup$ Commented Feb 25, 2017 at 5:27
  • $\begingroup$ I'm lost regarding how exactly the RHS of the first step is derived. $\endgroup$
    – RTF
    Commented May 2, 2023 at 23:59

If you want a solution that doesn't require derivatives or integrals, notice that \begin{eqnarray} 1+2x+3x^2+4x^3+\dots = 1 + x + x^2 + x^3 + \dots \\ + x + x^2+ x^3 + \dots\\ + x^2 + x^3 + \dots \\ +x^3 + \dots \\ + \dots \\ =1 + x + x^2 + x^3+\dots \\ +x(1+x+x^2+\dots) \\ +x^2(1+x+\dots)\\ +x^3(1+\dots)\\ +\dots \\ =(1+x+x^2+x^3+\dots)^2=\frac{1}{(1-x)^2} \end{eqnarray}

  • 28
    $\begingroup$ Your solution has a large gap. You will find it difficult to prove that the series converges without using as much technical machinery as the solution that goes through the finite sum before taking limits. $\endgroup$
    – user21820
    Commented Aug 16, 2015 at 5:39
  • 11
    $\begingroup$ @user21820: The proof as it stands (replacing the ellipses by a precise description of the general terms they stand for) is perfectly valid for if expressions are interpreted as formal power series in$~x$, in other words it shows that $\sum_{n\geq0}(n+1)X^n=(1-X)^{-2}$ in $\Bbb Z[[X]]$. With this established (or actually independently of it) is it very easy to show that the power series in both members have radius of convergence$~1$ (it suffices to observe that the coefficients increase, but less than exponentially), obtaining an identity of convergent power series within that radius. $\endgroup$ Commented Feb 24, 2017 at 5:50
  • 2
    $\begingroup$ @MarcvanLeeuwen: You say "very easy", but it is still much easier to just truncate at the finite terms because the remainder term goes to zero by elementary means. That is why I said very precisely "without using as much technical machinery as ...", which is practically none. Micah's answer as stated is extremely misleading to students, for the same reason that most people can't see the flaw in the proof of $1+2+3+\cdots = -\frac1{12}$? Nevertheless, if one wants to develop the general tool of generating functions for combinatorics, then yes we should develop formal power series. $\endgroup$
    – user21820
    Commented Feb 24, 2017 at 5:55
  • 4
    $\begingroup$ @MarcvanLeeuwen: Besides, I am appalled at the general lack of rigour in half the answers here, especially since the question itself clearly states that Wolfram Alpha cites convergence tests and asks for a simpler way, so anything that requires convergence tests is not simpler. $\endgroup$
    – user21820
    Commented Feb 24, 2017 at 6:06
  • 10
    $\begingroup$ @user21820: I don't understand what you mean. I don't see how the proof using truncation would go precisely, nor how it would easy because the crucial factoring of the sum as a product of two (equal) sums is only valid for the infinite sum. Your original comment does not sound as if it want to say (rather vacuously) "without using at least $0$ effort" either. Remains that you worry about misleading your students; I disagree. There is nothing wrong with the approach of proving something for formal power series first, then considering convergence afterwards. We should teach our students that $\endgroup$ Commented Feb 24, 2017 at 6:08

As indicated in other answers, you can reduce this to summing $\displaystyle{\sum_{n=1}^\infty na^n}$ with $|a|<1$ (by pulling out the constant $\frac{2}{3}$ and rewriting with $a=\frac{1}{3}$). This in turn can be reduced to summing geometric series by rearranging and factoring. Note that, assuming everything converges nicely (which it does):

$\begin{matrix} &a & + & 2a^2 & + & 3a^3 &+& 4a^4 &+& \cdots\\ =&a &+& a^2 &+& a^3 &+& a^4 &+& \cdots\\ +& & & a^2 &+& a^3 &+& a^4 &+& \cdots\\ +& & & & & a^3 &+& a^4 &+& \cdots\\ +& & & & & & & a^4 &+& \cdots\\ +& & & & & & & & & \vdots \end{matrix}$

Factoring out the lowest power of $a$ in each row yields

$\begin{align*} \sum_{n=1}^\infty na^n &= a(1+a^2+a^3+\cdots)\\ &+ a^2(1+a^2+a^3+\cdots)\\ &+ a^3(1+a^2+a^3+\cdots)\\ &+ a^4(1+a^2+a^3+\cdots)\\ &\vdots \end{align*}$

Each row in the last expression has the common factor $a(1+a+a^2+a^3+\cdots)$, and factoring this out yields

$\begin{align*}\sum_{n=1}^\infty na^n &=a(1+a+a^2+a^3+\cdots)(1+a+a^2+a^3+\cdots)\\ &=a(1+a+a^2+a^3+\cdots)^2.\end{align*}$

Now you can finish by summing the geometric series.

Eric Naslund's answer was posted while I was writing, but I thought that this alternative approach might be worth posting. I also want to mention that in general one should be careful about rearranging series as though they were finite sums. To be more formal, some of the steps above would require justification based on absolute convergence.


Factor out the $\frac{2}{3}$. Then write $$\sum_{n=1}^{\infty} \frac{n}{3^n} = \sum_{n=1}^{\infty} \frac{1}{3^n} + \sum_{n=2}^{\infty} \frac{1}{3^n} + \sum_{n=3}^{\infty} \frac{1}{3^n} + \cdots$$

It is easy to show that $$\sum_{n=m}^{\infty} \frac{1}{3^n} = \frac{3}{2} \left(\frac{1}{3} \right)^m$$ and so $$\sum_{n=1}^{\infty} \frac{n}{3^n} = \frac{3}{2} \sum_{n=1}^{\infty} \left( \frac{1}{3} \right)^n $$ which you can sum. Don't forget to put the $\frac{2}{3}$ back in.


My favorite proof of this is in this paper of Roger B. Nelsen

I also have the following method for $\sum_{n=1}^\infty {n\over 2^{n-1}}$ (one can use a similar method for $\sum_{n=1}^\infty {n\over3^n}$):

We first show that $\sum\limits_{n=7}^\infty {n\over 2^{n-1}} ={1\over4}$.

We start with a rectangle of width 1 and height $1/4$. Divide this into eights: enter image description here

Now divide each eighth-rectangle above in half and take 7 of them. This gives $A_1={7\over 2^6}$.

enter image description here There are $2\cdot8-7=9$ boxes left over, each having area $2^{-6}$.

Divide each remaining $16^{\rm th}$-rectangle in half and take 8 of them. This gives $A_2={7\over 2^6}+{8\over 2^7}$.

enter image description here There are $2\cdot9-8=10$ boxes left over, each having area $2^{-7}$.

Divide each remaining $32^{\rm nd}$-rectangle in half and take 9 of them. This gives $A_3={7\over 2^6}+{8\over 2^7}+{9\over 2^8}$.

enter image description here There are $2\cdot10-9=11$ boxes left over, each having area $2^{-8}$.

Divide each remaining $64^{\rm th}$-rectangle in half and take 10 of them. This gives $A_4={7\over 2^6}+{8\over 2^7}+{9\over 2^8}+{10\over2^9}$.

enter image description here There are $2\cdot11-9=12$ boxes left over, each having area $2^{-9}$.

At each stage, we double the number of remaining boxes, keeping the same leftover area, and take approximately half of them to form the next term of the series.

At the $n^{\rm th}$ stage, we have $$A_n= {7\over 2^6}+{8\over 2^7}+\cdots+{6+n\over2^{5+n}},$$

with leftover area $$ 2(n+7)-(n+6)\over 2^{n+5}.$$

It follows that, $$ {7\over2^6}+{8\over2^7}+{9\over2^8}+\cdots= {1\over4}. $$ Consequently, $$ \sum_{n=1}^\infty{n\over 2^{n-1}}= \sum_{n=1}^6 {n\over 2^{n-1}} +\sum_{n=7}^\infty{n\over 2^{n-1}} ={15\over 4}+{1\over4}=4. $$

You can also "Fubini" this (I think this is what Jonas is doing).



  1. You know (don't you?) the formula for $S(a) = \sum_{n=0}^\infty a^n$ for $|a| < 1$

  2. Take the derivative (with respect to $a$) of both sides to obtain a formula for $\sum_{n=1}^\infty n a^n$

  3. Show that your series can be put in that form.

  • 1
    $\begingroup$ Thanks for answering. 1) No I don't know that formula 2) Can you explain by what you mean by derive both sides? $\endgroup$
    – backus
    Commented Apr 3, 2011 at 21:59
  • 3
    $\begingroup$ 1. See here: en.wikipedia.org/wiki/Geometric_series 2) Derivation (or differentiation) is a mathematical operation (well, more than that) that is taught in Calculus - if you don't know about it, forget about my answer (and xen0m's). Perhaps you can solve your problem without knowing Calculus, by other methods... but normally you first learn calculus, then solve series. $\endgroup$
    – leonbloy
    Commented Apr 3, 2011 at 22:05

Note that $\int \{1 + 2x + 3x^2 + \cdots\} \, dx = x + x^2 + x^3 + \cdots + \text{const}$, i.e., a geometric series, which converges to $x/(1 - x)$ if $|x| < 1$. Therefore, $$\frac{d}{dx} \left(\frac{x}{1 - x}\right) = \frac{(1 - x)(1) - x(-1)}{(1 - x)^2} = \frac{1}{(1 - x)^2},$$ that is, $$1 + 2x + 3x^2 + \cdots = \frac{1}{(1 - x)^2}.$$

Another proof: Let $S = 1 + 2x + 3x^2 + \cdots$ with $|x| < 1$. Then $$xS = x + 2x^2 + 3x^3 + \cdots$$ so $$S - xS = (1- x)S = 1 + x + x^2 + \cdots = \frac{1}{1- x}.$$ Therefore: $S = (1 - x)^{-2}$.

  • $\begingroup$ +1, but wouldn't it have been simpler to say that it was the derivative of $1+x+x^2+...$, converging to $\frac1{1-x}$? $\endgroup$
    – Mike
    Commented Oct 29, 2012 at 22:46
  • $\begingroup$ I decided to start with what was given, so it is easier for the OP to see. $\endgroup$
    – glebovg
    Commented Oct 29, 2012 at 22:48
  • $\begingroup$ What I meant is that you chose $c=0$ while $c=1$ is a more well known series and is easier to take the derivative of. $\endgroup$
    – Mike
    Commented Oct 29, 2012 at 23:23
  • $\begingroup$ We could notice that the given series is converging to $\frac{1}{1-x}-1$ and take the derivative of that. $\endgroup$
    – inkievoyd
    Commented Nov 11, 2014 at 21:03
  • $\begingroup$ @inkievoyd That is exactly what I did. Note that $(1 - x)^{-1} - 1 = x(1 - x)^{-1}$. $\endgroup$
    – glebovg
    Commented Nov 12, 2014 at 11:47

You can find by differentiation. Just notice that $(x^n)' = nx^{n-1}$. By the theory of power series we obtain (by uniform convergence on any compact subset of $(-1,1)$) that $$ \left(\sum_{n=1}^\infty x^n\right)' = \sum_{n=1}^\infty (x^n)' = \sum_{n=1}^\infty n x^{n-1}. $$ The sum on the left hand side is equal to $\left(\frac{x}{1-x}\right)'$. You need to notice that your sum can be written in a similar way as $\sum_{n=1}^\infty nx^{n-1}$.

  • $\begingroup$ Thank you for helping, but I have never learned differentiation. $\endgroup$
    – backus
    Commented Apr 3, 2011 at 21:58
  • 2
    $\begingroup$ The sum $\sum x^n$ is equal to $\dfrac{1}{1-x}$ and therefore $\sum nx^n=x\left(\frac{1}{1-x}\right)'$ gives the correct result instead of $\left(\frac{x}{1-x}\right)'$. $\endgroup$
    – eyedropper
    Commented Mar 9, 2016 at 17:23

Consider the generating function $$g(x)=\sum_{n=0}^\infty{n+k-1\choose n}x^n={1\over (1-x)^k}.$$ If we let $k=2$, then $$\sum_{n=0}^\infty{n+1\choose n}x^n={1\over (1-x)^2}.$$ Since ${n+1\choose n}=(n+1)$ we can conclude that $$\sum_{n=0}^\infty{(n+1)x^n}={1\over (1-x)^2}.$$


Let be $$S_n(z)=\sum_{j=1}^{+\infty}j^nz^j\quad\text{for }z\in\Bbb{C}, |z|<1, n=0, 1, 2, \ldots $$ It's easy to prove that for $z\in\Bbb{C}, |z|<1$, the sums $S_n(z)$ satisfy the auto-convolutional recurrence relation $$ S_{n+1}(z)=S_n(z)+\sum_{k=0}^{n}\binom{n}{k} S_k(z)S_{n-k}(z)\qquad n=0, 1, 2, \ldots $$ Infact, performing the change index $q=j-i$ and using binomial theorem, we have $$ \begin{align} S_{n+1}(z)&=\sum_{j=1}^{+\infty}j^{n+1}z^j=\sum_{j=1}^{+\infty}j^{n}z^j+\sum_{i=1}^{+\infty}\sum_{j=i+1}^{+\infty}j^{n}z^j\\ &=S_n(z)+\sum_{i=1}^{+\infty}\sum_{q=1}^{+\infty}(i+q)^{n}z^{i+q}\\ &=S_n(z)+\sum_{i=1}^{+\infty}\sum_{q=1}^{+\infty}\sum_{k=0}^{n}\binom{n}{k}i^kq^{n-k}z^iz^q\\ &=S_n(z)+\sum_{k=0}^{n}\binom{n}{k}\sum_{i=1}^{+\infty}i^kz^i\sum_{q=1}^{+\infty}q^{n-k}z^q\\ &=S_n(z)+\sum_{k=0}^{n}\binom{n}{k} S_k(z)S_{n-k}(z) \end{align} $$

For $n = 0$ the sum $S_0(z)$ is the sum of geometric progression $$ S_0(z)=\sum_{j=1}^{+\infty}z^j=\frac{z}{1-z} $$ Using the recurrence we find $$ \begin{align} S_1(z)&=S_0(z)+S_0^2(z)=\frac{z}{(1-z)^2}\\ S_2(z)&=S_1(z)+2S_0(z)S_1(z)=\frac{z^2+z}{(1-z)^3}\\ S_3(z)&=S_2(z)+2S_0(z)S_2(z)+S_1^2(z)=\frac{z^3+4z^2+z}{(1-z)^4} \end{align} $$ and so on.

Using the founded results, for $a, b, z \in\Bbb{C}, z\neq 0,|z|<1$, putting $$\sigma(z;a,b)=\sum_{j=0}^{+\infty}(a+bj) z^j$$ one has $$ \sigma(z;a,b)=\sum_{j=0}^{+\infty}(a+bj) z^j=a[1+S_0(z)]+bS_1(z)=\frac{a+(b-a)z}{(1-z)^2} $$

So the required sum is $$ \sum_{n=0}^{+\infty}(n+1) x^n=\sigma(x;1,1)=\frac{1}{(1-z)^2} $$ and $$ \sum_{n=1}^{+\infty}\frac{2n}{3^{n+1}}=\frac{2}{3^2}\sigma\left(\frac{1}{3};1,1\right)=\frac{1}{2} $$

Note In alternative to the auto-convolution relation we can use another useful recursive relation for $z\in\Bbb{C}, |z|<1$, that is the linear recurrence $$ S_{n}(z)=\frac{z}{1-z}\left[1+\sum_{k=0}^{n-1}\binom{n}{k} S_k(z)\right]\qquad n=1, 2, \ldots $$


In fact, $$ \sum_{n=0}^{+\infty}(n+1)x^n = \sum_{n=0}^{+\infty}\frac{d}{dx}(x^{n+1})= \frac{d}{dx}\sum_{n=0}^{+\infty}x^{n+1} = \frac{d}{dx}\biggl(\frac{x}{1 - x}\biggr) = \frac{1}{(1 - x)^2} $$ For $x = \frac{1}{3}$, we have $$ \frac{9}{4} =\sum_{n=0}^{+\infty}(n+1)\frac{1}{3^n} = \sum_{m=1}^{+\infty}m\frac{1}{3^{m-1}} \quad \Rightarrow \quad \sum_{m=1}^{+\infty}\frac{m}{3^m} = \frac{3}{4} $$


Note that $n+1$ is the number ways to choose $n$ items of $2$ types (repetitions allowed but order is ignored), so that $n+1=\left(\!\binom2n\!\right)=(-1)^n\binom{-2}n$. (This uses the notation $\left(\!\binom mn\!\right)$ for the number of ways to choose $n$ items of $m$ types with repetition, a number equal to $\binom{m+n-1}n=(-1)^n\binom{-m}n$ by the usual definiton of binomial coefficients with general upper index.) Now recognise the binomial formula for exponent $-2$ in $$ \sum_{n\geq0}(n+1)x^n=\sum_{n\geq0}(-1)^n\tbinom{-2}nx^n =\sum_{n\geq0}\tbinom{-2}n(-x)^n=(1-x)^{-2}. $$ This is valid as formal power series in$~x$, and also gives an identity for convergent power series whenever $|x|<1$.

There is a nice graphic way to understand this identity. The terms of the square of the formal power series $\frac1{1-x}=\sum_{i\geq0}x^i$ can be arranged into an infinite matrix, with at position $(i,j)$ (with $i,j\geq0$) the term$~x^{i+j}$ . Now for given $n$ the terms $x^n$ occur on the $n+1$ positions with $i+j=n$ (an anti-diagonal) and grouping like terms results in the series $\sum_{n\geq0}(n+1)x^n$.

  • $\begingroup$ And notice that I didn't comment on your answer because you explicitly stated "formal power series" and also precisely stated the convergence radius. Now I do not believe this actually answers the question, for the reason I already gave you, namely that your answer requires tools that are not simpler than the convergence test used by WA, as requested in the question. But I have nothing wrong with your answer, since it is not mathematically incorrect or misleading. $\endgroup$
    – user21820
    Commented Feb 24, 2017 at 6:25

I assume that the $|x|$ to be less than $1$. Now, consider, $f(x)=\sum_{n=0}^{n=\infty} x^{n+1}$

This will converge only if $|x|<1$. Now, interesting thing here is, this is a geometric progression. The $f(x)=x/(1-x)$.

$f'(x)$ is the series you are interested in, right? Differentiate $x/(1-x)$ and you have your expression!


I first encountered this sum with the following problem:

Evaluate $$\bigg(\frac{1}{2}\bigg)^\dfrac{1}{3}\bigg(\frac{1}{4}\bigg)^\dfrac{1}{9}\bigg(\frac{1}{8}\bigg)^\dfrac{1}{27}\bigg(\frac{1}{16}\bigg)^\dfrac{1}{81}\dots$$

Which , of course simplified to $$\bigg(\frac{1}{2}\bigg)^{\dfrac{1}{3^1}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+\dfrac{4}{3^4}+\dots}=\bigg(\frac{1}{2}\bigg)^S$$

Getting back to your problem, now $$\sum_{n=1}^\infty \frac{2n}{3^{n+1}}=\frac{2}{3}\sum_{n=1}^\infty \frac{n}{3^n}=\frac{2}{3}S$$ Using a method similar to deriving geometric series suppose that $$S_k = \sum_{n=1}^k \frac{n}{3^{n}}$$ Then we have $$\begin{array}{lll} 3S_k-S_k &=& 1+\frac{2}{3^1}+\frac{3}{3^2}+\frac{4}{3^3}+\dots+\frac{k}{3^{k-1}}\\ &&-\frac{1}{3^1}-\frac{2}{3^2}-\frac{3}{3^3}\dots-\frac{k-1}{3^{k-1}}-\frac{k}{3^k}\\ 2S_k&=&\bigg(1+\frac{2-1}{3^1}+\frac{3-2}{3^2}+\frac{4-3}{3^3}+\dots+\frac{k-(k-1)}{3^{k-1}}\bigg)-\frac{k}{3^k}\\ &=&\frac{1-(\frac{1}{3})^{k}}{1-\frac{1}{3}} - \frac{k}{3^k}\\ 2S&=&\lim_{k\to\infty} \frac{1-(\frac{1}{3})^{k}}{1-\frac{1}{3}} - \frac{k}{3^k}\\ 2S&=&\frac{1}{1-\frac{1}{3}}=\frac{3}{2}\\ \frac{2}{3}S&=&\frac{1}{2}\\ \end{array}$$

by a similar method one can show, that if the series converges, that $$\sum_{n=0}^\infty (n+1)x^n = \frac{1}{(1-x)^2}$$


To avoid differentiating an infinite sum.

We start with the standard finite evaluation: $$ 1+x+x^2+...+x^n=\frac{1-x^{n+1}}{1-x}, \quad |x|<1. \tag1 $$ Then by differentiating $(1)$ we have $$ 1+2x+3x^2+...+nx^{n-1}=\frac{1-x^{n+1}}{(1-x)^2}+\frac{-(n+1)x^{n}}{1-x}, \quad |x|<1, \tag2 $$ and by making $n \to +\infty$ in $(2)$, using $|x|<1$, gives

$$ \sum_{n=0}^\infty(n+1)x^n=\frac{1}{(1-x)^2}. \tag3 $$


One method of evaluating $\sum_{n=0}^\infty(1+n)x^n$ can be like this, we take the generating function $$f = \sum_{n=0}^\infty x^n $$ then $$\sum_{n=0}^\infty (n+1)x^n = (xD + 1) f $$ $$ \frac{x}{(1-x)^2} + \frac{1}{1-x} = \frac{1}{(1-x)^2}$$ where $D$ means differentiation w.r.t. $x$.


No one like finite calculus notation? Unbelievable :(

I must add an answer in the form of finite calculus. You can read about this topic in the book Concrete Mathematics of Graham and Knuth, or this paper.

Finite calculus is analogous to the normal (infinitesimal) calculus where we use instead "discrete derivatives" and "discrete integrals" (actually just summations), and we can perform definite or indefinite sums in analogy to definite or indefinite integrals.

Analogously to the standard derivative the discrete derivative and the discrete (indefinite) integral can be written as

$$\Delta f(k):=f(k+1)-f(k),\quad\quad \sum f(k)\delta k=F(k)+C\tag{1}$$

for some $1$-periodic function $C$, and where we have too that

$$\sum_{k=a}^bf(k)=\sum\nolimits_a^{b+1}f(k)\delta k\tag{2}$$

And we have the summation by parts formula with this symbology represented by

$$\sum f(k)[\Delta g(k)]\delta k=f(k)g(k)-\sum \mathrm [E g(k)]f(k)\delta k\tag{3}$$

where $\mathrm E$ is the shift operator and is defined as $\mathrm E f(k):=f(k+1)$. By last, before to answer the question, it is not hard to check that

$$\Delta x^k=x^k(x-1),\quad\sum x^k\delta k=x^k(x-1)^{-1}+C\\\Delta (k+w)=1,\quad \sum (k+w)\delta k=\frac12 (k+w-1)(k+w)+C\tag{4}$$

Hence, using the above formulas, we have that

$$ \begin{align} \sum_{k=0}^\infty (k+1)x^k&=\sum\nolimits_0^\infty (k+1)x^k\delta k\\ &=\lim_{m \to \infty }\sum\nolimits_0^m (k+1)x^k\delta k\\ &=\lim_{m \to \infty }\big((k+1)x^k(x-1)^{-1}\big|_0^m-\sum\nolimits_0^m x^{k+1}(x-1)^{-1}\delta k\big)\\ &=\lim_{m \to \infty }\big((k+1)x^k(x-1)^{-1}-x^{k+1}(x-1)^{-2}\big)\big|_0^m\tag5 \end{align} $$

Then the above is finite when $|x|<1$, in this case we have that

$$ \sum_{k=0}^\infty (k+1)x^k=-\frac1{x-1}+\frac{x}{(x-1)^2}=\frac1{(x-1)^2}\tag6 $$


Solving $(an+b)-(a(n+1)+b)x=n+1$ for all $n$ gives $a=\frac1{1-x}$ and $b=\frac1{(1-x)^2}$. Therefore, $$ (n+1)x^n=\left(\frac1{(1-x)^2}+\frac{n}{1-x}\right)x^n-\left(\frac1{(1-x)^2}+\frac{n+1}{1-x}\right)x^{n+1}\tag1 $$ Using $(1)$ and telescoping series, we get $$ \sum_{k=0}^{n-1}(k+1)x^k=\frac1{(1-x)^2}-\left(\frac1{(1-x)^2}+\frac{n}{1-x}\right)x^n\tag2 $$ If $|x|\lt1$, then we get $$ \sum_{k=0}^\infty(k+1)x^k=\frac1{(1-x)^2}\tag3 $$

  • $\begingroup$ Hello robjohn, i need your help on that technique of generating valid telescoping series if you allow this ofc. $\endgroup$
    – Abr001am
    Commented Jun 6, 2018 at 20:29
  • $\begingroup$ @Abr001am: what is your question? $\endgroup$
    – robjohn
    Commented Jun 7, 2018 at 5:54
  • $\begingroup$ robjohn, you used this generating system of two equations that got you landed on a valid interleaving series, i tried as much with other series of different forms as often i return back from the starting point, feel like parcoursing a void circle, how can you tell if a series is reducible to a series of that sort ? $\endgroup$
    – Abr001am
    Commented Jun 9, 2018 at 15:42
  • $\begingroup$ @Abr001am: I think it is usually possible, but it may be as hard to find the telescoping terms as it is to find the closed form for the sum. $\endgroup$
    – robjohn
    Commented Jun 9, 2018 at 17:49

\begin{align} \sum_{n=0}^\infty (n+1)x^n &= \sum_{n=1}^\infty nx^{n-1} =\frac{d}{dx}\left( \sum_{n=0}^\infty x^{n}\right) =\frac{d}{dx}\left(\frac{1}{1-x}\right)=\frac{1}{(1-x)^2} \end{align} \begin{align} \tag*{$\Box$} \end{align}


What about the Cauchy product? $$ \sum_{n=0}^\infty(n+1)x^n=\sum_{n=0}^\infty\sum_{\ell=0}^nx^n=\left(\sum_{n=0}^\infty x^n\right)\left(\sum_{m=0}^\infty x^m\right)=\frac{1}{(1-x)^2}. $$


Suppose you play a repeatable game where you have probability $x$ of losing. Notice that

\begin{align} \sum\limits_{n=0}^{\infty} (n+1)x^n &= \sum\limits_{n=1}^{\infty} nx^{n-1} \\ &= \frac{1}{1-x} \sum\limits_{n=1}^{\infty} n (1-x)x^{n-1} \\ &= \frac{1}{1-x} \text{E}[n] \end{align}

where $\text{E}[n]$ is the expected number of times you have to play before your first win (you lose $n-1$ times and then win with probability $1-x$ in the $n$th game).


\begin{align} \text{E}[n] &= 1 + x \text{E}[n] \end{align}

because you either win in the first game, or you lose with probability $x$ and restart the process, with $\text{E}[n]$ more games to play. Then

\begin{align} \text{E}[n] &= \frac{1}{1-x} \end{align}

and thus

\begin{align} \sum\limits_{n=0}^{\infty} (n+1)x^n &= \frac{1}{(1-x)^2} \end{align}

(essentially we just re-derived the formula for the sum of a geometric series, but it's a fun way of doing it!)


Firstly, we need an infinite Geometric Series.

$$\sum_{n=0}^{\infty} x^{n}=\frac{1}{1-x} \quad \text { for }|x|<1 \tag*{(1)}$$ Multiplying the equation (1) by $x$ yields $$\sum_{n=0}^{\infty} x^{n+1}=\frac{x}{1-x}=\frac{1}{1-x}-1 \tag*{(2)} $$
Differentiating the equation (2) w.r.t. $x$ gives

$$\sum_{n=0}^{\infty}(n+1) x^{n}=\frac{1}{(1-x)^{2}} \tag*{(3)} $$ Re-indexing yields $$\sum_{n=1}^{\infty} n x^{n-1}=\frac{1}{(1-x)^{2}} \tag*{(4)} $$ Putting $x=\frac{1}{3} $ in (4) yields $$\sum_{n=1}^{\infty} \frac{n}{3^{n-1}} =\frac{1}{\left(1-\frac{1}{3}\right)^{2}} =\frac{9}{4} \tag*{(5)} $$
Multiplying (5) by $\displaystyle \frac{2}{9}$ conclude that $$ \boxed{\sum_{n=1}^{\infty} \frac{2 n}{3^{n+1}} =\frac{1}{2}} $$

  • $\begingroup$ There are similar earlier answers, e.g. here $\endgroup$ Commented Jan 9, 2022 at 11:25

This will get buried, but it's worth noting that given a (not necessarily complete) normed field $\mathbb{K}$, a natural $\text{n}$, an $\text{n}$-tuple $\left(\lambda_{0},\dots,\lambda_{\text{n}-1}\right)\in\mathbb{K}^{\text{n}}$ with entries of norm $<1$, and a polynomial $\Phi\in\mathbb{K}\left[x_{0},\dots,x_{\text{n}-1}\right]$, there's a general, finite (with complexity bounded by a function of the degree(s) of $\Phi$) formula that computes $$\underbrace{\sum_{x_{0}=0}^{\infty}\cdots\sum_{x_{\text{n}-1}=0}^{\infty}}_{\text{n sums}}\Phi\left(x_{0},\dots,x_{\text{n}-1}\right)\prod_{\text{i}=0}^{\text{n-1}} \lambda_{\text{i}}^{x_{\text{i}}}$$ as an element of $\mathbb{K}$. (So that the sum in particular converges absolutely—even if $\mathbb{K}$ fails to be complete!) The ideas behind its derivation are simple, but to avoid drowning ourselves in notation, let us first assume some (un-)conventions:

In what follows,

  • $\colon$ is shorthand for $\in$.

  • $\to$ forms the set of functions.

  • $\omega$ is the set of naturals (including $0$).

  • Each natural $\text{n}\colon\omega$ is identified with the set $\left\{0,\dots,\text{n}-1\right\}$.

Henceforth fix a normed field $\mathbb{K}$ and a natural $\text{n}$. (Everything we do in general will be motivated by the specialization of the argument that follows in the case that $\text{n}=1$.)

  • The set $\text{n}\to\omega$ is simultaneously identified with the evident set of such functions and the set of $\text{n}$-tuples of naturals (so that the two are in particular identified with one another in the evident way).

  • Given $c\colon\omega$, the tuple/function $c\colon \text{n}\to\omega$ is by definition the function that evaluates to $c$ everywhere.

  • Given $\mathscr{k}\colon\mathbb{K}$, the tuple/function $\mathscr{k}\colon \text{n}\to\omega$ is by definition the function that evaluates to $\mathscr{k}$ everywhere.

  • Given $\text{i}\colon\text{n}$, the tuple/function $1_{\text{i}}\colon \text{n}\to\omega$ is by definition the function that evaluates to $1$ at $\text{i}$ and $0$ elsewhere.

  • The addition of elements of $\text{n}\to\omega$ and the addition and subtraction of elements of $\text{n}\to\mathbb{K}$ is defined pointwise as usual.

  • Given a polynomial $\Phi\colon\mathbb{K}\left[x_{\text{i}}\right]_{\text{i}\colon\text{n}}$ and an element $x\colon\text{n}\to\omega$, the evaluation $\Phi\left(x\right)\colon\mathbb{K}$ is defined as $\Phi\left(x\left(\text{i}\right)\right)_{\text{i}\colon\text{n}}$ (so that, somewhat confusing notation, $\Phi$ is identified with a particular function $\Phi\colon\left(\text{n}\to\omega\right)\to\mathbb{K}$).

  • $\leq$ denotes the standard (product) partial order on $\text{n}\colon\omega$ when applied between elements thereof.

  • Given $\text{i}\colon\text{n}$, $\preccurlyeq_{\text{i}}$ denotes the total preorder on $\text{n}\colon\omega$ induced (from the standard total order on $\omega$) by evaluation at $\text{i}$.

  • Given $\lambda\colon\text{n}\to\mathbb{K}$ and $x\colon\text{n}\to\omega$, define the expoential $\lambda^{x}:=\prod_{\text{i}\colon\text{n}}\lambda\left(\text{i}\right)^{x\left(\text{i}\right)}$.

  • In analogy to the above, we might consider $\mathbb{K}\left[x_{\text{i}}\right]_{\text{i}\colon\text{n}}$ to be the free $\mathbb{K}$-vector space on $\left(x^{d}\right)_{d\colon\text{n}\to\omega}$, where $x^{d}$ is a shorthand for $\prod_{\text{i}\colon\text{n}}x_{\text{i}}^{d\left(\text{i}\right)}$.

  • Given $d\colon\text{n}\to\omega$, the function $\mathfrak{H}_{\text{n},d}\colon\left(\text{n}\to\omega\right)\to\mathbb{K}$ is defined so that $\mathfrak{H}_{\text{n},d}\left(x\right):=\prod_{\text{i}\colon\text{n}}\tbinom{x\left(\text{i}\right)}{d\left(\text{i}\right)}$. (I.e., as a product of binomial coefficients. Note that the outputs of $\mathfrak{H}_{\text{n},d}$ are all in the image of the canonical map $\omega\to\mathbb{K}$.)

  • Given $\text{i}\colon\text{n}$, the operator $$\text{S}_{\text{i}}\ \colon\ \left(\left(\text{n}\to\omega\right)\to\mathbb{K}\right)\to \left(\left(\text{n}\to\omega\right)\to\mathbb{K}\right)$$ is defined to map as $$\text{S}_{\text{i}}\ \colon\ \Psi\ \mapsto\ \left(x\ \mapsto\ \Psi\left(x+1_{\text{i}}\right)\right)\text{.}$$

  • Given $d\colon\text{n}\to\omega$, the operators $\text{S}^{d},\ \mathfrak{D}^{d}\colon \left(\left(\text{n}\to\omega\right)\to\mathbb{K}\right)\to \left(\left(\text{n}\to\omega\right)\to\mathbb{K}\right)$ are defined as $\text{S}^{d}:=\prod_{\text{i}\colon\text{n}} \text{S}_{\text{i}}^{d\left(\text{i}\right)}$ and $\mathfrak{D}^{d}:=\prod_{\text{i}\colon\text{n}}\left(\text{S}_{\text{i}}^{d\left(\text{i}\right)}-1\right)$ respectively. (Here the product denotes composition of (commuting) operators and $1\colon\left(\left(\text{n}\to\omega\right)\to\mathbb{K}\right)\to \left(\left(\text{n}\to\omega\right)\to\mathbb{K}\right)$ is the identity.)

  • Given $\Psi\colon\left(\text{n}\to\omega\right)\to\mathbb{K}$, the subset $\mathfrak{d}\left(\psi\right) \subseteq \text{n}\to\omega$ is defined to entail precisely those $d\colon\text{n}\to\omega$ for which $\mathfrak{D}^{d}\Psi\left(0\right)\neq 0$.

Also, every infinite sum in question will converge absolutely unless otherwise specified (i.e., wrt the net of finite subsets of the infinite indexing set over which it is taken), so we largely ignore analytic subtleties below.

In the above terms, the original question asks to compute $$\sum_{x\colon\ \text{n}\to\omega}\Phi\left(x\right)\lambda^{x}\text{,}$$ and we claim that given $\Phi\colon\mathbb{K}\left[x_{\text{i}}\right]_{\text{i}\colon\text{n}}$ and $\lambda\colon\text{n}\to\mathbb{K}$ with outputs of norm $<1$, $$\sum_{x\colon\ \text{n}\to\omega}\Phi\left(x\right)\lambda^{x}\ =\ \sum_{d\colon\ \mathfrak{d}\left(\Phi\right)}\sum_{d'\ \leq\ d}\left(-1\right)^{d-d'}\mathfrak{H}_{\text{n},d'}\left(d\right)\Phi\left(d'\right)\frac{\lambda^{d}}{\left(1-\lambda\right)^{d+1}}\text{,}$$ with $\mathfrak{d}\left(\Phi\right)$ a priori finite. (Actually, we prove something marginally stronger but in "practice" easier to work with, namely that the identity holds with the $\mathfrak{d}\left(\Phi\right)$ in the index of the first sum replaced with any finite superset thereof.)

Some fundamental/useful input from the discrete calculus

The crucial observations, essentially amounting to the multivariate discrete calculus, are that as follows:

Proposition: Given $\Xi\ \colon\left(\text{n}\to\omega\right)\to\mathbb{K}$, the sum $$\sum_{d\colon\text{n}\to\omega}\Xi\left(d\right)\mathfrak{H}_{\text{n},d}\left(x\right)$$ stabilizes in finitely many summands (not necessarily uniformly) at each argument $x\colon\text{n}\to\omega$, and so evaluates to a well-defined function $\left(\text{n}\to\omega\right)\to\mathbb{K}$.

Proof: In fact, $\mathfrak{H}_{\text{n},d}\left(x\right)$ is nonzero i(f)f $x\geq d$ (by the elementary properties of binomial coefficients), and thus for a given $x$ only finitely many of the summands are nonzero, as claimed. $\Box$

Proposition: Given $\Psi\colon\left(\text{n}\to\omega\right)\to\mathbb{K}$, there exists $\Xi\ \colon\left(\text{n}\to\omega\right)\to\mathbb{K}$ such that $$\Psi\left(x\right)\ =\ \sum_{d\colon\text{n}\to\omega}\Xi\left(d\right)\mathfrak{H}_{\text{n},d}\left(x\right)$$ in the above sense.

Proof: Fix a linearization (i.e., to the order isomorphism type of the standard well-order on $\omega$) $\tilde{\leq}$ extending the usual partial order $\leq$ on $\text{n}\to\omega$ (such $\tilde{\leq}$ exists by a standard explicit construction, e.g., the total-degree-then-lexicographical order). Then (by the elementary properties of binomial coefficients), if $d\tilde{\geq} x$, $$\mathfrak{H}_{\text{n},x}\left(x\right)\ =\ \begin{cases} 1\text{ if }d=x \\ 0\text{ otherwise}\end{cases}\text{.}$$ Thus we can build up our desired $\Xi$ inductively with respect to $\tilde{\leq}$ beginning with its minimal element (namely $0$), at each step choosing $$\Xi\left(x\right)\ =\ \Psi\left(x\right)-\sum_{d\tilde{<}x}\Xi\left(d\right)\mathfrak{H}_{\text{n},d}\left(x\right)\text{,}$$ leaving the values of $\Xi\left(x'\right)$ unaffected for $x'\tilde{<}x$. $\Box$

Proposition: Given $d_{0},d_{1}\colon\text{n}\to\omega$, $$\mathfrak{D}^{d_{1}}\mathfrak{H}_{\text{n},d_{0}}\ =\ \begin{cases} \mathfrak{H}_{\text{n},d_{0}-d_{1}}\text{ if }d_{0}\geq d_{1}\\ 0\text{ otherwise}\end{cases}$$ and in particular $$\mathfrak{D}^{d_{1}}\mathfrak{H}_{\text{n},d_{0}}\left(0\right)\ =\ \begin{cases} 1\text{ if }d_{0}=d_{1}\\ 0\text{ otherwise}\end{cases}\text{.}$$

Proof: We proceed by induction (namely on $\text{n}\to\omega$ via the $\text{n}$ pointwise "successor" relations) on $d_{1}$. The base case of $d_{1}=0$ is all but tautological. As for the inductive step, it suffices to verify for $\text{i}\colon\text{n}$ that if $$\mathfrak{D}^{d_{1}}\mathfrak{H}_{\text{n},d_{0}}\ =\ \begin{cases} \mathfrak{H}_{\text{n},d_{0}-d_{1}}\text{ if }d_{0}\geq d_{1}\\ 0\text{ otherwise}\end{cases}\text{,}$$ then $$\mathfrak{D}^{d_{1}+1_{\text{i}}}\mathfrak{H}_{\text{n},d_{0}}\ =\ \begin{cases} \mathfrak{H}_{\text{n},d_{0}-d_{1}-1_{\text{i}}}\text{ if }d_{0}\geq d_{1}+1_{\text{i}}\\ 0\text{ otherwise}\end{cases}\text{.}$$ Indeed, it suffices to consider the case that $d_{1}\leq d_{0}$, as else is trivial. In this case, \begin{align*} \mathfrak{D}^{d_{1}+1_{\text{i}}}\mathfrak{H}_{\text{n},d_{0}}\ &=\ \mathfrak{D}^{1_{\text{i}}}\mathfrak{D}^{d_{1}}\mathfrak{H}_{\text{n},d_{0}}\\ &=\ \left(\text{S}_{\text{i}}-1\right)\mathfrak{H}_{\text{n},d_{0}-d_{1}}\\ &=\ \left(\text{S}_{\text{i}}-1\right)\prod_{\text{i}'\colon\text{n}} \left(x\mapsto\binom{x}{d_{0}\left(\text{i}'\right)-d_{1}\left(\text{i}'\right)}\right)\\ &=\ \prod_{\substack{\text{i}'\colon\text{n} \\ \text{i}'\neq\text{i}}} \left(x\mapsto\binom{x}{d_{0}\left(\text{i}'\right)-d_{1}\left(\text{i}'\right)}\right)\ \cdot\ \left(\text{S}_{\text{i}}-1\right)\left(x\mapsto\binom{x}{d_{0}\left(\text{i}\right)-d_{1}\left(\text{i}\right)}\right)\\ &=\ \prod_{\substack{\text{i}'\colon\text{n} \\ \text{i}'\neq\text{i}}} \left(x\mapsto\binom{x}{d_{0}\left(\text{i}'\right)-d_{1}\left(\text{i}'\right)}\right)\ \cdot\ \left(x\mapsto\begin{cases} \binom{x}{d_{0}\left(\text{i}\right)-d_{1}\left(\text{i}\right)-1}\text{ if }d_{0}\left(\text{i}\right)-d_{1}\left(\text{i}\right)\geq 1\\ 0\text{ otherwise}\end{cases}\right)\\ &=\ \begin{cases} \mathfrak{H}_{\text{n},d_{0}-d_{1}-1_{\text{i}}}\text{ if }d_{0}\geq d_{1}+1_{\text{i}}\\ 0\text{ otherwise}\end{cases}\text{.} \end{align*}

The second part of the claim is then just by evaluation. $\Box$

Proposition: Given $\Psi\colon\left(\text{n}\to\omega\right)\to\mathbb{K}$, $$\Psi\left(x\right)\ =\ \sum_{d\colon\text{n}\to\omega}\mathfrak{D}^{d}\Psi\left(0\right)\cdot\mathfrak{H}_{\text{n},d}\left(x\right)$$ in the above sense.

Proof: This follows by applying the previous proposition to the penultimate proposition. (I.e., we use the former to compute the $\Xi$ from the latter as $\Xi\left(d\right)=\mathfrak{D}^{d}\Psi\left(0\right)$.) $\Box$

Proposition: Given $\Phi\colon\mathbb{K}\left[x_{\text{i}}\right]_{\text{i}\colon\text{n}}$, the set $\mathfrak{d}\left(\Phi\right)$ is contained in the downward closure under $\leq$ of the set of $d\colon\text{n}\to\omega$ of degrees for which the coefficient of $x^{d}$ in $\Phi$ is nonzero (so is in particular finite).

Proof: Remark that (by the elementary properties of binomial coefficients) for $\text{i}$, $d$ as above, $\mathfrak{D}^{1_{\text{i}}}\left(x\mapsto x^{d}\right)$ is $0$ if $d\left(\text{i}\right)=0$ and a $\mathbb{K}$-linear combination of terms $x^{d'}$ otherwise. The claim now follows by induction as before. $\Box$

Proposition: Given $\Phi\colon\mathbb{K}\left[x_{\text{i}}\right]_{\text{i}\colon\text{n}}$ and finite $D\supseteq \mathfrak{d}\left(\Phi\right)$, $$\Phi\left(x\right)\ =\ \sum_{d\colon\ D}\sum_{d'\ \leq\ d}\left(-1\right)^{d-d'}\mathfrak{H}_{\text{n},d}\left(d\right)\Phi\left(d'\right)\mathfrak{H}_{\text{n},d}\left(x\right)$$ identically in $x\colon\text{n}\to\omega$.

Proof: By the previous and penultimate propositions, as well as a classical extension of the binomial theorem, \begin{align*} \Phi\left(x\right)\ &=\ \sum_{d\colon D}\mathfrak{D}^{d}\Phi\left(0\right)\cdot\mathfrak{H}_{\text{n},d}\left(x\right)\\ &=\ \sum_{d\colon D}\left(\prod_{\text{i}\colon\text{n}}\left(\text{S}^{d\left(\text{i}\right)\cdot 1_{\text{i}}}-1\right)\Phi\left(0\right)\right)\mathfrak{H}_{\text{n},d}\left(x\right)\\ &=\ \sum_{d\colon D}\left(\sum_{d'\leq d}\left(-1\right)^{d-d'}\mathfrak{H}_{\text{n},d'}\left(d\right)\text{S}^{d'}\right)\Phi\left(0\right)\cdot\mathfrak{H}_{\text{n},d}\left(x\right)\\ &=\ \sum_{d\colon D}\sum_{d'\leq d}\left(-1\right)^{d-d'}\mathfrak{H}_{\text{n},d'}\left(d\right)\Phi\left(d'\right)\mathfrak{H}_{\text{n},d}\left(x\right) \end{align*} as claimed. $\Box$

Back to the main question

Lemma: Given $\lambda\colon\text{n}\to\mathbb{K}$ with outputs of norm $<1$, and $d\colon\text{n}\to\omega$, $$\sum_{x\colon\ \text{n}\to\omega}\mathfrak{H}_{\text{n},d}\left(x\right)\lambda^{x}\ =\ \frac{\lambda^{d}}{\left(1-\lambda\right)^{d+1}}\text{.}$$

Proof: We proceed by induction (namely on $\text{n}\to\omega$ via the $\text{n}$ pointwise "successor" relations). The base case of $x=0$, which is classical, is that $$\sum_{x\colon\ \text{n}\to\omega}\mathfrak{H}_{\text{n},0}\left(x\right)\lambda^{x}\ =\ \frac{1}{\left(1-\lambda\right)^{1}}\text{.}$$ As for the inductive step for the successor relation pertaining to $\text{i}\colon\text{n}$, we have by hypothesis for a given $x$ that $$\sum_{x\colon\ \text{n}\to\omega}\mathfrak{H}_{\text{n},d}\left(x\right)\lambda^{x}\ =\ \frac{\lambda^{d}}{\left(1-\lambda\right)^{d+1}}\text{,}$$ from which we deduce that \begin{align*} \frac{\lambda^{d+1_{\text{i}}}}{\left(1-\lambda\right)^{d+1_{\text{i}}+1}}\ &=\ \frac{\lambda\left(\text{i}\right)}{1-\lambda\left(\text{i}\right)}\sum_{x\colon\ \text{n}\to\omega}\mathfrak{H}_{\text{n},d}\left(x\right)\lambda^{x}\\ &=\ \sum_{x'\colon\ \text{n}\to\omega}\mathfrak{H}_{\text{n},d}\left(x'\right)\lambda^{x'}\frac{\lambda\left(\text{i}\right)}{1-\lambda\left(\text{i}\right)}\\ &=\ \sum_{x'\colon\ \text{n}\to\omega}\mathfrak{H}_{\text{n},d}\left(x'\right)\lambda^{x'}\left(\sum_{\text{k}\colon\omega}\lambda^{\left(\text{k}+1\right) 1_{\text{i}}}\right)\\ &=\ \sum_{x'\colon\ \text{n}\to\omega}\sum_{x\succ_{\text{i}}x'}\mathfrak{H}_{\text{n},d}\left(x'\right)\lambda^{x}\\ &=\ \sum_{x\colon\ \text{n}\to\omega}\sum_{x'\prec_{\text{i}}x}\mathfrak{H}_{\text{n},d}\left(x'\right)\lambda^{x}\\ &=\ \sum_{x\colon\ \text{n}\to\omega}\mathfrak{H}_{\text{n},d+1_{\text{i}}}\left(x\right)\lambda^{x}\text{,} \end{align*} as needed. (The absolute convergences and equality of the infinite sums are justified by the absolute convergence and equality of the sums from which they derive in turn, the first by hypothesis.) $\Box$

Result: Given $\Phi\colon\mathbb{K}\left[x_{\text{i}}\right]_{\text{i}\colon\text{n}}$, $\lambda\colon\text{n}\to\mathbb{K}$ with outputs of norm $<1$, and finite $D\supseteq \mathfrak{d}\left(\Phi\right)$, $$\boxed{\sum_{x\colon\ \text{n}\to\omega}\Phi\left(x\right)\lambda^{x}\ =\ \sum_{d\colon\ D}\sum_{d'\ \leq\ d}\left(-1\right)^{d-d'}\mathfrak{H}_{\text{n},d'}\left(d\right)\Phi\left(d'\right)\frac{\lambda^{d}}{\left(1-\lambda\right)^{d+1}}}\text{.}$$

Proof: By the above, we have that \begin{align*} \sum_{d\colon\ \mathfrak{z}\left(\Phi\right)}\sum_{d'\ \leq\ d}\left(-1\right)^{d-d'}\mathfrak{H}_{\text{n},d'}\left(d\right)\Phi\left(d'\right)\frac{\lambda^{d}}{\left(1-\lambda\right)^{d+1}}\ &=\ \sum_{d\colon\ \mathfrak{d}\left(\Phi\right)}\sum_{d'\ \leq\ d}\left(-1\right)^{d-d'}\mathfrak{H}_{\text{n},d'}\left(d\right)\Phi\left(d'\right)\left(\sum_{x\colon\ \text{n}\to\omega}\mathfrak{H}_{\text{n},d}\left(\text{x}\right)\lambda^{x}\right)\\ &=\ \sum_{x\colon\ \text{n}\to\omega}\left(\sum_{d\colon\ \mathfrak{d}\left(\Phi\right)}\sum_{d'\ \leq\ d}\left(-1\right)^{d-d'}\mathfrak{H}_{\text{n},d'}\left(d\right)\Phi\left(d'\right)\mathfrak{H}_{\text{n},d}\left(\text{x}\right)\right)\lambda^{x}\\ &=\ \sum_{x\colon\ \text{n}\to\omega}\Phi\left(x\right)\lambda^{x}\text{,} \end{align*} as claimed. (The absolute convergences and equality of the infinite sums are as before justified by the absolute convergence and equality of the sums from which they derive in turn.) $\blacksquare$

Sample computation

Suppose, for instance, that we wish to compute $$\sum_{x=0}^{\infty}\sum_{y=0}^{\infty}\frac{x^{3}-3xy+2y^{2}}{\left(-4\right)^{x}5^{y}}\text{.}$$ In the language of the above result, this is the computation $$\sum_{x\colon\ \text{n}\to\omega}\Phi\left(x\right)\lambda^{x}$$ with $$\text{n}\ =\ 2$$ $$\Phi\ \colon\ \left(x_{0},x_{1}\right)\ \mapsto\ x_{0}^{3}-3x_{0}x_{1}+2x_{1}^{2}$$ $$\lambda\ =\ \left(-\tfrac{1}{4},\ \tfrac{1}{5}\right)$$ Now, we have by the usual bound that $$\mathfrak{d}\left(\Phi\right)\ \subseteq\ \left\{\left(0,0\right), \left(0,1\right), \left(0,2\right), \left(1,0\right), \left(1,1\right), \left(2,0\right), \left(3,0\right)\right\}\text{,}$$ so the formula reduces the infinite sum to the $19$-term sum $$\left(-1\right)^{\left(0,0\right)-\left(0,0\right)}\mathfrak{H}_{2,\left(0,0\right)}\left(\left(0,0\right)\right)\Phi\left(\left(0,0\right)\right)\frac{\left(-\tfrac{1}{4},\ \tfrac{1}{5}\right)^{\left(0,0\right)}}{\left(\tfrac{5}{4},\ \tfrac{4}{5}\right)^{\left(0,0\right)+1}}$$ $$+\ \left(-1\right)^{\left(0,1\right)-\left(0,0\right)}\mathfrak{H}_{2,\left(0,0\right)}\left(\left(0,1\right)\right)\Phi\left(\left(0,0\right)\right)\frac{\left(-\tfrac{1}{4},\ \tfrac{1}{5}\right)^{\left(0,1\right)}}{\left(\tfrac{5}{4},\ \tfrac{4}{5}\right)^{\left(0,1\right)+1}}\ +\ \left(-1\right)^{\left(0,1\right)-\left(0,1\right)}\mathfrak{H}_{2,\left(0,1\right)}\left(\left(0,1\right)\right)\Phi\left(\left(0,1\right)\right)\frac{\left(-\tfrac{1}{4},\ \tfrac{1}{5}\right)^{\left(0,1\right)}}{\left(\tfrac{5}{4},\ \tfrac{4}{5}\right)^{\left(0,1\right)+1}}$$ $$+\ \left(-1\right)^{\left(0,2\right)-\left(0,0\right)}\mathfrak{H}_{2,\left(0,0\right)}\left(\left(0,2\right)\right)\Phi\left(\left(0,0\right)\right)\frac{\left(-\tfrac{1}{4},\ \tfrac{1}{5}\right)^{\left(0,2\right)}}{\left(\tfrac{5}{4},\ \tfrac{4}{5}\right)^{\left(0,2\right)+1}}\ +\ \left(-1\right)^{\left(0,2\right)-\left(0,1\right)}\mathfrak{H}_{2,\left(0,1\right)}\left(\left(0,2\right)\right)\Phi\left(\left(0,1\right)\right)\frac{\left(-\tfrac{1}{4},\ \tfrac{1}{5}\right)^{\left(0,2\right)}}{\left(\tfrac{5}{4},\ \tfrac{4}{5}\right)^{\left(0,2\right)+1}}\ +\ \left(-1\right)^{\left(0,2\right)-\left(0,2\right)}\mathfrak{H}_{2,\left(0,2\right)}\left(\left(0,2\right)\right)\Phi\left(\left(0,2\right)\right)\frac{\left(-\tfrac{1}{4},\ \tfrac{1}{5}\right)^{\left(0,2\right)}}{\left(\tfrac{5}{4},\ \tfrac{4}{5}\right)^{\left(0,2\right)+1}}$$ $$+\ \left(-1\right)^{\left(1,0\right)-\left(0,0\right)}\mathfrak{H}_{2,\left(0,0\right)}\left(\left(1,0\right)\right)\Phi\left(\left(0,0\right)\right)\frac{\left(-\tfrac{1}{4},\ \tfrac{1}{5}\right)^{\left(1,0\right)}}{\left(\tfrac{5}{4},\ \tfrac{4}{5}\right)^{\left(1,0\right)+1}}\ +\ \left(-1\right)^{\left(1,0\right)-\left(1,0\right)}\mathfrak{H}_{2,\left(1,0\right)}\left(\left(1,0\right)\right)\Phi\left(\left(1,0\right)\right)\frac{\left(-\tfrac{1}{4},\ \tfrac{1}{5}\right)^{\left(1,0\right)}}{\left(\tfrac{5}{4},\ \tfrac{4}{5}\right)^{\left(1,0\right)+1}}$$ $$+\ \left(-1\right)^{\left(1,1\right)-\left(0,0\right)}\mathfrak{H}_{2,\left(0,0\right)}\left(\left(1,1\right)\right)\Phi\left(\left(0,0\right)\right)\frac{\left(-\tfrac{1}{4},\ \tfrac{1}{5}\right)^{\left(1,1\right)}}{\left(\tfrac{5}{4},\ \tfrac{4}{5}\right)^{\left(1,1\right)+1}}\ +\ \left(-1\right)^{\left(1,1\right)-\left(0,1\right)}\mathfrak{H}_{2,\left(0,1\right)}\left(\left(1,1\right)\right)\Phi\left(\left(0,1\right)\right)\frac{\left(-\tfrac{1}{4},\ \tfrac{1}{5}\right)^{\left(1,1\right)}}{\left(\tfrac{5}{4},\ \tfrac{4}{5}\right)^{\left(1,1\right)+1}}\ +\ \left(-1\right)^{\left(1,1\right)-\left(1,0\right)}\mathfrak{H}_{2,\left(1,0\right)}\left(\left(1,1\right)\right)\Phi\left(\left(1,0\right)\right)\frac{\left(-\tfrac{1}{4},\ \tfrac{1}{5}\right)^{\left(1,1\right)}}{\left(\tfrac{5}{4},\ \tfrac{4}{5}\right)^{\left(1,1\right)+1}}\ +\ \left(-1\right)^{\left(1,1\right)-\left(1,1\right)}\mathfrak{H}_{2,\left(1,1\right)}\left(\left(1,1\right)\right)\Phi\left(\left(1,1\right)\right)\frac{\left(-\tfrac{1}{4},\ \tfrac{1}{5}\right)^{\left(1,1\right)}}{\left(\tfrac{5}{4},\ \tfrac{4}{5}\right)^{\left(1,1\right)+1}}$$ $$+\ \left(-1\right)^{\left(2,0\right)-\left(0,0\right)}\mathfrak{H}_{2,\left(0,0\right)}\left(\left(2,0\right)\right)\Phi\left(\left(0,0\right)\right)\frac{\left(-\tfrac{1}{4},\ \tfrac{1}{5}\right)^{\left(2,0\right)}}{\left(\tfrac{5}{4},\ \tfrac{4}{5}\right)^{\left(2,0\right)+1}}\ +\ \left(-1\right)^{\left(2,0\right)-\left(1,0\right)}\mathfrak{H}_{2,\left(1,0\right)}\left(\left(2,0\right)\right)\Phi\left(\left(1,0\right)\right)\frac{\left(-\tfrac{1}{4},\ \tfrac{1}{5}\right)^{\left(2,0\right)}}{\left(\tfrac{5}{4},\ \tfrac{4}{5}\right)^{\left(2,0\right)+1}}\ +\ \left(-1\right)^{\left(2,0\right)-\left(2,0\right)}\mathfrak{H}_{2,\left(2,0\right)}\left(\left(2,0\right)\right)\Phi\left(\left(2,0\right)\right)\frac{\left(-\tfrac{1}{4},\ \tfrac{1}{5}\right)^{\left(2,0\right)}}{\left(\tfrac{5}{4},\ \tfrac{4}{5}\right)^{\left(2,0\right)+1}}$$ $$+\ \left(-1\right)^{\left(3,0\right)-\left(0,0\right)}\mathfrak{H}_{2,\left(0,0\right)}\left(\left(3,0\right)\right)\Phi\left(\left(0,0\right)\right)\frac{\left(-\tfrac{1}{4},\ \tfrac{1}{5}\right)^{\left(3,0\right)}}{\left(\tfrac{5}{4},\ \tfrac{4}{5}\right)^{\left(3,0\right)+1}}\ +\ \left(-1\right)^{\left(3,0\right)-\left(1,0\right)}\mathfrak{H}_{2,\left(1,0\right)}\left(\left(3,0\right)\right)\Phi\left(\left(1,0\right)\right)\frac{\left(-\tfrac{1}{4},\ \tfrac{1}{5}\right)^{\left(3,0\right)}}{\left(\tfrac{5}{4},\ \tfrac{4}{5}\right)^{\left(3,0\right)+1}}\ +\ \left(-1\right)^{\left(3,0\right)-\left(2,0\right)}\mathfrak{H}_{2,\left(2,0\right)}\left(\left(3,0\right)\right)\Phi\left(\left(2,0\right)\right)\frac{\left(-\tfrac{1}{4},\ \tfrac{1}{5}\right)^{\left(3,0\right)}}{\left(\tfrac{5}{4},\ \tfrac{4}{5}\right)^{\left(3,0\right)+1}}\ +\ \left(-1\right)^{\left(3,0\right)-\left(3,0\right)}\mathfrak{H}_{2,\left(3,0\right)}\left(\left(3,0\right)\right)\Phi\left(\left(3,0\right)\right)\frac{\left(-\tfrac{1}{4},\ \tfrac{1}{5}\right)^{\left(3,0\right)}}{\left(\tfrac{5}{4},\ \tfrac{4}{5}\right)^{\left(3,0\right)+1}}$$ So that (computing each term) \begin{align*} &\sum_{x=0}^{\infty}\sum_{y=0}^{\infty}\frac{x^{3}-3xy+2y^{2}}{\left(-4\right)^{x}5^{y}}\\ &=\ 0+0+\tfrac{1}{2}+0-\tfrac{1}{4}+\tfrac{1}{2}+0-\tfrac{1}{5}+0+\tfrac{1}{10}+\tfrac{1}{20}+0-\tfrac{2}{25}+\tfrac{8}{25}+0-\tfrac{3}{125}+\tfrac{24}{125}-\tfrac{27}{125}\\ &=\ \boxed{\frac{223}{250}}\text{.} \end{align*} (Which conforms to the value I got when I approximated it...)

  • $\begingroup$ You should have written the sample computation as a question explaining your method. $\endgroup$
    – Bob Dobbs
    Commented Mar 8 at 16:38

For $|x|<1,$ recall the geometric series identity $\sum_{k=1}^\infty x^{k-1}=\frac{1}{1-x},$ so by Fubini's theorem,

$$\small \sum_{k=1}^\infty kx^{k-1}=\sum_{k=1}^\infty \sum_{j=1}^\infty {\bf1}_{j\leq k} x^{k-1}=\sum_{j=1}^\infty\sum_{k=1}^\infty {\bf1}_{j\leq k} x^{k-1}=\sum_{j=1}^\infty\sum_{k=j}^\infty x^{k-1}=\sum_{j=1}^\infty x^{j-1}\sum_{k=1}^\infty x^{k-1}=\sum_{j=1}^\infty \frac{x^{j-1}}{1-x}=\frac{1}{(1-x)^2}.$$ which formalizes this proof.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .