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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 $$\sum_{n=0}^\infty (n+1)x^n?$$

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    $\begingroup$ I believe this is an arithmo-geometric series. You can find information here: artofproblemsolving.com/wiki/… $\endgroup$
    – Jason Kim
    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
    Mar 13 at 17:16

23 Answers 23

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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 $(1-r)^{k+1}$.

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    $\begingroup$ @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? $\endgroup$ Jun 5, 2014 at 16:38
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    $\begingroup$ Small nitpick: "equality holds for any $r$" should be "equality holds for any $r\neq1$" $\endgroup$ Feb 25, 2017 at 5:27
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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}

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    $\begingroup$ 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
    Aug 16, 2015 at 5:39
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    $\begingroup$ @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. $\endgroup$ Feb 24, 2017 at 5:50
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    $\begingroup$ @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
    Feb 24, 2017 at 5:55
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    $\begingroup$ @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
    Feb 24, 2017 at 6:06
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    $\begingroup$ @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 $\endgroup$ Feb 24, 2017 at 6:08
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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.

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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.

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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).

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Hints

  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.

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    $\begingroup$ 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
    Apr 3, 2011 at 21:59
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    $\begingroup$ 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
    Apr 3, 2011 at 22:05
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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: $S = (1 - x)^{-2}$.

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  • $\begingroup$ +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
    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
    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
    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
    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
    Nov 12, 2014 at 11:47
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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 $\sum_{n=1}^\infty nx^{n-1}$.

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  • $\begingroup$ Thank you for helping, but I have never learned differentiation. $\endgroup$
    – backus
    Apr 3, 2011 at 21:58
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    $\begingroup$ 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
    Mar 9, 2016 at 17:23
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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 $$\sum_{n=0}^\infty{(n+1)x^n}={1\over (1-x)^2}.$$

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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 $$ S_{n}(z)=\frac{z}{1-z}\left[1+\sum_{k=0}^{n-1}\binom{n}{k} S_k(z)\right]\qquad n=1, 2, \ldots $$

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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 $$ \frac{9}{4} =\sum_{n=0}^{+\infty}(n+1)\frac{1}{3^n} = \sum_{m=1}^{+\infty}m\frac{1}{3^{m-1}} \quad \Rightarrow \quad \sum_{m=1}^{+\infty}\frac{m}{3^m} = \frac{3}{4} $$

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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 $\sum_{n\geq0}(n+1)x^n$.

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  • $\begingroup$ 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
    Feb 24, 2017 at 6:25
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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!

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

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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

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

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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. $x$.

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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

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

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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 $$ \sum_{k=0}^\infty(k+1)x^k=\frac1{(1-x)^2}\tag3 $$

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  • $\begingroup$ Hello robjohn, i need your help on that technique of generating valid telescoping series if you allow this ofc. $\endgroup$
    – Abr001am
    Jun 6, 2018 at 20:29
  • $\begingroup$ @Abr001am: what is your question? $\endgroup$
    – robjohn
    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
    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
    Jun 9, 2018 at 17:49
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\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}

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

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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).

Now

\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!)

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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 $$ \boxed{\sum_{n=1}^{\infty} \frac{2 n}{3^{n+1}} =\frac{1}{2}} $$

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

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