# How can I evaluate $\sum_{n=0}^\infty (n+1)x^n$

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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Similar to this. Maybe duplicate. – leo Sep 21 '13 at 17:27
It is the other way around. The linked question might be duplicate of this one. – leo Sep 21 '13 at 17:34

No need to use Taylor Series, this can be derived in a similar way to the formula for geometric series. Lets 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*}

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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@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? – Dmitry Kazakov Jun 5 '14 at 16:38

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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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. – user21820 Aug 16 at 5:39

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 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 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:

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

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 $A_2={7\over 2^6}+{8\over 2^7}$.

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 $A_3={7\over 2^6}+{8\over 2^7}+{9\over 2^8}$.

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 $A_4={7\over 2^6}+{8\over 2^7}+{9\over 2^8}+{10\over2^9}$.

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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Thanks for answering. 1) No I don't know that formula 2) Can you explain by what you mean by derive both sides? – Backus Apr 3 '11 at 21:59
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. – leonbloy Apr 3 '11 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: $S = (1 - x)^{-2}$.

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+1, but wouldn't it have been simpler to say that it was the derivative of $1+x+x^2+...$, converging to $\frac1{1-x}$? – Mike Oct 29 '12 at 22:46
I decided to start with what was given, so it is easier for the OP to see. – glebovg Oct 29 '12 at 22:48
Justin, observe that this only holds for $|x| < 1$, otherwise your sum simplifies to $\infty$. – glebovg Oct 29 '12 at 22:54
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. – Mike Oct 29 '12 at 23:23
We could notice that the given series is converging to $\frac{1}{1-x}-1$ and take the derivative of that. – inkievoyd Nov 11 '14 at 21:03

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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Thank you for helping, but I have never learned differentiation. – Backus Apr 3 '11 at 21:58

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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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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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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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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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 $$\sum_{n=0}^\infty (n+1)x^n = (xD + 1) f$$ $$\frac{x}{(1-x)^2} + \frac{1}{1-x} = \frac{1}{(1-x)^2}$$here D means differentiation w.r.t x

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Do the derivative from the normal geometric series, and then you can get what you want.

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Please elaborate on this. – Marconius Oct 15 at 22:10
Your idea has been suggested by many others a few year's ago. I don't see how your answer add anything new. – John Ma Oct 27 at 22:26

## protected by John MaOct 27 at 22:27

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