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

Question

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

Answer 1

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

Answer 2

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.

Answer 3

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 $\sum_{n=1}^\infty nx^{n-1}$.

Answer 4

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.

Answer 5

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.

Answer 6

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

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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, $$ \sum_{n=1}^\infty{n\over 2^{n-1}}= \sum_{n=1}^6 {n\over 2^{n-1}} +\sum_{n=7}^\infty{n\over 2^{n-1}} ={15\over 4}+{1\over4}=4. $$

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

Answer 7

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: $S = (1 - x)^{-2}$.

Answer 8

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}

Answer 9

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

Answer 10

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

Answer 11

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.