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 $$\sum_{n=0}^\infty (n+1)x^n?$$
$\endgroup$ 223 Answers
$\begingroup$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 $(1-r)^{k+1}$.
$\endgroup$ 2 $\begingroup$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}
$\endgroup$ 6 $\begingroup$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.
$\endgroup$ $\begingroup$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.
$\endgroup$ 0 $\begingroup$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).
$\endgroup$ 0 $\begingroup$Hints
You know (don't you?) the formula for $S(a) = \sum_{n=0}^\infty a^n$ for $|a| < 1$
Take the derivative (with respect to $a$) of both sides to obtain a formula for $\sum_{n=1}^\infty n a^n$
Show that your series can be put in that form.
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: $S = (1 - x)^{-2}$.
$\endgroup$ 6 $\begingroup$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 $\sum_{n=1}^\infty nx^{n-1}$.
$\endgroup$ 2 $\begingroup$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 $$\sum_{n=0}^\infty{(n+1)x^n}={1\over (1-x)^2}.$$
$\endgroup$ $\begingroup$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 $$ S_{n}(z)=\frac{z}{1-z}\left[1+\sum_{k=0}^{n-1}\binom{n}{k} S_k(z)\right]\qquad n=1, 2, \ldots $$
$\endgroup$ 0 $\begingroup$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 $$ \frac{9}{4} =\sum_{n=0}^{+\infty}(n+1)\frac{1}{3^n} = \sum_{m=1}^{+\infty}m\frac{1}{3^{m-1}} \quad \Rightarrow \quad \sum_{m=1}^{+\infty}\frac{m}{3^m} = \frac{3}{4} $$
$\endgroup$ $\begingroup$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 $\sum_{n\geq0}(n+1)x^n$.
$\endgroup$ 1 $\begingroup$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!
$\endgroup$ $\begingroup$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 $$\sum_{n=0}^\infty (n+1)x^n = \frac{1}{(1-x)^2}$$
$\endgroup$ $\begingroup$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
$$ \sum_{n=0}^\infty(n+1)x^n=\frac{1}{(1-x)^2}. \tag3 $$
$\endgroup$ $\begingroup$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. $x$.
$\endgroup$ $\begingroup$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
$$ \sum_{k=0}^\infty (k+1)x^k=-\frac1{x-1}+\frac{x}{(x-1)^2}=\frac1{(x-1)^2}\tag6 $$
$\endgroup$ $\begingroup$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 $$ \sum_{k=0}^\infty(k+1)x^k=\frac1{(1-x)^2}\tag3 $$
$\endgroup$ 4 $\begingroup$$\endgroup$ $\begingroup$\begin{align} \sum_{n=0}^\infty (n+1)x^n &= \sum_{n=1}^\infty nx^{n-1} =\frac{d}{dx}\left( \sum_{n=0}^\infty x^{n}\right) =\frac{d}{dx}\left(\frac{1}{1-x}\right)=\frac{1}{(1-x)^2} \end{align} \begin{align} \tag*{$\Box$} \end{align}
What about the Cauchy product?$$ \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}. $$
$\endgroup$ 1 $\begingroup$Suppose you play a repeatable game where you have probability $x$ of losing. Notice that
\begin{align} \sum\limits_{n=0}^{\infty} (n+1)x^n &= \sum\limits_{n=1}^{\infty} nx^{n-1} \\ &= \frac{1}{1-x} \sum\limits_{n=1}^{\infty} n (1-x)x^{n-1} \\ &= \frac{1}{1-x} \text{E}[n] \end{align}
where $\text{E}[n]$ is the expected number of times you have to play before your first win (you lose $n-1$ times and then win with probability $1-x$ in the $n$th game).
Now
\begin{align} \text{E}[n] &= 1 + x \text{E}[n] \end{align}
because you either win in the first game, or you lose with probability $x$ and restart the process, with $\text{E}[n]$ more games to play. Then
\begin{align} \text{E}[n] &= \frac{1}{1-x} \end{align}
and thus
\begin{align} \sum\limits_{n=0}^{\infty} (n+1)x^n &= \frac{1}{(1-x)^2} \end{align}
(essentially we just re-derived the formula for the sum of a geometric series, but it's a fun way of doing it!)
$\endgroup$ $\begingroup$Firstly, we need an infinite Geometric Series.
$$\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}}
$$
For $|x|<1,$ recall the geometric series identity $\sum_{k=1}^\infty x^{k-1}=\frac{1}{1-x},$ so by Fubini's theorem,
$$\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.
$\endgroup$
