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How can you derive that $$ \sum\limits_{n=1}^{\infty} \frac{n}{3^n} = \frac{3}{4} \, ?$$ I suspect some clever use of the geometric series will do, but I don't know how.

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    $\begingroup$ Compare to the same expression where you substitute $n-1$ for $n$. $\endgroup$
    – user65203
    Feb 6, 2017 at 7:55
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    $\begingroup$ See this. $\endgroup$ Feb 6, 2017 at 8:36

5 Answers 5

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We can proceed with generating functions. The trick initially seems a little contrived, but it can be made rigorous, and I think that the more you see it the more it makes sense. The first implication comes from differentiation. $$\frac{1}{1-x}=\sum_{n=0}^{\infty} x^n \implies \frac{1}{(1-x)^2}=\sum_{n=0}^{\infty}n \cdot x^{n-1}$$

So we can multiply by $x$ to get

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

Substituting $x=1/3$, we obtain

$\frac{1/3}{(2/3)^2}=3/4$

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Let $$S_n= \sum_{n=1}^{\infty} \frac{n}{3^n}\tag1$$ $$\frac13S_n=\sum_{n=1}^{\infty} \frac{n}{3^{n+1}}=\sum_{n=1}^{\infty} \frac{n+1}{3^{n+1}}-\sum_{n=1}^{\infty}\frac1{3^{n+1}}\tag2$$

$(1)-(2)$, $$\begin{align}\frac23S_n&=\sum_{n=1}^{\infty} \frac{n}{3^n}-\sum_{n=1}^{\infty} \frac{n+1}{3^{n+1}}+\sum_{n=1}^{\infty}\frac1{3^{n+1}}\\ &=\sum_{n=1}^{\infty} \frac{n}{3^n}-\sum_{n=2}^{\infty} \frac{n}{3^{n}}+\sum_{n=1}^{\infty}\frac1{3^{n+1}}\\ &=\frac13+\sum_{n=1}^{\infty}\frac1{3^{n+1}}\\ &=\sum_{n=1}^{\infty}\frac1{3^{n}}=\frac12\end{align}$$

Thus, $$S_n=\frac34$$

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  • $\begingroup$ 1. No $S_n$ actually depends on $n$. 2. The justification of the step $(1)-(2)$ requires to know beforehand the series converges. $\endgroup$
    – Did
    Aug 11, 2018 at 16:55
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Take a geometric series with term $r$ and take derivative with respect to $r$ both sides:

$\frac{d}{dr}\sum_{n=0}^{\infty}r^n=\frac{d}{dr}\frac{1}{1-r}.$

Derivate,

$\sum_{n=0}^{\infty}nr^{n-1}=\frac{1}{(1-r)^2}.$

At the left hand side, for $n=0$ the term is zero, so we can start the sum at $n=1$, and also multiply $r$ both sides to balance the exponent:

$r\sum_{n=1}^{\infty}nr^{n-1}=r\frac{1}{(1-r)^2}.$

Now you have

$\sum_{n=1}^{\infty}nr^{n}=\frac{r}{(1-r)^2}.$

For $r=\frac{1}{3}$,

$\sum_{n=1}^{\infty}n\left(\frac{1}{3}\right)^{n}=\frac{\frac{1}{3}}{\left(1-\frac{1}{3}\right)^2}.$

Operate,

$\sum_{n=1}^{\infty}n\frac{1}{3^n}=\frac{\frac{1}{3}}{\frac{4}{9}}=\frac{3}{4}.$

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Consider the geometric series $\sum_{n=0}^\infty x^k=\frac{1}{1-x}$ for $0<x<1$.

Take the derivative with respect to $x$ on both sides. This leads to

\begin{align*} \sum_{n=1}^{\infty}nx^{n-1}&=\frac{1}{(1-x)^2}\qquad |\cdot x\\ \sum_{n=1}^\infty nx^n&=\frac{x}{(1-x)^2} \end{align*}

Setting $x=\frac{1}{3}$ gives the desired result.

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$$\sum_{n=1}^{\infty} \frac{n}{3^n} = \sum_{n=1}^{\infty}\sum_{k=n}^{\infty}\frac{1}{3^n}\frac{1}{3^{k-n}}=\\ =\sum_{n=1}^{\infty}\frac{\left(\frac{1}{3}\right)^n}{1-\frac{1}{3}}=\frac{3}{2}\sum_{n=1}^{\infty}\left(\frac{1}{3}\right)^n =\\ = \frac{3}{2}\frac{\frac{1}{3}}{1-\frac{1}{3}} = \frac{3}{2}\frac{1}{2}=\frac{3}{4}$$

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