Prove that $\sum\limits_{n=1}^{\infty} \frac{n}{3^n} = \frac{3}{4}$ 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. 
 A: 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
$\frac{1/3}{(2/3)^2}=3/4$
A: 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,
$$S_n=\frac34$$
A: 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,
$\sum_{n=1}^{\infty}n\frac{1}{3^n}=\frac{\frac{1}{3}}{\frac{4}{9}}=\frac{3}{4}.$
A: 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.
A: $$\sum_{n=1}^{\infty} \frac{n}{3^n} = \sum_{n=1}^{\infty}\sum_{k=n}^{\infty}\frac{1}{3^n}\frac{1}{3^{k-n}}=\\
=\sum_{n=1}^{\infty}\frac{\left(\frac{1}{3}\right)^n}{1-\frac{1}{3}}=\frac{3}{2}\sum_{n=1}^{\infty}\left(\frac{1}{3}\right)^n =\\
= \frac{3}{2}\frac{\frac{1}{3}}{1-\frac{1}{3}} = \frac{3}{2}\frac{1}{2}=\frac{3}{4}$$
