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.

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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

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

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,
$$S_n=\frac34$$

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,

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

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.

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