find $\sum_{i=1}^\infty \frac{1}{n3^n}$ How to find $$\sum_{i=1}^\infty \frac{1}{n3^n}$$
Don't know how to start, any hints
A rigorous proof is also welcome

Consider
\begin{align*}
\sum_{n=1}^\infty\frac{x^n}{n}.
\end{align*}
This series converges for all $|x|<1$. Differentiation wrt $x$ yields
\begin{align*}
\sum_{n=1}^\infty x^{n-1}=\sum_{n=0}^\infty x^n=\frac{1}{1-x},
\end{align*}
and integrating again gives
\begin{align*}
\sum_{n=1}^\infty\frac{x^n}{n}=\int\frac{1}{1-x}dx+c=-\log(1-x)+c.
\end{align*}
When we plug in $x=0$ we obtain $c=0$, so we eventually have
\begin{align*}
\sum_{n=1}^\infty\frac{x^n}{n}=-\log(1-x),\qquad |x|<1.
\end{align*}
Now, for $x=\frac{1}{3}$ we obtain
\begin{align*}
\sum_{n=1}^\infty\frac{1}{n3^n}=\log\left(\frac{3}{2}\right).
\end{align*}
