# 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

• Integrate an appropriate series. – David Mitra Dec 29 '14 at 18:13
• ... or differentiate this one with $1/3$ replaced by $x$. – Robert Israel Dec 29 '14 at 18:14

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*}
• Couldn't you write $dx$ instead of wrt $x$? – Mohammad Zuhair Khan Jan 12 '18 at 4:16
• "Differentiation dx" does not make sense imo. Here we take the derivative of a function with respect to (=wrt) its variable $x$, and that's why I've written "Differentiation wrt x". – sranthrop Jan 12 '18 at 12:36