Find the sum of the series $\sum_{n=1}^{\infty}\frac{1}{n3^n}.$ I m stuck in finding the sum $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n3^n}.$ Please give me some hint.
 A: $$
\sum_{n=1}^\infty \frac{1}{n 3^n} = \sum_{n=1}^\infty \frac {x^n} n \quad \text{where } x = \frac 1 3. \tag 1
$$
\begin{align}
\frac d {dx} \sum_{n=1}^\infty \frac{x^n} n  \overset{\text{ ? }}{=} \sum_{n=1}^\infty \frac d {dx}\,\frac{x^n} n = \sum_{n=1}^\infty x^{n-1} = \frac 1 {1-x}.
\end{align}
So the sum is an antiderivative of $\dfrac 1 {1-x}$, evaluated at $x=1/3$.  Which antiderivative of that function is it?  It is the one that is equal to $0$ when $x=0$, because the expression in $(1)$ above is equal to $0$ when $x=0$.
PS suggested in "Dr. MV" 's comment: $\displaystyle \frac d {dx} \sum_{n=1}^\text{something} g_n(x) = \sum_{n=1}^\text{something} \frac d {dx} g_n(x)$ when $\text{“something''}$ (the number of terms being added) is finite.  But this equality doesn't always hold when $\text{“something''}$ is $\infty$. That is why the $\text{“?''}$ surmounts the $\text{“}=\text{''}$ above.  However it is true when the sum is a convergent power series, provided $x$ is in the interior (not on the boundary) of the disk of convergence.
A: Here, we use a standard "trick" for evaluating certain types of series.  The key of the development relies on the observation that
$$\int_0^x t^{n-1}\,dt=\frac{x^n}{n} \tag 1$$
Then, summing both sides of $(1)$ over $n$, we obtain for $|x|<1$
$$\begin{align}
\sum_{n=1}^{\infty}\frac{x^n}{n}&=\sum_{n=1}^{\infty}\int_0^x t^{n-1}\,dt\\\\
&=\lim_{N\to \infty}\sum_{n=1}^N \int_0^x t^{n-1}\,dt\\\\
&=\lim_{N\to \infty} \int_0^x \sum_{n=1}^N t^{n-1}\,dt\\\\
&=\lim_{N\to \infty}\int_0^x\frac{1-t^N}{1-t}\,dt\\\\
&=\int_0^x \lim_{N\to \infty}\left(\frac{1-t^N}{1-t}\right)\,dt\\\\
&=\int_0^x \frac{1}{1-t}\,dt\\\\
&=-\log(1-x)
\end{align}$$
where the Dominated Convergence Theorem guarantees the validity of the third to last equality.  Now, set $x=1/3$.
A: $$\sum_{n = 1}^{+\infty} \frac{1}{n3^n} = \sum_{n = 1}^{+\infty}\frac{3^{-n}}{n}$$
This has the form of the well known series
$$\sum_{k = 1}^{+\infty}\frac{a^{-k}}{k} =  - \log\left(\frac{a-1}{a}\right)$$
Thence
$$\sum_{n = 1}^{+\infty} \frac{1}{n3^n} =  -\log\left(\frac{3-1}{3}\right) = - \log\frac{2}{3} = \log\frac{3}{2}$$
