Evaluate: $\sum_{n=1}^{\infty}\frac{1}{n k^n}$
Solution 1:
$$\frac{1}{k} \sum_{n=1}^{\infty} \frac{1}{nk^{n-1}} =\frac{1}{k} \sum_{n=1}^{\infty} \int^1_0 (x/k)^{n-1} dx=\frac{1}{k} \int^1_0 \sum_{n=1}^{\infty} (x/k)^{n-1} dx = \int^1_0 \frac{1}{k-x} dx= \log \left( \frac{k}{k-1} \right).$$
Solution 2:
Hint: for $0<x<1$, we have $\sum_{n=0}^\infty x^n = \frac{1}{1-x}$
Now integrate with respect to $x$, and then let $x=\frac{1}{k}$