Evaluate: $\sum_{n=1}^{\infty}\frac{1}{n k^n}$ How to evaluate this series for $k > 1$?
$$\sum_{n=1}^{\infty}\frac{1}{n k^n}$$
For $k = 2$, I tried to evaluate $\displaystyle \sum_{n = 0}^\infty  \int_{1}^{2} x^{-(n+1)}dx =   \int_{1}^{2} \sum_{n = 0}^\infty x^{-(n+1)}dx = \int_1^{2}\frac{1}{x(x-1)}dx$ $\displaystyle = \int_{1}^{2}\frac{1}{x-1}dx - \int_{1}^{2}\frac 1 x dx = \sum_{n=1}^\infty \left( \frac{1}{n} - \frac{1}{n2^n}\right)$
But both $\displaystyle \int_{1}^{2}\frac{1}{x-1}dx$ and $\displaystyle \sum_{n=1}^\infty\frac1n$ diverges. The answer is $\ln 2$, are these divergent terms equal?

$$\frac{1}{k} \sum_{n=1}^{\infty} \frac{1}{nk^{n-1}} =\frac{1}{k}  \sum_{n=1}^{\infty} \int^1_0 (x/k)^{n-1} dx=\frac{1}{k} \int^1_0 \sum_{n=1}^{\infty} (x/k)^{n-1} dx = \int^1_0 \frac{1}{k-x} dx= \log \left( \frac{k}{k-1} \right).$$
