# Convergence of series to log(1/(1-k)) [closed]

In my textbook, there is a series going from x=1 to infinity, whereby each added term is k^x/x, where k is a constant between 0 including and 1 excluding. Now my textbook says that this series converges to log(1/(1-k)).

Why?

## closed as off-topic by Chappers, TheGeekGreek, Foobaz John, C. Falcon, Trevor GunnJun 11 '17 at 2:31

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$f(a)=\displaystyle \sum_{n=1}^\infty \dfrac{a^n}{n}$. Take $f'(a) =$.....a geometric series . Can you continue?

Let $0<k<1$ then we have

$$\sum_{x=0}^{\infty}k^x=\frac{1}{1-k}$$

I use another symbol besides $k$, say $s$ and integrate from $0$ to $k$

$$\int_{0}^{k}\sum_{x=0}^{\infty}s^x\mathbb{d}x=\int_{0}^{k}\frac{1}{1-s}\mathbb{d}s$$

$$\sum_{x=0}^{\infty}\int_{0}^{k}s^x\mathbb{d}s=-\log|1-s||_{0}^{k}$$

$$\sum_{x=0}^{\infty}\frac{s^{x+1}}{x+1}|_{0}^{k}=-(\log|1-k|-\log|1|)$$

$$\sum_{x=0}^{\infty}\frac{k^{x+1}}{k+1}=-\log(1-k)=\log(\frac{1}{1-k})$$

changing the index $x+1->x$

$$\sum_{x=1}^{\infty}\frac{k^{x}}{k}=\log(\frac{1}{1-k})$$