Convergence of $\Sigma_{k=1}^{\infty}(\frac{z}{k}+\log(1-\frac{z}{k}))$

Question

I would like to prove that $\Sigma_{k=1}^{\infty}(\frac{z}{k}+\log(1-\frac{z}{k}))$ converges for every complex $z$.

I know that $\log(1-x)=\Sigma_{n=1}^{\infty}-\frac{x^n}{n}$, but I stuck on the double summation here so maybe it is not the right approach. Any help is appreciated and thanks in advance!

Answer 1

Fix $z \in \mathbb{C}$ and let $N$ be a positive integer such that $N > 2|z|$.

For all $k \geqslant N$ we have $|z|/k < 1$ and $\log(1 - z/k)$ is analytic in the disk $D(0,N/2)$, whence

$$\log \left(1 - \frac{z}{k} \right) = -\sum_{n=1}^\infty \frac{z^n}{nk^n}. $$

Hence,

$$\left|\frac{z}{k} + \log \left(1 - \frac{z}{k} \right)\right| \leqslant \sum_{n=2}^\infty \frac{|z|^n}{nk^n} \leqslant \frac{|z|^2}{k^2} \sum_{n=0}^\infty \frac{|z|^n}{N^n} = \frac{|z|^2}{1 - |z|/N}\frac{1}{k^2} < \frac{2|z|^2}{k^2}.$$

The last inequality follows from $|z|/N < 1/2$ and $1 - |z|/N > 1/2$. Hence, we have

$$\left|\frac{z}{k} + \log \left(1 - \frac{z}{k} \right)\right| < \frac{2|z|^2}{k^2}.$$

Therefore, the series converges by the comparison test.

Answer 2

Let $k>|z|$. Then $$\frac z k+\log\left(1-\frac z k\right)=-\frac{z^2}{2k^2}-\frac{z^3}{3k^3}-\cdots=O((z/k)^2).$$ For a fixed $z$, then the sum is convergent by comparison to $\sum_k 1/k^2$. The convergence is uniform on any compact subset of $\Bbb C$.