How to prove that $\,\,\sum_{n=1}^{\infty}\frac{1}{n^{1+i}}\,$ diverges

Question

I am stuck on the following question :

Prove that $$\sum_{n=1}^{\infty}\frac{1}{n^{1+i}}$$ diverges, where $i=\sqrt{-1}$

I am not sure how to progress with it. Can someone explain? Thanks and regards to all.

Answer 1

Although this is essentially the same question as Divergence of $\sum\limits_{n=1}^{\infty} \frac{\cos(\log(n))}{n}$ and Divergence of $ \sum_{n=1}^{\infty} \frac{1}{n^{i+1}}$, maybe one more explanation will help. The argument of $n^i = \exp(i\log n)$ changes slowly when $n$ is large. More precisely, within the range $\{n: k\le n\le 2k \}$ the argument changes by at most $\log 2\le \pi/4$. When $|\arg z|\le \pi/4$, we have $\operatorname{Re}z\ge |z|/\sqrt{2}$. Hence, $$ \operatorname{Re} e^{i\log k}\sum_{n=k}^{2k} \frac{1}{n^{1+i} } = \operatorname{Re} \sum_{n=k}^{2k} \frac{e^{i (\log k -\log n)}}{n } \ge \frac{1}{\sqrt{2}}\sum_{n=k}^{2k} \frac{1}{n } \ge \frac{1}{2\sqrt{2}} $$ where the last inequality uses the fact that each of $2k+1$ terms is at least $1/2k$. Thus, the series fails the Cauchy criterion for convergence.