Is the series: $\,\sum_{n=1}^{\infty}\frac{\mathrm{e}^{-in}}{n}\,$ divergent?

Question

According to mathematica, the complex series $\displaystyle\sum_{n=1}^{\infty}\frac{e^{-in}}{n}$ does not converge.

I know that the factor $\dfrac{1}{n}$ in the above series is diverging, but I don't know how to relate this when a complex number is multiplied.

Answer 1

This is not divergent. It converges conditionally, but it does not converge absolutely.

Let $$ s_n=\mathrm{e}^{-i}+\mathrm{e}^{-2i}+\cdots+\mathrm{e}^{-ni}=\frac{\mathrm{e}^{-i}-\mathrm{e}^{-ni}}{1-\mathrm{e}^{-i}}. $$ Clearly $$ \lvert s_n\rvert\le \frac{2}{1-\cos(1)}=M. $$ Then $$ \sum_{k=1}^n\frac{\mathrm{e}^{-ki}}{k}=\sum_{k=1}^n\frac{s_k-s_{k-1}}{k}= \sum_{k=1}^n\frac{s_k}{k}-\sum_{k=2}^n\frac{s_{k-1}}{k}= \sum_{k=1}^n\frac{s_k}{k}-\sum_{k=1}^{n-1}\frac{s_{k}}{k+1}=\frac{s_n}{n}+ \sum_{k=1}^{n-1}\frac{s_k}{k(k+1)}. $$ It is now clear that each of the two sequences converges: $$ \left|\frac{s_n}{n}\right|\le\frac{M}{n}\to 0, $$ while $$ \sum_{k=1}^{n-1}\frac{s_k}{k(k+1)} $$ is dominated by $$ \sum_{k=1}^{n-1}\frac{M}{k(k+1)} $$ and hence it also converges.