Convergence of $\sum_{n=1}^{\infty}\frac{e^{in}}{n}$

Question

How to analyze the convergence of the series $$\sum_{n=1}^{\infty}\frac{e^{in}}{n}$$ ?

Which test should I use?

I'm lost in this topics. Any hint/help will be appreciated.

Answer 1

In the Dirichlet's test as described here, take $a_n = \frac{1}{n}$ and $b_n = e^{in}$. Then check the required conditions for the series $\sum a_nb_n$ to converge.

$$a_{n+1} = \frac{1}{n+1} \leq \frac{1}{n} = a_n$$ $$\lim\limits_{n\rightarrow\infty}a_n = \lim\limits_{n\rightarrow\infty}\frac{1}{n} = 0$$ $$\left|\sum_{n=1}^{N} b_n\right| = \left|\sum_{n=1}^{N} e^{in}\right| \leq \left|\frac{1}{\sin(1/2)}\right| = M$$

The answer uses $\sum\limits_{n=1}^{N}e^{in} = e^{i(N-1)/2}\frac{\sin(N/2)}{sin(1/2)}$ as shown in the answer by @Salahamam_Fatima.

Answer 2

hint

$$e^{in}=\cos (n)+i\sin (n) $$

with Dirichlet test for $$\sum \frac {\cos (n)}{n} $$and $$\sum \frac {\sin (n)}{n} $$ using $$e^i+e^{2i}+...e^{ni}=$$ $$e^{i(n-1)/2}\frac {\sin (n/2)}{\sin (1/2)} $$

Answer 3

Multiply with $1-e^i\ne 0$ to get the series $$ e^i-\sum_{n=2}^\infty \frac{e^{in}}{n(n-1)} $$ which is absolutely convergent.

In principle this is the direct application of the proof method of the Dirichlet test.