How to prove that this complex series is Conditionally Converagent?

Question

How can I prove that this series is conditionally convergent,

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

I tried to write $\exp{in}= \sin(n) + i \cos(n)$

then the series splits into two series with general terms $a_n= \sin(n)/n$ and $b_n= \cos(n)/n$

How can I prove that this series are convergent but the series of their absolute values are divergent?

Taking absolute values, you get the harmonic series $\sum 1/n$, so it cannot be absolutely convergent. — Aaron, Mar 16 '16 at 16:58
Yes. it absolutely diverges, the problem is to proof that it converges conditionally, my problem is that i cannot find that this series converges, i don't know how — programmer0, Mar 16 '16 at 17:02
after summation by part, it becomes absolutely convergent , this is because $\displaystyle \sum_{n=0}^N e^{in} = \frac{1-e^{i(N+1)}}{1-e^i}$ is bounded independently of $N$ — reuns, Mar 16 '16 at 18:36

C. Dubussy · Accepted Answer · 2016-03-16T17:56:44.350

2

Let us use Dirichlet's criterion. Of course $\frac{1}{n}$ decreases to $0$. Now consider $\left| \sum_{n=p}^q e^{in} \right|$ for all $q>p$. If we can bound this sum with a constant (independent of $p$ and $q$) then we will be able to conclude that the series is conditionally convergent. We have $$\left| \sum_{n=p}^q e^{in} \right| = \left|e^{ip}\sum_{n=0}^{q-p}e^{in}\right| = \left| \frac{1-e^{i(q-p+1)}}{1-e^i} \right| \leq 2 \left| \frac{1}{1-e^i} \right|.$$ The final inequality holds thanks to Minkowski and the constant does not depend on $p,q$, hence your series is convergent.

edited Mar 16 '16 at 17:56

answered Mar 16 '16 at 17:19

C. Dubussy

9,120

This is usually called Dirichlet's criterion. – Julián Aguirre Mar 16 '16 at 17:35
You're right, I've edited. – C. Dubussy Mar 16 '16 at 17:56
@JuliánAguirre in my opinion, that so called "Dirichlet's criterion" should be better called the "summation by part" criterion – reuns Mar 16 '16 at 18:39

How to prove that this complex series is Conditionally Converagent?

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