# How to prove that this complex series is Conditionally Converagent?

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?

• @LeGrandDODOM - $\exp{in}= i \sin(n) + \cos(n)$ – Liebe Mar 16 '16 at 16:56
• @Liebe it's up to him to fix this. – Gabriel Romon Mar 16 '16 at 16:57
• Taking absolute values, you get the harmonic series $\sum 1/n$, so it cannot be absolutely convergent. – Aaron Mar 16 '16 at 16:58
• This may help: math.stackexchange.com/a/13494/169852 – Bungo 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

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.