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?

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.