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I've done the following exercise:

Is series $\displaystyle\sum^{\infty}_{n=1}\frac{\cos(nx)}{n^\alpha}$, for $\alpha>0$, convergent?

My approach:

We're going to use the Dirichlet's criterion for convergence of series. Let $\displaystyle\ \{a_n\}=\frac{1}{n^{\alpha}}$ and $\{b_n\}=\cos(nx)$.

We see that $\{a_n\}$ is decreasing and has limit $0$. We have to see now that $$\sup_{N}{\left| \sum_{n=1}^{N}{\cos(nx)} \right|}<\infty.$$

$$\displaystyle \sum_{n=0}^{N}e^{inx}=\sum_{n=0}^{N}(e^{ix})^{n}= \frac{1-e^{ix(n+1)}}{1-e^{ix}}, \text{if x} \neq 2k\pi. $$

So $${\left| \sum_{n=0}^{N}{\cos(nx)} \right|}={\left| \Re\left(\sum_{n=0}^{N}{e^{inx}}\right) \right|}={\left| \Re\left(\frac{1-e^{ix(N+1)}}{1-e^{ix}}\right) \right|} \leq {\left| \frac{1-e^{ix(N+1)}}{1-e^{ix}} \right|}\leq {\frac{\left|1\right|+\left| e^{i(N+1)x} \right|}{\left|1-e^{{ix}} \right|}}=\frac{2}{\left|1-e^{{ix}} \right|}.$$

So, by Dirichlet, if $x\neq 2k\pi$, the series is convergent.

What happens if $x= 2k\pi$?

$$\displaystyle\sum^{\infty}_{n=1}\frac{\cos(n(2k\pi))}{n^\alpha}=\displaystyle\sum^{\infty}_{n=1}\frac{1}{n^\alpha}$$

and we know that this series converges when $\alpha>1$ and diverges if $\alpha\leq 1$.

Have I done any mistake/s? Is my approach correct? Thank you.

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  • 3
    $\begingroup$ Looks perfectly correct. $\endgroup$ – Daniel Fischer Oct 7 '15 at 18:49
  • $\begingroup$ Your analysis is correct! Well done. $\endgroup$ – Mark Viola Oct 7 '15 at 18:51
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    $\begingroup$ Good job, I guess you don't need any answers $\endgroup$ – user2566092 Oct 7 '15 at 19:00
  • $\begingroup$ $\Re\,\mathrm{Li}_{\alpha}\left(\,\mathrm{e}^{\,\mathrm{i}x}\right) = \mathrm{Ci}_{\alpha}\left(x\right)$. $\endgroup$ – Felix Marin Mar 4 '17 at 6:21
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Your approach is correct, and you didn't do any mistakes. Please accept this (or another) answer, so that this question will be complete.

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