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Does this series converge

$$\sum_{n=1}^\infty \frac{\cos(n)}{n}$$

someone has told me that I have to apply Dirichlet's test but I don't know how to calculate the sum

$$\left|\sum_{n=1}^{N} \cos(n)\right|$$

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    $\begingroup$ See this question: Computing the trigonometric sum $ \sum_{j=1}^{n} \cos(j) $ $\endgroup$ – Winther Nov 23 '16 at 0:18
  • $\begingroup$ @Winther How does that affect this question? Does it imply anything? $\endgroup$ – VermillionAzure Nov 23 '16 at 0:39
  • $\begingroup$ @VermillionAzure It implies convergence. Dirichlet's test says that $\sum a_n b_n$ converges when $\sum b_n$ ($= \sum \cos(n)$) is bounded and $a_n$ is decreasing with $a_n\to 0$. The question linked to above (which is also done in the answer below) shows that the $\cos$-sum is bounded. $\endgroup$ – Winther Nov 23 '16 at 0:41
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It is the finite sums that you have to bound (it does not imply that the series converges). You have \begin{align} \left|\sum_{n=1}^N\cos n\right| & =\left|\sum_{n=1}^N \operatorname{Re}\,e^{in}\right| =\left|\operatorname{Re}\,\sum_{n=1}^N e^{in}\right| =\left|\operatorname{Re}\,\frac{e^{i}-e^{i(N+1)}}{1-e^i}\right| \\[10pt] & \leq\left|\frac{e^{i}-e^{i(N+1)}}{1-e^i}\right| \leq\frac2{|1-e^i|} =\frac2{\sqrt{(1-\cos1)^2+\sin^21}}=\frac{\sqrt 2 }{\sqrt{1-\cos1}} \end{align}

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  • $\begingroup$ Thanks for the edits, Michael! $\endgroup$ – Martin Argerami Nov 23 '16 at 3:14
  • $\begingroup$ @Winther: nothing that three keystrokes cannot solve. Thanks! $\endgroup$ – Martin Argerami Nov 23 '16 at 3:18
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Multiply with $2\sin(\frac12)$ to get \begin{align} 2\sin(\frac12)\sum_{n=1}^N\frac{\cos n}{n} &=\sum_{n=1}^N\frac{\sin(n+\frac12)-\sin(n-\frac12)}{n} \\ &=\sin(\frac12)+\sum_{n=1}^{N-1}\frac{\sin(n+\frac12)}{n(n+1)}-\frac{\sin(N+\frac12)}{N+1} \end{align} The sum in the middle is obviously absolutely convergent for $N\to\infty$, from where the convergence of the original series follows.

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