# Convergence of $\sum_{n=0}^\infty \frac{\cos(n)}{n}$

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

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

• See this question: Computing the trigonometric sum $\sum_{j=1}^{n} \cos(j)$ – Winther Nov 23 '16 at 0:18
• @Winther How does that affect this question? Does it imply anything? – VermillionAzure Nov 23 '16 at 0:39
• @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. – Winther Nov 23 '16 at 0:41

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.