Convergence of averaged sine function I have stumbled upon those two problems which I got a little stuck on that is show convergence or divergence for the series

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

and

$$\sum_{n=1}^{+\infty}\frac{\sin(n)}{n}.$$
The following answer is not entirely rigorous as written, but you can actually sum these series explicitly. Moreover I will proceed a bit more generically by considering $\{\cos n\phi,\sin n\phi\}$ for some real $\phi\in(0,\pi]$. (The case $\phi=0$ would give the divergent harmonic series, and other real $\phi$ can be determined by symmetry &amp; periodicity.) Recalling Euler's formula $e^{i \phi}=\cos\phi+i \sin \phi$, observe that

$$f(\phi):=\sum_{n=1}^\infty \frac{\cos n\phi}{n}+i \sum_{n=1}^\infty \frac{\sin n\phi}{n}=\sum_{n=1}^\infty \frac{e^{i n\phi}}{n}=\sum_{n=1}^\infty \frac{z^n}{n}$$

where $z:=e^{i \phi}$. We may recognize this last expression in terms of the Taylor series of the logarithm as $f(\phi)=-\ln(1-z)=-\ln(1-e^{i\phi}).$ (To check this, compute $f'(z)$ term by term and sum the resulting geometric series.) This can be simplified by observing that 
$$1-e^{i \phi}=e^{i\phi/2}\left(e^{-i\phi/2}-e^{i\phi/2}\right)=-2i e^{i\phi/2}\sin\frac{\phi}{2} = \left(2\sin\frac{\phi}{2}\right) e^{i(\phi-\pi)/2}.$$ and so $f(\phi)=-\ln\left(2\sin\dfrac{\phi}{2}\right)-i\left(\dfrac{\phi-\pi}{2}\right)$. As a check, note that $f(\pi)=\sum\limits_{n=1}^\infty \dfrac{1}{n}(-1)^n=-\ln 2$. With this in hand, we have $f(1)=-\ln\left(2\sin\dfrac{1}{2}\right)+i\left(\dfrac{\pi-1}{2}\right)$; to get the cosine and sine series, simply take the real and imaginary parts respectively. So we have summed the series.
We know that $(-1) \le \cos n$

So $(-1)/n \le \cos n/n$

Summation $(-1)/n$ is divergent. So by Comparison test of first type, summation $\cos n/n$ is divergent.