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
$$\sum_{n=1}^{+\infty}\frac{\sin(n)}{n}.$$
 A: 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 & 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.
A: Use Dirichlet's test. The sequence $a_n = \dfrac 1n$ decreases to zero, and the series $\displaystyle \sum_{n=1}^\infty \cos n$ has bounded partial sums (this takes a little trigonometry to justify). This implies $\displaystyle \sum_{n=1}^{\infty}\frac{\cos n}{n}$ converges.
The other series is treated similarly.
A: We can use Fourier analysis to evaluate $\sum_{n=1}^{\infty} \sin(n)/n$ as follows.
The key is the Poisson summation formula: if $f(x)$ has Fourier transform $\hat{f}(u) = \int_{-\infty}^{\infty} f(x) e^{-2\pi i u t} dt$, then
$$\sum_{n=-\infty}^{\infty} f(n) = \sum_{k=-\infty}^{\infty} \hat{f}(k)$$
If we define
$$f(x) = \begin{align}
1 & \text{ if }|x| < \frac{1}{2\pi} \\
0 & \text{ otherwise} \\
\end{align}$$
then
$$\begin{align}
\hat{f}(u) &= \int_{-1/(2\pi)}^{1/(2\pi)} e^{-2\pi i u t} dt \\
&= \frac{e^{i u} - e^{- i u}}{2\pi i u}\\
&= \frac{1}{\pi} \frac{\sin(u)}{u}
\end{align}$$
Observe that
$$\sum_{n=-\infty}^{\infty} f(n) = 1$$
since only the $n=0$ index lies within the interval $[-1/(2\pi), 1/(2\pi)]$.
Therefore the Poisson summation formula gives us
$$1 = \sum_{n=-\infty}^{\infty} f(n) = \sum_{k=-\infty}^{\infty} \hat{f}(k) = 
\frac{1}{\pi}\sum_{k=-\infty}^\infty \frac{\sin(k)}{k}$$
Rearranging, we conclude that
$$\sum_{k=-\infty}^\infty \frac{\sin(k)}{k} = \pi$$
Since $\sin(k)/k$ is even, we have
$$\pi = \sum_{k=-\infty}^\infty \frac{\sin(k)}{k} = 1 + 2\sum_{k=1}^{\infty} \frac{\sin(k)}{k}$$
so
$$\sum_{k=1}^{\infty} \frac{\sin(k)}{k} = \frac{\pi - 1}{2}$$
A: 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.  
