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

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}

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.