Convergence of $\sum_{k=1}^\infty \frac{\cos(2\pi kx)}{k}$ How to find which $x\in[0,1]$ the sum $$\sum_{k=1}^\infty\frac{\cos(2\pi kx)}{k}$$ converge using the identity $$\sum_{k=1}^n a_kb_k=A_nb_n+\sum_{k=1}^{n-1}A_k(b_k-b_{k+1}), \qquad A_k=\sum_{j=1}^k a_j.$$
$x\in\{0,1\}$ clearly makes divergent as $$\sum_{k=1}^\infty\frac{1}{k}>\infty.$$
Using the identity $$\sum_{k=1}^n a_kb_k=A_nb_n+\sum_{k=1}^{n-1}A_k(b_k-b_{k+1}), \qquad A_k=\sum_{j=1}^k a_j,$$ we have \begin{align*}\sum_{k=1}^n\frac{\cos(2\pi kx)}{k}&=\sum_{k=1}^n\cos(2\pi kx)\cdot\frac{1}{k}\\&=\sum_{k=1}^n a_kb_k\\&=A_nb_n+\sum_{k=1}^{n-1}A_k(b_k-b_{k+1})\\&=A_n\frac{1}{n}+\sum_{k=1}^{n-1}A_k\frac{1}{k(k+1)}.\end{align*}
Since $$\lvert A_k \rvert\le\sum_{j=1}^k \lvert a_j\rvert\le\sum_{j=1}^k 1=k$$ we have \begin{align*} \biggl|\sum_{k=1}^n\frac{\cos(2\pi kx)}{k}\biggr| &\le \lvert A_n\rvert \frac{1}{n}+\sum_{k=1}^{n-1}\lvert A_k\rvert \frac{1}{k(k+1)} \\&\le n\frac{1}{n}+\sum_{k=1}^{n-1}k \frac{1}{k(k+1)} \\&= 1+\sum_{k=1}^{n-1} \frac{1}{k+1}\end{align*} which does not help much.
Any suggestions how to make an upper bound that is more 'limited' so that I can find the $x$ for which it converges?

Fix $x\in(0,1)$. Using
$$ \sin x\cos y=\frac{1}{2}\bigg[\sin(x+y)+\sin(x-y)\bigg], \sin x-\sin y=2\sin(\frac{x+y}2)\cos(\frac{x-y}2) $$
one has
$$ \sin(\pi x)\sum_{k=1}^n\cos(2\pi kx)=\frac12\sum_{k=1}^n\bigg[\sin(\pi(2k+1)x)-\sin(\pi(2k-1)x)\bigg]=\frac12\bigg[\sin(\pi(2n+1)x)-\sin(\pi x)\bigg]=\sin(\pi(n+1) x)\cos(\pi n x) $$
and hence
$$ \sum_{k=1}^n\cos(2\pi kx)=\frac{\sin(\pi(n+1) x)\cos(\pi n x)}{\sin(\pi x)}. $$
So
$$ \bigg|\sum_{k=1}^n\cos(2\pi kx)\bigg|=\bigg|\frac{\sin(\pi(n+1) x)\cos(\pi n x)}{\sin(\pi x)}\bigg|\le\frac{1}{\sin(\pi x)}. $$
