sequences and series - Prove that $sum_{n=1}^infty{cos(nx) over n^2}$ converges for all real $x$.

Prove that $$\sum_{n=1}^\infty{\cos(nx) \over n^2}$$ converges for all real $x$.

We know that for all $t\in\mathbb R$ that $|\cos(t)| \le 1$. Does it suffice to show that $$\sum_{k=1}^n\left|{1\over k^2}\cos(kx)\right| = \sum_{k=1}^n{1\over k^2}|\cos(kx)| \le \sum_{k=1}^n{1\over k^2}(1) \le \sum_{k=1}^\infty{1\over k^2} < \infty?$$

If we can show absolute convergence, then this will imply convergence of the original infinite series. However, does this implication still hold for all $x \in \mathbb R$? My intuition says yes, but we know that for all $n\in\mathbb N$ that $$\sum_{n=1}^k\cos(nx) = {\sin\left[\left(n+{1\over2}\right)x\right] - \sin({x\over2})\over2\sin({x\over2})},$$ is bounded where $x$ is not an integer multiple of $2\pi$. I understand the sequence of partial sums could be different between different series, but different enough that an unbounded sequence for some values of $x$ becomes bounded for all $x$?

Answer

$$(\forall n>0)\;\;(\forall x\in \mathbb R \; ) \;|\frac{\cos(nx)}{n^2}|\leq \frac{1}{n^2}$$

$\implies$

$$\sum \frac{\cos(nx)}{n^2}$$

normally convergent at $\mathbb R$ since $\sum \frac{1}{n^2} \;$ is absolutly convergent.

$\implies$

it converges uniformly at $\mathbb R$.

$\implies$

$$(\forall x\in \mathbb R )\; \sum_{n=1}^{+\infty}\frac{\cos(nx)}{n^2} \in \mathbb R$$