Divergence of $\sum\frac{\cos(\sqrt{n}x)}{\sqrt{n}}$

I have difficulties in showing the series $f(x)=\sum_{n=1}^\infty \frac{\cos(\sqrt{n}x)}{\sqrt{n}}$ is divergent at every real numbers $x$.

However I cannot find any elementary methods to do this. Can anyone help me on this?

Thanks.

Solutions Collecting From Web of "Divergence of $\sum\frac{\cos(\sqrt{n}x)}{\sqrt{n}}$"

Since you asked for elementary methods, here is one. Intuitively the idea for your first series is that $\sqrt{n}x$ increases more and more slowly and that one can find larger and larger intervals of integers $n$ such that, on each of these intervals the cosines are uniformly larger than a positive constant, a fact which brings you back to the evaluation of finite sums of $1/\sqrt{n}$.

More precisely, once you solved the case $x=0$, assume without loss of generality that $x$ is positive and choose your favorite angle whose cosine is positive, for example $\cos(\pi/3)=1/2$. Hence $\cos(\sqrt{n}x)\ge1/2$ for every $n$ such that there exists an integer $k$ such that
$$
2k\pi-\pi/3\le\sqrt{n}x\le2k\pi+\pi/3.
$$
This condition translates as $n\in I_k$ where $I_k$ is an integer interval of width of order $(wk)$ around an integer of order $(ck)^2$, where $c=2\pi/x$ and $w=8\pi^2/(3x^2)$.

The sum of $\cos(\sqrt{n}x)/\sqrt{n}$ over $I_k$ is at least $1/2$ times the sum over $I_k$ of $1/\sqrt{n}$. This last sum is of order $1/(ck)$ times the number of terms in $I_k$, which is of order $(wk)$. Thus the sum of $\cos(\sqrt{n}x)/\sqrt{n}$ over $I_k$ is at least $w/(2c)+o(1)$.

When a series converges, each of its partial sums for $n$ in an interval $[n_0,n_1]$ is as small as one wants provided $n_0$ is large enough (this in fact characterizes convergent series). We disproved this property, hence the series diverges.