# 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.

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It may be helpful to consider the cases when $x=0$ and when $x\neq 0$ separately. Also, when you speak of elementary methods, do these include basic tests such as the the divergence test, integral test, comparison test, ration & root tests, alternating series test? Note that for $x=0$, $f(x)$ behaves something like $\frac{1}{n}$. I'm not sure if that helps, I hope it does. –  fdart17 Mar 14 '11 at 19:00
@Fdart17: Yes, that includes all of the above test as well as Abel's Test, Dirichlet's Test, etc. commonly encountered in a basic course in analysis. But all the above tests seem not to apply to this case. –  digiboy1 Mar 15 '11 at 5:10
Did you get something out of the answer below? –  Did Apr 7 '11 at 8:05

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.

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I think it should be $w=8\pi/(3x^2)$ (without a square on $\pi$)? –  joriki Mar 15 '11 at 6:44
@joriki No: $(2k\pi\pm\pi/3)^2$ is something $\pm4k\pi^2/3$ hence one asks $nx^2$ to be in an interval whose width is twice $4k\pi^2/3$. –  Did Mar 15 '11 at 6:51
Sorry, don't know where my mind was -- I didn't see that there's a $\pi$ in both terms :-) –  joriki Mar 15 '11 at 6:56