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.
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.
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.