Given series $\sum_{n=1}^{\infty}\frac {\cos(n^2)}n$.

It is easy to prove it does not converge absolutely.

I need to prove that it converges сonditionally.

I thought about using Dirichlet's test because $1/n$ series is monotone and $\lim_{n\rightarrow\infty} 1/n = 0$.

So the thing i need to prove is $\left|\sum_{1}^k \cos(n^2)\right| < M,\ \forall k$.

If there was $\cos(n)$ instead of $n^2$, it would be easy to prove this statement by multiplication and division by $\cos(0.5)$ and then using some trigonometric formula so that $\left|\sum_{1}^k \cos(n^2)\right| = \left|\frac{cos(0.5)-cos(n-0.5)}{\cos(0.5)}\right| < 2$ or something like that. But this approach seems to be impossible for $\cos(n^2)$.

Maybe there is a way to prove it using the fact that $\int \cos(x^2) = \sqrt{2/\pi}$?

Or any simpler way?

  • $\begingroup$ Your bounds would need to go from $1$ to $\infty$ since $\frac{\cos(n^2)}{n}$ is undefined at $0$. $\endgroup$ – Kaj Hansen May 25 '14 at 23:02
  • $\begingroup$ Sure, I'll fix it, thanks. $\endgroup$ – Daniel May 25 '14 at 23:03

It is sufficient to show that $$\left|\sum_{n=0}^{N}\cos(n^2)\right|\leq C\sqrt{N}\log N\tag{1}$$ to ensure convergence by partial summation. Consider that: $$\left\|\sum_{n=1}^{N}e^{in^2}\right\|^2 = \left(\sum_{n=1}^{N}e^{in^2}\right)\cdot\left(\sum_{n=1}^{N}e^{-in^2}\right)=N+\sum_{d=1}^{N-1}\sum_{r=1}^{N-d}2\cos(2dr+d^2),\tag{2}$$ and that: $$(2)\ll \sum_{d=1}^{N-1}\min\left(N-d,\left\|\frac{d}{\pi}\right\|^{-1}\right)\tag{3}$$ (where $\|x\|$ denotes the distance of $x$ from the closest integer) by the usual arguments about simple exponential sums. If now we take $\frac{a}{q}$ as a good rational approximation of $\frac{1}{\pi}$, $\left|\frac{a}{q}-\frac{1}{\pi}\right|<\frac{1}{3Nq}$, it is not difficult to see that: $$ (3)\ll \sum_{\substack{d=1\\q\nmid d}}^{N}\left\|\frac{a d}{q}\right\|^{-1}\ll(N+q)\log q\ll N\log N,\tag{4}$$ hence $(1)$ holds and the series $\sum_{n=1}^{+\infty}\frac{\cos n^2}{n}$ converges.

| cite | improve this answer | |
  • $\begingroup$ This is also known as "Weyl differencing trick". $(1)$ is just a consequence of the general en.wikipedia.org/wiki/Weyl%27s_inequality . $\endgroup$ – Jack D'Aurizio May 25 '14 at 23:31
  • $\begingroup$ I do not understand how to prove convergence then. What kind of convergence test that is? $\endgroup$ – Daniel May 25 '14 at 23:59
  • $\begingroup$ You can call it "generalized Dirichlet test", but is just partial summation. What we have is: $$\sum_{n=1}^{N}\frac{\cos(n^2)}{n}=\frac{1}{N}\sum_{n=1}^{N}\cos(n^2)+\sum_{n=1}^{N-1}\frac{1}{n(n+1)}\sum_{k=1}^{n}\cos(n^2)\ll\sum_{n=1}^{N}\frac{\sqrt{n}\log n}{n^2}.$$ $\endgroup$ – Jack D'Aurizio May 26 '14 at 2:14
  • $\begingroup$ Hi, great solution! I have a question: can you give some estimates for sums of form $\sum_{n=1}^N exp( i f(n))$ where $f$ is not a polynomial, but a bit more complicated, say $f(n) = \sqrt{n}$. ? $\endgroup$ – orangeskid Aug 18 '17 at 11:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.