Convergence of $\sum \frac{ \cos\sqrt n} {\sqrt n}$

Question

As you undestand from the topic title, I am wondering how to determine whether the series $$ \sum_{n=1}^{\infty} \frac{ \cos\sqrt n} {\sqrt n} $$ converges or not(it diverges actually).

You know, the problem is, if it would be something like

$$ \sum_{n=1}^{\infty} \frac{ \cos n} {\sqrt n} $$

Then the task can be easily solved using Dirichlet's test: we just need to show that $|\sum_{n=1}^{N} \cos n| \le K$ for all $N$ having fixed $K$. This was described here

Also, if it was like

$$ \sum_{n=1}^{\infty} \frac{ \cos\sqrt n} {n} $$

then we are in much more complicated situation. Every reasonable solution I found on the internet involves approximating the series with an integral. This type of task implies very cute mathematical background(which I don't have:))

In my school we didn't study improper integrals(even just integrals). And actually our problem distincts from two others described above: $$ \sum_{n=1}^{\infty} \frac{ \cos\sqrt n} {\sqrt n} $$

This is why I believe there must be a solution which is simpler than using improper integrals.

We studied lots of different convergence tests at school, but they seems useless here. Well, actually I think we should apply them, but after some mathematical magic. Also I tried to expand terms using Taylor's formula, but it didn't help.

Any ideas?

Yep, I know that. And we also studied it to prove Dirichlet's and Abel's tests. — Levon Minasian, Feb 22 '21 at 18:56
Yes, I do think that it will be useful. It seems to me like upper bounds on $\left|\sum_{n<x}e^{\sqrt{n}i}\right|$ won't be too hard to obtain; I'll add an answer when I figure it out — Milo Moses, Feb 22 '21 at 18:58
Do you think we can show that the $\sum \cos \sqrt{n}$ is bounded to use Dirichlet's test? It seems like it is unbounded: https://math.stackexchange.com/questions/982150/proof-that-sum-limits-i-1n-cos-sqrti-is-unbounded — Levon Minasian, Feb 22 '21 at 19:01
I doubt that it is bounded; I would expect some bound of height at most $n^{\epsilon}$, which gives the series a term size of $\frac{n^{\epsilon}}{n}/\sqrt{n}=\frac{1}{n^{3/2-\epsilon}}$ which goes to $0$ (for $\epsilon<1/2$) — Milo Moses, Feb 22 '21 at 19:03

Adam Latosiński · Accepted Answer · 2021-02-22T19:27:53.660

4

Let $M\in\mathbb{N}$. Let us count how many $n\in\mathbb{N}$ satisfy $2\pi M - \frac{\pi}{4} \le \sqrt{n} \le 2\pi M + \frac{\pi}{4}$.

This condition is satisfied by $n \in [N_1,N_2]$ where $$ N_1 = \lceil (2\pi M - \frac{\pi}{4})^2\rceil = \lceil 4\pi^2 M^2 - \pi^2M - \frac{\pi^2}{16}\rceil $$ $$ N_2 = \lfloor (2\pi M + \frac{\pi}{4})^2\rfloor = \lfloor 4\pi^2 M^2 + \pi^2M - \frac{\pi^2}{16} \rfloor $$ We have $$ N_1 \le 4\pi^2 M^2 - \pi^2M - \frac{\pi^2}{16} + 1 $$ $$ N_2 \ge 4\pi^2 M^2 + \pi^2M - \frac{\pi^2}{16} -1 $$ so $$ N_2-N_1 \ge 2\pi^2M -2 $$ For $n\in[N_1,N_2]$ we have $$ 2\pi M - \frac{\pi}{4} \le \sqrt{n} \le 2\pi M + \frac{\pi}{4}$$ $$ \cos \sqrt{n} \ge \frac{1}{\sqrt{2}}$$ $$ \frac{\cos \sqrt{n}}{\sqrt{n}} \ge \frac{1}{\sqrt{2}(2\pi M + \frac{\pi}{4})}$$ so $$ \sum_{n=N_1}^{N_2} \frac{\cos \sqrt{n}}{\sqrt{n}} \ge \frac{N_2-N_1+1}{\sqrt{2}(2\pi M + \frac{\pi}{4})} \ge \frac{2\pi^2M -1}{\sqrt{2}(2\pi M + \frac{\pi}{4})} $$ For large $M$ this bound tends to $\frac{\pi}{\sqrt{2}} > 0$ and from this we can conclude that the sequence of partial sums of the series isn't a Cauchy sequence, and therefore it's not convergent.

This proof relies on the fact that for large $n$ the square rootsts of natural numbers get close together, and therefore $\cos\sqrt{n}$ changes slowly - there are long stretches of natural numbers for which $\cos\sqrt{n}$ is positive (followed by long stretches for which it is negative, but it is enough to consider the 'positive ones'). During these long stretches the partial sums of the series significantly increase, and a lower bound can be put on this increase, thus proving that the partial sums do not create a Cauchy sequence.

edited Feb 22 '21 at 19:27

answered Feb 22 '21 at 19:08

Adam Latosiński

10,558

So, you just showed that for arbitrariliy large $N_1$ the sum is bounded from bottom? – Levon Minasian Feb 22 '21 at 19:16
Why do you wanna check how many $n$ satisfy the inequality? – Levon Minasian Feb 22 '21 at 19:17
No. I showed that there exists $\epsilon

0$ such that for arbitraily large $N$ there exist $N_1, N_2>N$ such that $a_{N_1}+a_{N_1+1} + \dots a_{N_2} > \epsilon$. That is the negation of the Cauchy condition for the sequence of the partial sums.

– Adam Latosiński Feb 22 '21 at 19:19
I need to check how many $n$ satisfy this inequality to get a lower bound on the number of terms in the sum in the last inequality. – Adam Latosiński Feb 22 '21 at 19:21
Adam, shouldn't there be $ \frac{\cos \sqrt{n}}{\sqrt{n}} \ge \frac{1}{\sqrt{2}(2\pi M + \frac{\pi}{4})}$ ? – Levon Minasian Feb 22 '21 at 19:26
Yes, you're right about that. Doesn't change the conclusion though. – Adam Latosiński Feb 22 '21 at 19:28
Looks like you're correct. Is it okay I'll accept your answer a little bit later? I want to deep dive into it. – Levon Minasian Feb 22 '21 at 19:31

Convergence of $\sum \frac{ \cos\sqrt n} {\sqrt n}$

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