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Looking for exercises of divergent/convergent series, I stumbled upon this problem, that seems fairly solvable, but I'm a bit stuck... We want to prove that

$$ \sum_{n=1}^{\infty} \frac {|\alpha+\sin(n^2)|}n $$

diverges, but I don't want to use summation by parts and then $\sum_{n \leq N } \sin(n^2) \leq \sqrt{N}$, because it's too advanced for the students now. Is there a simple way to prove that $n^2$ avoids frequently some (narrow) zones of $[0, 2 \pi]$ (we did already Dirichlet approx. theorem, if it can help).

Jack D'Aurizio
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StheW
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1 Answers1

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The statement is trivial if $|\alpha|>1$, so let us assume $|\alpha|\leq 1$.

Assume by contradiction that the given series is convergent. Kronecker's lemma hence implies that the mean value of $\left|\alpha+\sin(n^2)\right|\geq 0$ is zero, hence $\sin(n^2)$ is concentrated around $-\alpha$ and $e^{in^2}$ is concentrated around $\pm\sqrt{1-\alpha^2}+i\alpha$. Let us set $b_n = e^{in^2}$: it is trivial that $b_n\overline{b_{n+1}}$ is equidistributed on the unit circle, but the previous point contradicts that.

Of course this is just a simplified version of Van Der Corput's trick, and summation by parts is hidden into Kronecker's lemma, but we do not need explicit bounds on $\sum_{n=1}^{N}\sin(n^2)$.

Jack D'Aurizio
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  • what do you mean by $sin(n^2)$ concentrated around $-\alpha$? I think one has to be more precise... My strategy wanted to be something like: there is M such that for every $n$, among $sin(n^2), sin((n+1)^2) ,\ldots, sin((n+M)^2)$ there is at least one that is far from $\alpha$, using the differentiation trick directly... – StheW Jan 26 '18 at 14:44
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    By concentrated I mean this: for any $\varepsilon>0$, the density of the set of integers $n$ such that $|\sin(n^2)+\alpha|\geq \varepsilon$ converges to zero as $\varepsilon\to 0^+$. I will keep thinking if we can simplify this argument further. – Jack D'Aurizio Jan 26 '18 at 14:50