Does the series $\sum \limits_{n=1}^{\infty} \sin(n^2)$ converge? I think I've tried everything, I have no more ideas.
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If the series $\sum\limits_n\sin(n^2)$ was convergent, then $\sin(n^2)$ would converge to zero hence, for every $n$ large enough, $n^2$ would be some multiple of $\pi$ plus or minus some small error. In particular, for every $n$ large enough, $$(n+1)^2-2n^2+(n-1)^2=2,$$ would be nearly a multiple of $\pi$, which is impossible. Hence the series $\sum\limits_n\sin(n^2)$ diverges.
(More rigorously perhaps, if $|n^2-\pi k_n|\lt\frac14$ for some integers $k_n$ and for three consecutive integers $n=N-1$, $N$ and $N+1$, then the identity $(N+1)^2-2N^2+(N-1)^2=2$ and the triangular inequality imply that $$\left|2-\pi(k_{N+1}-2k_N+k_{N-1})\right|\lt1\cdot\tfrac14+2\cdot\tfrac14+1\cdot\tfrac14=1,$$ which is impossible since $|2-\pi\ell|\gt1$ for every integer $\ell$.)
The same ideas of discrete differentiation and that the distance between each nonzero rational number and $\pi\mathbb Z$ is positive can be adapted to show the following:
If $P$ is a polynomial with at least one coefficient not a rational multiple of $\pi$, then the sequence $(\sin(P(n)))_n$ does not converge to zero hence the series $\sum\limits_n\sin(P(n))$ diverges.
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i did not understand why was there a need for it to converge to 0 and not anything else – avz2611 Oct 04 '14 at 20:10
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your last statement is a consequence of the equidistribution of $P(n)$ mod $2\pi$ or you have another way to prove it ? – Gabriel Romon Aug 26 '16 at 07:25
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Does the sum term tend to 0?
It changes sign all the way, so I do not know. Anyway, it can tend to 0 and still diverge
– Mark Oct 04 '14 at 19:34