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Prove that $\displaystyle P(n):|\sum_{k=1}^{n} \sin(k) \sin(k^2)| \leq 1, \forall n\geq 1$ is true

What I've tried: For $n=1 \implies \left|\sin(1)\sin(1^2)\right| \leq1$ is true.

Suppose that $P(n)$ is true.

Then, we need to show that $P(n+1): |\displaystyle\sum_{k=1}^{n+1} \sin(k) \sin(k^2)| \leq 1$ is true $\forall n\geq 1$.
$$\left|\displaystyle\sum_{k=1}^{n+1} \sin(k) \sin(k^2)\right| \\= \left|\sum_{k=1}^{n} \sin(k) \sin(k^2) + \sin(n+1)\sin((n+1)^2) \right| \\ \leq \left|\sum_{k=1}^{n} \sin(k) \sin(k^2)\right| + \left|\sin(n+1)\sin((n+1)^2 \right| \\ \leq 1 + \left|\sin(n+1)\right|\left| \sin((n+1)^2)\right| \\\leq 1 + 1*1 = 1+1 = 2$$
But how can I obtain that it's $\leq 1$ ?

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  • $\begingroup$ Formatting tip: Do not write sin(k) in equations because it looks ugly: $sin(k)$ looks like multiplication between $s,i,n$ and $k$. Instead, use \sin(k) to have it appear normally as $\sin(k)$. Also, replace <= with \leq. $\endgroup$ Sep 5, 2022 at 14:06
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    $\begingroup$ Sorry! I've edited. Think now is good $\endgroup$ Sep 5, 2022 at 14:08
  • $\begingroup$ It is sufficient to prove that $P(\infty) \leq 1$. Have you tried Poisson summation? $\endgroup$
    – Ari.stat
    Sep 5, 2022 at 14:19
  • $\begingroup$ @Ari.stat Why is that sufficient? The series does not appear monotonous to me. $\endgroup$ Sep 5, 2022 at 14:20
  • $\begingroup$ Oops, I am sorry. I did not see the abs. $\endgroup$
    – Ari.stat
    Sep 5, 2022 at 14:28

1 Answer 1

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$$S=\sum_{k=1}^n\sin k\sin (k^2)=\frac12\left(\sum_{k=1}^n\cos\left({k^2-k}\right)-\cos\left(k^2+k\right)\right)$$$$= \frac12\Bigg((\cos0-\cos 2)+(\cos 2-\cos 6)+(\cos 6-\cos 12)+…+(\cos(n^2-n)-\cos(n^2+n))\Bigg)$$$$=\frac12(\cos 0-\cos(n^2+n))$$$$=\frac12(1-\cos(n^2+n))$$

Now, $1\geq\cos(n^2+n)\geq-1$ so $0\leq 1-\cos(n^2+n)\leq 2$ so $0\leq S\leq 1$ and hence $|S|=S\leq 1$.

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