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

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 $$\leq 1$$ ?

• 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. Sep 5, 2022 at 14:06
• Sorry! I've edited. Think now is good Sep 5, 2022 at 14:08
• It is sufficient to prove that $P(\infty) \leq 1$. Have you tried Poisson summation? Sep 5, 2022 at 14:19
• @Ari.stat Why is that sufficient? The series does not appear monotonous to me. Sep 5, 2022 at 14:20
• Oops, I am sorry. I did not see the abs. Sep 5, 2022 at 14:28

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