# How to show this series converges $\sum\limits_{n=1}^\infty n^{-1/2}\sin(n)\sin(n^2)$ [closed]

I want to see if this series converges or not: $$\sum_{n=1}^\infty n^{-1/2}\sin(n)\sin(n^2).$$
I tried comparison tests but nothing. I saw that integral criteria works but I don't know how to show that.

Thank you

• Convergence would follow if sums of the form $\sum_{n=1}^M \sin(n)\sin(n^2)$ had an upper bound independent of $M$. This is certainly known for sums $\sum_{n=1}^M \sin(n)$. I'm sure it's also true for these sums, but I don't have an immediate proof. – Todd Trimble Nov 16 '20 at 1:34
• I'm pretty sure using basic trig and quadratic identities allows us to reduce @ToddTrimble 's method to whether $\sum_{n = 1}^M \sin((n + a)^2)$ and $\sum_{n = 1}^M \cos((n + a)^2)$ are bounded. That said, I suspect that while the question may not be immediately obvious, it seems unlikely to be asked with research-level understanding. – user44191 Nov 16 '20 at 4:15
• Sounds like one of those "it's either trivial, or depends on the irrationality measure of pi being 2" questions :D – Dror Speiser Nov 16 '20 at 7:33
• Well, indeed the sum telescopizes nicely, but the joy is immediately killed by the suspicion of homeworkness – Pietro Majer Nov 16 '20 at 7:52
• Thank you everyone for help. I'm training for some exams and I found it. I lost like 3 hours trying to solve it. Now I got it, thank you – ursuv2 Nov 16 '20 at 22:12

As indicated by Todd Trimble in comments, we can use the Dirichlet test; here, since $$\sin(n)\sin(n^2)=\frac12\big( \cos n(n-1) - \cos n(n+1) \big)$$ we have a telescopic sum $$\sum_{n=1}^M \sin(n)\sin(n^2)=\frac12-\frac12 \cos M(M+1)=\sin^2\Big(\frac{M(M+1)}2\Big),$$ that does not exceed $$1$$ in absolute value.