# Determine convergence / divergence of $\sum \sin \frac{\pi}{n^2}$

Determine convergence / divergence of $$\sum \sin \frac{\pi}{n^2}$$

let $a_n= \sin \frac{\pi}{n^2}$

I attempted the integral test but on the interval $[1, \sqrt{2})$ it is increasing and decreasing on $(\sqrt{2}, \infty)$. so the integral test in only applicable for the decreasing part. + the integral computation seems to lead to 3 pages of steps...

I believe the comparison test would be the most reasonable test, graphically I observed that $a_n$ behaves like $b_n=1/n$ when $n$ is large.

$\sum b_n = \infty$; but, since $b_n > a_n \implies$ inconclusive for divergence/convergence.

I attempted the limit comparison, and ratio test, but inconclusive. I am uncertain if I am doing them properly.

How could I bound below $a_n$ to proceed with the comparison test? Is there a more appropriate method? How would you proceed?

• Are you summing over $\lim_{x\to \infty} 1\leq n \leq x$? Aug 7 '17 at 23:58
• @Romain Presumably yes, since the OP mentions the integral test along $[1,\infty)$. Aug 8 '17 at 0:15
• @romain If $0 \leq x < 1$ is added, $a_n$ oscillates $-1$ to $1$ on that interval. Can you use the alternating test with $a_n =(-1)^n b_n$? But what would $b_n$ be?
– rei
Aug 8 '17 at 2:16

Converges because $\sin\left(\frac{\pi}{n^2}\right)\leq\frac{\pi}{n^2}$ and $\sum\limits_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$

HINT: For all $x\gt 0$, $$\sin x\lt x$$ and you know that the sum $$\sum \frac{1}{n^2}$$ converges. Can you use this to prove convergence of your sum?

• thx for the input. How would you justify that $\sin x< x$?
– rei
Aug 8 '17 at 2:33
• @rei Write $f(x)=x$ and $g(x)=\sin x$. Note that $f(0)=g(0)=0$ and $|g'(x)|=|\cos x| \leq 1=f'(x)$ so that $\sin x \leq x$ for $x \geq 0$. Aug 8 '17 at 5:45

Hint: $\lim_{x\rightarrow 0}{{\sin(x)}\over x}=1$ this implies that $\sin({\pi\over n^2})\simeq {\pi\over n^2}$ use the comparison test.

It is not true that $a_n$ behaves like $1/n$ for large $n$, because $$|\sin(x)| \le |x|$$ for all $x$ (and in fact, $|\sin x| \approx |x|$ for values of $x$ near zero). In particular, this means that

$$a_n \le \pi/n^2.$$

If you prefer, the limit comparison test can be applied with $b_n = \pi/n^2$ and using the sharpness of the small angle approximation mentioned above.

• :D Ya beat me to it, and (+1) I think it is a very useful inequality you have there. Aug 8 '17 at 0:00