# convergence $\sum_{n=1}^{\infty}\frac{1}{n}\sin\left(\frac{\pi}{n^{2}}\right)$

I'm checking convergence of the series $$\sum_{n=1}^{\infty}\frac{1}{n}\sin\left(\frac{\pi}{n^{2}}\right)$$ using the integral test.

I calculated the integral $$\int_{1}^{+\infty}\frac{1}{x}\sin\left(\frac{\pi}{x^{2}}\right)dx$$ using substitution $$u=\frac{\pi}{x^{2}}$$,

I got: $$\frac{1}{2}\int_{0}^{\pi}\frac{\sin u}{u}du$$

but I don't know what to do next

thanks for any help, and sorry if I have English mistakes.

• Maybe you will find the inequality $|\sin(x)| \leq |x|$ useful. Nov 5, 2021 at 10:51
• Exactly what is the statement of the integral test that's supposed to apply here? (The fact that $\frac1x\sin\left(\frac\pi{x^2}\right)$ is not monotone bothers me...) Nov 5, 2021 at 10:58
• What David C. Ullrich said. The integral test can only be applied when the underlying function is positive and decreasing, which is not the case with your function. This time you cannot use a hammer, because the job is to drive a screw into the wall instead of a nail. Nov 5, 2021 at 11:25
• @DavidC.Ullrich apart from the first term, the sequence is monotone decreasing and positive Nov 5, 2021 at 11:35
• @DanielG I don't think you should attempt to use the integral test. Just use the fact that if a series is absolutely convergent, then it is convergent. Nov 5, 2021 at 12:03

It can be proven that $$|\sin\alpha|\le|\alpha|$$ for any real $$\alpha$$, so we have

$$\sum_{n\ge1}\frac1n\left|\sin\left(\pi\over n^2\right)\right|\le\sum_{n\ge2}{\pi\over n^3}=\pi\zeta(3)-\pi$$

Thus the series converges absolutely.

In fact, you can plug the aforementioned inequality into the integral to get a similar result.

• I thought it could not be so simple. Nov 7, 2021 at 15:27

Using the following fact:

$$-1 \leq \sin(x) \leq 1 \Rightarrow |\sin(x)| \leq 1$$

and it becomes evident that:

$$\sum_{n=1}^{\infty}\frac{1}{n}\sin\left(\frac{\pi}{n^{2}}\right) \leq \sum_{n=1}^{\infty} \frac{1}{n}\left |\sin \left(\frac{\pi}{n^{2}}\right)\right| \le \sum_{n = 1}^{\infty} \frac{1}{n}$$

But the p series, since $$p \leq 1$$ the test isn't conclusive, thus we need something else. It's possible to prove that:

$$|\sin(\pi/n) - \sin(\pi/(n+1))| \leq C /n^2$$ for some constant $$C > 0$$, then, it's clear that, it happens the following:

$$\Big|\sum_{n=k}^l\sin(\pi/2n)-\sin(\pi/(2n+1))\Big|\leq \frac{C}{4}\sum_{n=k}^ln^{-2}\xrightarrow{k,l\rightarrow \infty}0,$$

so the sequence is Cauchy. Well, since

$$\sum_{n=1}^{\infty}\frac{1}{n}\sin\left(\frac{\pi}{n^{2}}\right) \leq \sum_{n=1}^{\infty} \frac{1}{n}\left |\sin \left(\frac{\pi}{n^{2}}\right)\right| \leq \sum_{n=1}^{\infty} \frac{1}{n}\left |\sin \left(\frac{\pi}{n^{}}\right)\right|$$

And since, for som $$k, l \to \infty$$, we have that the difference between the terms will be lesser than $$\frac{C}{n^2} * \frac{1}{n(n-1)}$$, but by the argument that $$\sum_{n = 2}^{\infty} \frac{C}{n^3(n-1)} \leq \sum_{n = 2}^{\infty} \frac{C}{n^2}$$

converges (because of the $$p$$ series convergence the $$1/n^2$$ series), then we can say that the original series will converge.

The integral test says that if the integral converges, then so does the series. Using this fact, we can say that the series $$\sum_{n=1}^{\infty}\frac{1}{n}\sin\left(\frac{\pi}{n^2}\right)$$ converges if the following integral also converges: $$\int_{1}^{\infty}\frac{1}{x}\sin\left(\frac{\pi}{x^2}\right)\mathrm{d}x$$ If we let $$t=\frac{\pi}{x^2}$$ we get the following integral: $$\frac{1}{2}\int_{0}^{\pi}\frac{\sin(u)}{u}\mathrm{d}u$$ Considering the function $$\frac{\sin(u)}{u}$$ in $$(0,\pi)$$, it is easy to notice that it is monotone decreasing, continuous and finite both as $$u\to 0$$ and at $$u=\pi$$. Therefore, the above integral also has a finite value.