Ernest Ryland

2022-01-17

Prove that the series $\sum _{i=1}^{\mathrm{\infty}}\frac{1}{\sqrt{n}}\mathrm{sin}\left(\frac{1}{\sqrt{n}}\right)$ is divergent

Navreaiw

Beginner2022-01-18Added 34 answers

For $0\le x\le \frac{\pi}{2}$ , we have

$\frac{2}{\pi}x\le \mathrm{sin}\left(x\right)\le x$

Therefore use$\frac{2}{\pi}x\le \mathrm{sin}\left(x\right),x=\frac{1}{\sqrt{n}}$ , and get after multiplication $\frac{2}{\pi}\frac{1}{n}\le \mathrm{sin}\left(\frac{1}{\sqrt{n}}\right)$ .

Therefore use

Paul Mitchell

Beginner2022-01-19Added 40 answers

A variant is to use

Now write

And still conclude by comparison with

Note: a faster way is to use equivalents, the limit 1 means