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

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 $\frac{2}{\pi }\frac{1}{n}\le \mathrm{sin}\left(\frac{1}{\sqrt{n}}\right)$.

Paul Mitchell

A variant is to use $\underset{u\to 0}{lim}\frac{\mathrm{sin}u}{u}=1$ and in particular this term is small enough.
Now write $\frac{1}{\sqrt{n}}\mathrm{sin}\left(\frac{1}{\sqrt{n}}\right)=\frac{1}{n}×\frac{\mathrm{sin}\left(\frac{1}{\sqrt{n}}\right)}{\frac{1}{\sqrt{n}}}>\frac{1}{n}×\frac{12}{}$ for n large enough.
And still conclude by comparison with $\sum \frac{1}{n}$
Note: a faster way is to use equivalents, the limit 1 means $\frac{1}{\sqrt{n}}\mathrm{sin}\left(\frac{1}{\sqrt{n}}\right)\sim \frac{1}{n}$

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