Prove that the series ∑i=1∞1nsin⁡(1n) is divergent

For ${\displaystyle 0\le x\le \frac{\pi }{2}}$, we have
${\displaystyle \frac{2}{\pi }x\le \sin \left(x\right)\le x}$
Therefore use ${\displaystyle \frac{2}{\pi }x\le \sin \left(x\right),x=\frac{1}{\sqrt{n}}}$, and get after multiplication ${\displaystyle \frac{2}{\pi }\frac{1}{n}\le \sin \left(\frac{1}{\sqrt{n}}\right)}$.

A variant is to use ${\displaystyle \underset{u\to 0}{lim}\frac{\sin u}{u}=1}$ and in particular this term is $>\frac{1}{2}\text{ for }|u|\ll 1$ small enough.
Now write ${\displaystyle \frac{1}{\sqrt{n}}\sin \left(\frac{1}{\sqrt{n}}\right)=\frac{1}{n}\times \frac{\sin \left(\frac{1}{\sqrt{n}}\right)}{\frac{1}{\sqrt{n}}}>\frac{1}{n}\times \frac{12}{}}$ for n large enough.
And still conclude by comparison with ${\displaystyle \sum \frac{1}{n}}$
Note: a faster way is to use equivalents, the limit 1 means ${\displaystyle \frac{1}{\sqrt{n}}\sin \left(\frac{1}{\sqrt{n}}\right)\sim \frac{1}{n}}$