TITLE: Convergence of the series $\sum_{1}^\infty \left(1-n\sin(\frac{1}{n})\right)$
QUESTION [7 upvotes]: I have to study the convergence of the series $$\sum_{n=1}^\infty \left(1-n\sin\left(\frac{1}{n}\right)\right).$$ I know I should study the limit $$\lim_{n \to \infty}n^\alpha\left( 1-n\sin\left(\frac{1}{n}\right)\right).$$ 
But I have some troubles with this limit (and so the convergence of the series). Any ideas?

REPLY [5 votes]: You may observe that, using the Taylor expansion of $\sin x$ near $0$, you get, as $n \to \infty$:
$$
\sin \left(\frac1n\right)=\frac1n+\mathcal{O}\left(\frac1{n^3}\right)
$$ giving
$$
1-n\sin \left(\frac1n\right)=\mathcal{O}\left(\frac1{n^2}\right)
$$ and the initial series $\displaystyle \sum_{n\geq1}\left(1-n\sin \left(\frac1n\right)\right)$ is convergent as is $\displaystyle \sum_{n\geq1}\frac1{n^2}.$