<s>[INST] For what values of $p$ does the series $\sum_{n=1}^{\infty}\left(1-n\sin{\frac{1}{n}}\right)^p$ converge? For what values of $p$ is the series $$\sum_{n=1}^{\infty}\left(1-n\sin{\frac{1}{n}}\right)^p \quad \text{converge}?$$

This is my professors solution: Note that 
$$1-n\sin{\frac{1}{n}}=\frac{1}{6n^2}+O(1/n^4),$$
so the series is a positive serie for every $p$. Furthermore:
$$\frac{\left(1-n\sin{\frac{1}{n}}\right)^p}{\frac{1}{n^{2p}}}=\frac{\left(\frac{1}{6n^2}+O(1/n^4)\right)^p}{\frac{1}{n^{2p}}}\rightarrow\frac{1}{6^p}, \ \text{as} \ n\rightarrow\infty.$$
Since $\sum_{k=1}^{\infty}1/n^a$ converges iff $a>1$ it follows that our series converges iff $p>1/2$ according to the limit comparison test.
Questions:
Can someone elaborate on the steps here and break down things a bit?


*

*How did he know that he should choose $1/n^{2p}$ to compare?

*How does that final limit tend to $1/6^p$?

 [/INST] For point 1
Note that for $n$ large $$a_n\sim \frac{1}{n^{2p}}$$
thus we choose this “tail” for the ratio test.
For point 2
$$\frac{\left(\frac{1}{6n^2}+O\left(\frac{1}{n^4}\right)\right)^p}{\frac{1}{n^{2p}}}=\left(\frac{n^2}{6n^2}+O\left(\frac{n^2}{n^4}\right)\right)^p=\left(\frac{1}{6}+O\left(\frac{1}{n^2}\right)\right)^p$$
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