# 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?

1. How did he know that he should choose $1/n^{2p}$ to compare?
2. How does that final limit tend to $1/6^p$?
• The limit didn't "tend" to $1/6^p,$ it "is" $1/6^p.$ – zhw. Jan 12 '18 at 1:21

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

$$\frac{\left(\frac{1}{6n^2}+O\left(\frac{1}{n^4}\right)\right)^p}{\frac{1}{n^{2p}}}=\left(\frac{n^2}{6n^2}+O\left(\frac{n^2}{n^4}\right)\right)^p=\left(\frac{1}{6}+O\left(\frac{1}{n^2}\right)\right)^p$$

$$1-n\sin(1/n) = 1/(6n^2) + \text { something much smaller}.$$
This suggests strongly that $(1-n\sin(1/n))^p$ should be compared to $(1/(6n^2))^p,$ or just $(1/n^2)^p$ if you like, as the $6$ is just an annoying constant.
• Yes, I have always been annoyed by the number $6$ too. (+1) ... I hope that the number $1$ doesn't annoy you. – Mark Viola Jan 12 '18 at 3:55
• @MarkViola No, I'm fine with $1.$ Especially $+1.$ – zhw. Jan 12 '18 at 21:41
• And yet $1=e^{i\color{red}{6}\pi}=\frac{\color{red}{6}}{\color{red}{6}}$. And so on. – Mark Viola Jan 12 '18 at 21:51
• Maybe I should have written $1=e^{i((+1)+(+1))\pi}=\frac{+1}{+1}$ – Mark Viola Jan 12 '18 at 23:25