# Comparison test to determine convergence

I have tried using comparison test to determine whether a series converges or diverges. However, am not sure on whether i'm right.

Question: $$\sum_{n=1}^{\infty} \frac{tan(\frac{1}{n})}{\sqrt{n}}$$

My solution:

$$\sum_{n=1}^{\infty} \frac{tan(\frac{1}{n})}{\sqrt{n}} < \frac{1}{\sqrt{n}}$$

So it converges as for the series $\frac{1}{n^{p}}$, p>0 means it converges. And the above converges as $\frac{1}{\sqrt{n}}$ is the same as $\frac{1}{n^{(\frac{1}{2})}}$ and p = $\frac{1}{2}$, thus as p is more than 0, it converges.

Am i right?

Any and all suggestions and help is appreciated! Thank you for your time, guys :)

Edit: Sorry guys, made a mistake with the p-series. Re-read my notes again.Thank you for pointing that out :)

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A $p$-series $\sum \frac 1 n^p$ converges if and only if $p > 1$< not $p > 0$. –  user61527 Mar 24 '14 at 21:05

## 1 Answer

There is a lot of difficulty here.

First: the series $$\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}$$ actually diverges -- $p$-series only converge for $p>1$, not $p>0$.

With this in mind, this particular comparison is no good. However, do you know about limit comparison? If you do, then it might help to remember that $$\lim_{x\to0}\frac{\sin x}{x}=1,$$ so that $$\lim_{x\to0}\frac{\tan x}{x}=\lim_{x\to0}\frac{\sin x}{x}\cdot\frac{1}{\cos(x)}=1$$ as well.

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Sorry, just re-read my notes again after seeing the comment, it states that p<= 0 is divergent, so i assumed if p>0 is converging and i realised, i interpreted it incorrectly. Thank you! –  Phantom Mar 24 '14 at 21:12