Test for convergence

**boquist** · August 15th 2010, 08:46 PM

Did I go about this correctly?
Since b_n is p series and p=3/2 which is greater than 1, it is convergent and so is a_n due to the comparison test.

It makes sense to me but I need confirmation since Im not too comfortable with this material.

Thank you!

**Prove It** · August 15th 2010, 09:15 PM

I don't agree with your upper bound...

$-\frac{\pi}{2} < \arctan{n} < \frac{\pi}{2}$ for all $n$ , so surely

$-\frac{\frac{\pi}{2}}{n\sqrt{n}} < \frac{\arctan{n}}{n\sqrt{n}} < \frac{\frac{\pi}{2}}{n\sqrt{n}}$ (we can assume $n\sqrt{n} > 0$ since we are summing from $n = 1$ to $\infty$ ...

So you will need to choose $\frac{\frac{\pi}{2}}{n\sqrt{n}}$ as your upper bound.

The rest of your logic looks fine.

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