Thread: Test for convergence

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!

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

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

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


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

The rest of your logic looks fine.