TITLE: Is the series $\sum_{n=1}^\infty \frac{n^2}{2^{\sqrt n}}$ convergent?
QUESTION [6 upvotes]: Is this series convergent?
$$\sum_{n=1}^\infty \frac{n^2}{2^{\sqrt n}}$$
Edit: I'm sorry I wrote a double negative. This is the series.
(thanks Brian M. Scott for editing)

REPLY [5 votes]: Note that $2^{\sqrt{n}}=e^{(\log 2)\sqrt{n}}$. 
By looking at the power series expansion of $e^x$, we can see that for positive $x$, we have $e^x\gt \frac{x^8}{8!}$. Thus 
$$e^{\log 2\sqrt{n}}\gt \frac{(\log 2)^8}{8!}n^4.$$
So the $n$-th term of our sequence is $\lt \frac{8!}{(\log 2)^8}\frac{1}{n^2}$.
Comparison with the convergent series $\sum\frac{1}{n^2}$ shows that our series converges.