Convergence of $\sum_{n = 0}^{\infty} \frac{n}{e^{\sqrt n}}$ 
Is the following series convergent or divergent?
  $$\sum_{n = 0}^{\infty} \frac{n}{e^{\sqrt n}}$$

I find that both the ratio and root tests fail for this example. And I do not know what series I can use for the comparison test.
I would like to avoid the integral test. Could anyone help?
 A: Assuming that you mean $\sqrt n$ instead of $\sqrt k$, you can use


*

*$e^x = \sum_{k=0}^{\infty}\frac{x^k}{k!}$.


Hence,
$$e^{\sqrt{n}} \geq \frac{\left(\sqrt{n}\right)^6}{6!} = \frac{n^3}{6!}$$
It follows
$$\sum_{n = 1}^{\infty} \frac{n}{e^{\sqrt n}} \leq \sum_{n = 1}^{\infty} \frac{n}{\frac{n^3}{6!}} = 6!\sum_{n = 1}^{\infty} \frac{1}{n^2}< \infty$$
A: We can write the sum as
$$\sum_{m=1}^{\infty}\,\sum_{m^2\le n<(m+1)^2}\frac{n}{e^{\sqrt n}}$$ $$< \sum_{m=1}^{\infty}((m+1)^2-m^2)\frac{(m+1)^2}{e^{m}} < \sum_{m=1}^{\infty}\frac{(m+1)^4}{e^m},$$
and it's standard that the last series converges.
A: For any $a>1$, there is an integer $N>0$ so that $n>N\Rightarrow a^n>n^3.$ 
There are many ways to see this. You can note that, by L'Hospital, $\underset{x\to \infty}\lim\frac{a^x}{x^3}=\underset{x\to \infty}\lim\frac{(\ln a)^3a^x}{6}=\infty$ 
or that $a^n-n^3>\sum_{k=0}^{\infty}\frac{(n\ln a)^k}{k!}-n^3>-\frac{5n^3}{6}+\frac{(\ln a)^4n^4}{12}\to \infty$.
Now compare your series to $\sum \frac{1}{n^2}.$
