Convergence of $\sum2^{-\sqrt{k}} $ So I am going to determine whether this series converges or not:
$$\sum_{k=0}^\infty 2^{-\sqrt{k}} $$

Since this chapter is about the ratio test, I applied that test to this series.
I end up with this limit $$\lim_{k \to \infty} 2^{\sqrt{k}-\sqrt{k+1}}$$

I'm stuck here, don't know how to calculate this limit. I could simplify to: $$\sqrt{k}-\sqrt{k+1} = \frac{1}{\sqrt{k}+\sqrt{k+1}} $$
But I doubt this will help me.

Could anyone help me?
The integral test works. We calculate \begin{align*} \int^\infty_0 2^{-\sqrt x} &amp;= \int^\infty_0 e^{-\sqrt{x}\log 2} dx \\
&amp;= \frac{ 2 }{(\log 2)^2}\int^\infty_0 te^{-t} dt \,\,\,\,\,\,\, [t = \sqrt x \log 2].
\end{align*} The latter converges (easily shown via integration by parts) so the sum converges as well.

EDIT: As suggested in the above comments, a comparison should work as well. For sufficiently large $k$, we see $\sqrt{k} \ge 2\log_2 k$. This follows since asymptotically $\sqrt{k} \gg \log_2 k$. For such $k$, we have $$2^{-\sqrt k} \le 2^{-2\log_2 k} = \frac 1 {k^2}.$$
Hint:

$$\lim\limits_{k \to \infty} 2^{\sqrt{k} -\sqrt{k+1}} = 2^{\lim\limits_{k \to \infty} \sqrt{k} -\sqrt{k+1}}$$

Can you finish from here?