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?
 A: The integral test works. We calculate \begin{align*} \int^\infty_0 2^{-\sqrt x} &= \int^\infty_0 e^{-\sqrt{x}\log 2} dx \\
&= \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 $$2^{-\sqrt k} \le 2^{-2\log_2 k} = \frac 1 {k^2}.$$ 
A: It is not difficult to prove, if $k
 $ is sufficiently large, $k\geq K
 $ say, that $$\sqrt{2^{k}}\geq2k
 $$so if we use the Cauchy condensation test we note that $$\sum_{k\geq K}\frac{2^{k}}{2^{\sqrt{2^{k}}}}\leq\sum_{k\geq K}\left(\frac{1}{2}\right)^{k}$$ and so we can conclude that the series converges.
A: Let's compare with a series you probably know converges $\sum 1/n^2$. 
$$
L=\lim_{n\rightarrow \infty}\frac{2^{-\sqrt{n}}}{1/n^2}=
\lim_{n\rightarrow \infty}\frac{n^2}{2^{\sqrt{n}}}
$$
Since this is hard to look at, take $t=\sqrt{n}$, then 
$$
\lim_{t\rightarrow \infty}\frac{t^4}{2^{t}}=
0
$$
By l'hospital's a few times, or just knowing that exponential growth beats polynomial growth for any degree polynomial. So your series converges by comparison to $\sum 1/n^2$
A: Since @User8128 gave the solution using the integral test (probably the simplest for this problem), let me give another way.
For the case where the ratio test is inconclusive as here $$u_k=2^{-\sqrt{k}}\implies \frac {u_{k+1}}{u_k}=2^{\sqrt{k}-\sqrt{k+1}}=1-\frac{\log (2)}{2} \frac 1 {\sqrt{k}} +O\left(\frac{1}{k}\right)$$ Raabe's test is very convenient.
Considering  $$R=k \left(\frac {u_k}{u_{k+1}} -1\right)=k\left(2^{\sqrt{k+1}-\sqrt{k}}-1\right)$$ and expanding again for large values of $k$, you will find $$R=\frac{\log ^2(2)}{8}+\frac{\log (2)}{2} \sqrt{k} +O\left(\frac{1}{k^{1/2}}\right)$$ So $R>1$ then the series will be absolutely convergent. 
A: 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?
