Test for convergence $\sum_{n=1}^{\infty} \frac{1}{2^\sqrt{n}}$ 
Possible Duplicate:
convergence of a series involving $x^\sqrt{n}$ 

Test for convergence
$$\sum_{n=1}^{\infty} \frac{1}{2^\sqrt{n}}$$
My first thought was to use the ratio test but it's inconclusive since it yields $1$.
Are there some easy means to test the sum for convergence? Thanks!
 A: A quick answer can be given by comparison with the series $\sum_{n\geq 1}1/n^2$.
Check that $\lim_{n\rightarrow +\infty} n^2 /2^\sqrt{n}=0$.
Deduce that there is a positive constant $K$ such that $\frac{1}{2^\sqrt{n}}\leq\frac{K}{n^2}$ for all $n\geq 1$.
Conclude by comparison.
A: $$2^{-\sqrt n}\leqslant\int_{n-1}^n\exp(-\sqrt x\log 2).$$
The integral $\int_1^{+\infty}\exp(-\sqrt x\log 2)dx$ is convergent, using the substitution $s=\sqrt x$. We conclude by integral test, after having checked we can use it. 
A: Hint: For $n>2^{10}$, we have that $$\sqrt{n}\geq 2\log_2(n),$$  and so for $n>2^{10}$ $$2^{\sqrt{n}}\geq n^2.$$  Now try using the comparison test.
A: This can be handled with a higher order ratio test, Raabe's test. 
Let $a_n = 1/2^{\sqrt{n}}$. 
Then 
$$\begin{eqnarray*}
n\left(\left|\frac{a_{n+1}}{a_n}\right| - 1\right) 
&=& n\left(2^{\sqrt{n}-\sqrt{n+1}} - 1\right) \\
&<& n\left(2^{-1/(2\sqrt{n+1})}-1\right) \\
&<& -\frac{n}{4\sqrt{n+1}}\log 2.
\end{eqnarray*}$$
In the second to last line we use the square root inequality 
$\sqrt{n}-\sqrt{n+1} < -1/(2\sqrt{n+1})$ for $n\ge 0$. 
In the last line we use the exponential inequality 
$e^{-x} < 1-x/2$ for $0<x<2+W(-2/e^2) = 1.59362\cdots$. 
Now we need only show that 
$$\lim_{n\to\infty} 
-\frac{n}{4\sqrt{n+1}}\log 2 < -1.$$
A: So, you could use the comparison test, that one would yield results.
