Apparently it can be proven with a comparison but I've tried to compare it to $\frac{1}{n^{p}}$ with not results.
I've also tried comparing $\sqrt{n}$ with $\ln{n}$ but $\frac{1}{2^{\ln n}}$ diverges so that doesn't give me anything useful.
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Related : http://math.stackexchange.com/questions/386878/does-the-series-sum-n-1-inftyx-sqrt-n-converge-pointwise-if-it-then – lab bhattacharjee Jun 03 '13 at 10:28
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For the comparison, it is enough to show that after a while $2^{\sqrt{n}}\gt n^2$.
Equivalently, we want to prove that after a while $(\ln 2)\sqrt{n}\gt 2\ln n$. There are various ways to do this. For example by using L'Hospital's Rule we can show that $$\lim_{x\to \infty} \frac{2\ln x}{(\ln 2)\sqrt{x}}=0.$$
André Nicolas
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I swear I tried the exact same thing and for some reason the $>$ was a $<$ instead. Thanks for the solution. – Luka Horvat Jun 03 '13 at 10:53
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$$\sum_{n=1}^{\infty}\frac{1}{2^{\sqrt{n}}}= \sum_{k=1}^{\infty}\sum_{n=k^2+1}^{(k+1)^2}\frac{1}{2^{\sqrt{n}}}<\sum_{k=1}^{\infty}\sum_{n=k^2+1}^{(k+1)^2}\frac{1}{2^k} <\sum_{k=1}^{\infty}\frac{2k+1}{2^k}<\infty$$
Mher
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