proving convergence of $\sum a^\sqrt{n} $ for $ 0 < a<1 $ my textbook says it can be proven in three steps :
1) for very large $ n $, $ a^{2^{n/2}} < 3^{-n} $
2) $ \sum 2^{n}a^{2^{n/2}} < \infty $
3) $ \sum_{n=1} a^{\sqrt{n}} <  \sum_{n=0} 2^{n}a^{2^{n/2}} < \infty  $
step 1) is trivial since $a^{2^{n/2}}$ decreases exponentially.
step 2) is proven from step 1. the convergence of  $ \sum (\frac{2}{3})^n $ is a proof.
I'm stuck at step 3). I think i'm missing something critical here - from my naive understanding, the exponential decrease of $ a^{2^{n/2}} $ should outweigh everything. how can the convergence of  $ \sum a^{\sqrt{n}} $ for $ (0<a<1) $ can be proven?

We have $\int a^{\sqrt x}\,dx=\dfrac2{(\log a)^2}\,a^{\sqrt x}(\sqrt x\,\log a-1)\xrightarrow{x\to\infty}0$ since $\log a<0$ (verify!). Therefore the series $\sum a^{\sqrt n}$ converges by integral test. 
