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?

The idea is:
$$
\sum_{n=2^k}^{2^{k+1}-1} a^{\sqrt{n}} \leq \sum_{n=2^k}^{2^{k+1}-1} a^{\sqrt{2^k}} = 2^ka^{\sqrt{2^k}}
$$
This is known as Cauchy Condensation Test.

In this situation, the bound given by the Integral Test is somewhat similar:
$$
\sum_{n=1}^\infty a^{\sqrt{n}} \leq \int_0^\infty a^{\sqrt{x}}\,dx = 2\int_0^\infty a^t t\, dt < \infty.
$$
