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?
 A: 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:
$$
\sum_{n=1}^\infty a^{\sqrt{n}} \leq \int_0^\infty a^{\sqrt{x}}\,dx = 2\int_0^\infty a^t t\, dt < \infty.
$$
A: 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. 
A: iAnother way to look at it is this:
Consider the infinite sum:
$\sum^{n}_{i=0}a^{p}$
If p > 1 this converges for all a such that abs(a) < 1 (basic principle behind geometric series)
We note that given your particular equation sqrt(i) > 1 from i = 2 onwards. Which in other words means that:
$\sum^{n}_{i=2}a^{i^{1/2}}$
Converges for abs(a) < 1.
Now if we just evaluate the expression for i = 0 and 1 (which amounts to the values of 1 and a. 
Then we can write:
$\sum^{n}_{i=0}a^{i^{1/2}} = 1 + a + \sum^{n}_{i=2}a^{i^{1/2}}$
Of which the all 3 terms are known to converge if abs(a) < 1 thereby implying that:
$\sum^{n}_{i=0}a^{i^{1/2}}$ converges for abs(a) < 1
