What is convergence interval of $ \sum_{n=1}^\infty a^{\sqrt{n}} \cdot x^n$ where $a \in (0,1)?$ 
What is convergence interval of $ \sum_{n=1}^\infty a^{\sqrt{n}} \cdot x^n$ where $a \in (0,1)?$

So far, I´ve got radius
$$ R^{-1} = \lim_{n\to\infty} \left|a^{\sqrt{n+1}-\sqrt{n}}\right| = 1.$$
Thus interval of convergence is $(-1,1)$. But now, I need to investigate convergence of
 $$ \sum_{n=1}^\infty a^{\sqrt{n}} $$ and $$ \sum_{n=1}^\infty  a^{\sqrt{n}} \cdot (-1)^n. $$ I can't see, what criterion I should use.
 A: Two hints:


*

*If a series converges, then its terms converge to zero.

*If a positive decreasing sequence converges to zero, then the sum of its terms taken with alternating signs converges.
Also, keywords: comparison test.
edit
Ok, seems, the hint about comparison test was a bit inevident. We will assume that $a\in (0,1).$
Let us show that there exists a positive constant $C>0$ such that $\forall n\ge  1$ we have $$a^{\sqrt{n}}\le \frac{C}{n^2}.$$ It is sufficient to show that $$\forall n\ge 1 \quad \ln C -2\ln n - \sqrt{n}\ln a\ge 0.$$
We know that $\lim _{n\to \infty}(-\sqrt{n}\ln a-2\ln n)=+\infty$ (because $a\in(0,1)$), therefore we can take $$\ln C=\inf_{n\in\Bbb N}(-\sqrt{n}\ln a-2\ln n)>-\infty.$$
Armed with this inequality, we can now say that $$\sum_{n\ge 1}a^{\sqrt{n}}\le \sum_{n\ge 1}\frac{C}{n^2}\le \frac{\pi^2C}{6}.$$
edit 2
To reiterate upon the comparison and geometric series.
We know that $$\sum_{k^2\le n<(k+1)^2}a^{\sqrt{n}}\le ((k+1)^2-k^2)a^{k},$$
therefore
$$\sum_{n\ge 1}a^{\sqrt{n}}\le \sum_{k\ge 1}  ((k+1)^2-k^2)a^{k} = \sum_{k\ge 1}  ( 2k+1)a^{k}, $$and the latter geometric series converges.
A: The integral test works. Let $b=\ln a$. Then $b<0$ and $a^{\sqrt{x}}=e^{b\sqrt{x}}$ is continuous, positive and decreasing. So, we compute
$$I=\int_1^{\infty}e^{b\sqrt{x}}dx$$
Let $u=\sqrt{x}$, so $du=dx/2\sqrt{x}=dx/2u$. When $x=1$, $u=1$ and as $x\to\infty$, $u\to\infty$. Therefore,
\begin{align*}
I&=2\int_1^{\infty}ue^{bu}du=\lim_{N\to\infty}{2}(u/b-1/b^2)e^{bu}|_1^{N}
=\lim_{N\to\infty}2(N/b-1/b^2)e^{bN}-2(b-1)e^b/b^2.\\
\end{align*}
Since $b<0$, $\lim_{N\to\infty}(N/b-1/b^2)e^{bN}=0$, and $I=2(1-b)e^b/b^2$.
Therefore, by the integral test, 
$$\sum_{n=1}^\infty a^{\sqrt{n}}$$
converges.
It follows that $\sum (-1)^na^{\sqrt{n}}$ is absolutely convergent, so it is convergent as well.
