For which $x$ does $\sum_{n=1}^\infty n^{\alpha}x^{\sqrt{n}}$ converge? I want to know for which values of $x$ this series 
$$\sum_{n=1}^\infty n^{\alpha}x^{\sqrt{n}}$$ converges with $\alpha \in  \mathbb{R}$$.
This series is defined for $  x 
 \ge 0$.
The necessary condition for the convergence $a_n \sim 0$ is satified for $0 \le x < 1$
and now I'm trying to apply some criteria for the convergence and either the 
root criterion or the ratio one are inconclusive.
Can you suggest something to solve the exercize?
 A: hint
For $ x>0,$ The general term is
$$u_n=e^{\sqrt{n}\Bigl(\frac{\alpha \ln(n)}{\sqrt{n}}+\ln(x)\Bigr)}$$
If $x>1$, its limit is not zero, thus the series diverges.
If $ x=1$, the series is a Riemann one which converges only if $ -\alpha>1$.
If $ x<1$,
the series converges because
$$\lim_{n\to +\infty}n^2u_n=0$$
(For $ n$ great enough $0<n^2u_n<1$ and the comparison test ).
instead of $n^2$ , you can take $n^{\beta}$ for any $\beta>1$.
A: Suppose $\alpha >0.$ Then clearly the series diverges for $x\ge 1,$ simply because the $n$th term doesn't tend to $0.$
Staying with $\alpha >0,$  let $0\le x <1.$ Let's group terms:
$$\tag 1 \sum_{n=1}^{\infty}n^\alpha x^{\sqrt n} = \sum_{m=1}^{\infty}\sum_{n=m^2}^{(m+1)^2-1}n^\alpha x^{\sqrt n}.$$
In the inside sum there are $2m$ terms, each of which is bounded above by $[(m+1)^2]^\alpha x^m.$ Therefore $(1)$ is bounded above by
$$\sum_{m=1}^{\infty}2m[(m+1)^2]^\alpha x^m.$$
This series converges because we have polynomial growth in $m$ vs. exponential decay in $x^m.$ If I seem to be hand-waving, use the ratio test to see this.
I'll leave $\alpha \le 0$ to you for now.
