Convergence of $\sum a^{1/x_n}$ for $a$ in $(0,1)$ and $\sum x_n$ a positive convergent series 
Let $\sum x_n$ be a convergent series of positive real numbers and $0<a<1 $,  then is the series $\sum a^{1/{x_n}}$ convergent ? 

I have only figured out that $\lim a^{1/{x_n}}=0$.
 A: Put $b=-\log a >0$, $f(y)=y\exp(-by)$. It is easy to see that $f$ is continuous on $I=[0,+\infty[$, and that $f(y)\to 0$ if $y\to +\infty$. Hence $f$ is bounded, say by $M$, on $I$. We have $f(1/x_n)\leq M$, hence $a^{1/x_n}\leq M x_n$, and it is easy to finish. 
A: Assume that $\sum x_n <1$, then pick $a_n= \inf\{k | x_m<\frac{1}{n}, \forall m\geq k     \}$ for $n\geq 1$. Notice that 
$a_n \leq n$, and that $a_n$ is increasing.
\begin{align*}
\sum _{n=1}^{\infty} a^{\frac{1}{x_n}} &= \sum_{k=1}^{\infty}\sum _{n=a_k}^{a_{k+1}} a^{\frac{1}{x_n}} \\
&\leq \sum_{k=1}^{\infty}\sum _{n=a_k}^{a_{k+1}} a^{n}\\
&= \sum_{k=1}^{\infty}(a_{k+1}  -a_k+1) a^{k} \\
&\leq \sum_{k=1}^{\infty}(k+1) a^{k}
\end{align*}
Now using the root test we can see that $\sum_{k=1}^{\infty}(k+1) a^{k}$ convergences, and by the comparison tst we conclude the result.
A: as $0<a<1$ , so $1/a >1$ so $\lim _{x \to \infty } xa^x=\lim_{x \to \infty }x/a^{-x}=0$ ; now as $x_n>0$ and $\lim x_n=0$ , so $\lim \dfrac 1{x_n}=\infty$ , so by sequential criteria of limit at infinity , $0=\lim _{x \to \infty } xa^x=\lim_{n \to \infty} (\dfrac 1{x_n})a^{\dfrac 1{x_n}}=\lim_{ n\to \infty } \dfrac {a^{1/x_n}}{x_n}$ , so by Comparison test , $\sum a^{1/x_n}$ converges 
