Convergence of $\sum a^{1/x_n}$ for $a$ in $(0,1)$ and $\sum x_n$ a positive convergent series

Question

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 $\lim a^{1/{x_n}}=0$.

Answer 1

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.

Answer 2

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.

Answer 3

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