Finding supremum of a set of elements each of which gives a bounded series

Question

Let $S=\big\{x\in \mathbb{R} :x\ge 0 ,\sum\limits_1^\infty {x^{\sqrt{n}}} < \infty\big\}$. Find the supremum of $S$. It is clear that if $x\ge 1$,the given series is unbounded. Does every x less than 1 belong to S? How do I find the supremum ? Help please.

Answer 1

Claim: $S=[0,1)$. Proof: If $x<1$ then $x^{\sqrt n}=e^{-\sqrt n t}$ where $t=-\ln x$. Use the fact that $e^{x} \geq \frac {x^{4}} {4!}$ to see that $x^{\sqrt n} \leq \frac {24} {n^{2}t^{4}}$. Recall that $\sum \frac 1 {n^{2}} <\infty$.

Of course supremum of $[0,1)$ is $1$.