I need to find the supremum of the set $S$

Question

Let $S:=\{x\ge 0,\sum_{n=1}^{\infty} x^{\sqrt{n}}<\infty\}$; I need to find the supremum of the set $S$.

Could any one tell me where to start?

Answer 1

If $x\geq1$, then $\lim_{n\rightarrow\infty}x^{\sqrt(n)}$ does not go to zero, which prevents the sum to be finite. Hence $S\subseteq[0,1)$. Also, for each $x<1$, \begin{eqnarray*} \lim_{n\rightarrow\infty}\frac{x^\sqrt{n+1}}{x^\sqrt{n}}&=&\lim_{n\rightarrow\infty}\frac{x^\sqrt{n+1} \log x}{x^\sqrt{n}2\sqrt{n+1}\log x}\\ &<&\lim_{n\rightarrow\infty}\frac{1}{2\sqrt{n+1}}\\ &=&0. \end{eqnarray*} Thus the sum is finite for each $x<1$. So supremum of $S$ is 1.

Answer 2

We have for all $t\geqslant 0$ that $\exp(t)\geqslant \frac{t^4}{24}$, hence if $x<1$, we have $$x^{\sqrt n}=\exp(\sqrt n\log x)\leqslant \frac{24}{n^2(-\log x)^4},$$ which gives convergence of the series (why?).

What if $x=1$?