$\text{Sup}\{x\geq0:\sum\limits_{n=1}^\infty x^{\sqrt n}<\infty \}$

Question

Possible Duplicate:
convergence of a series involving $x^\sqrt{n}$

How to find $\text{Sup}\{x\geq0:\sum\limits_{n=1}^\infty x^{\sqrt n}<\infty \}?$

Answer 1

If $x\ge 1$, this series diverges obviously. If $0\le x<1$, by Cauchy consendation test, this series converges iff $$\sum_{n=1}^\infty 2^n x^{\sqrt{2}^n}$$ converges. And $$\lim_{n\to\infty} \sqrt[n]{2^n x^{\sqrt{2}^n}}= \lim_{n\to\infty}2x^{\frac{\sqrt{2}^n}{n}}=0.$$ So by root test, $\sum_{n=1}^\infty 2^n x^{\sqrt{2}^n}$ converges.

Answer 2

Let $X$ be the set : $$X=\{x \geq 0 / \sum_{n=1}^{+\infty} x^{\sqrt n} < +\infty \}$$ We have : $X = [0,1[$. and $\sup(X)=1$. In fact : If $x \geq 1$ the serie $\sum x^{\sqrt n} $ diverges since $\displaystyle \lim_{n \to +\infty} x^{ \sqrt n} \neq 0$ (the limit is $1$ if $x=1$ and $+\infty$ if $x > 1$ )

If $x=0$ the sum is $0$.

If $0 < x < 1$ for all $n \in \mathbb N$ there is a unique $p_n \in \mathbb N$ such that $p_n \leq \sqrt n < p_n+1$ then $x^{\sqrt n} \leq x^{p_n}$ and then we have for all $n \in {\mathbb N}^*$ : $$\sum_{n=1}^{N} x^{\sqrt n} \leq \sum_{n=1}^N x^{p_n} \leq \sum_{n=1}^{p_N} x^n \leq \sum_{n=1}^{+\infty} x^n=\frac{1}{1-x} $$

Answer 3

For $x \geq 1$, the terms don't go to zero so the series diverges. When $0 \leq x < 1$ we can write $$\sum\limits_{n=1}^\infty x^{\sqrt n} = \sum_{k = 1}^{\infty}\sum_{n = k^2}^{(k+1)^2 - 1}x^{\sqrt n}$$ In the inner sum, there are $2k + 1$ terms, each of which is bounded by $x^{\sqrt{k^2}} = x^k$, since $x^{\sqrt{n}}$ decreases in $n$ when $0 \leq x < 1$. Hence the sum is bounded by $$\sum_{k = 1}^{\infty}(2k+1)x^k$$ This can readily be shown to be convergent using the ratio or root test, so we have convergence when $0 \leq x < 1$.