TITLE: Convergence of series $\sum_{n=1}^{\infty}{\frac{1}{\sqrt{1}+\sqrt{2}+...+\sqrt{n}}}$
QUESTION [5 upvotes]: $\sum_{n=1}^{\infty}{\frac{1}{\sqrt{1}+\sqrt{2}+...+\sqrt{n}}}$ $\\$
I tried a comparison test, Leibniz, Cauchy, Cauchy-integral and D'Alembert criteria but nothing gives me the result. Any hints would be helpful!

REPLY [4 votes]: Use AM-GM inequality and we get $\frac{\sqrt{1}+\sqrt{2}+\dots +\sqrt{n}}{n}>\sqrt[n]{(n!)^{\frac{1}{2}}}$ and then use the result $(n!)^2>n^n$, when $n=3,4,\dots$(Why?).
Note that $\sum_{n=1}^{\infty}\frac{1}{n^s}$ is finite if $s>1$