Convergence of series $\sum_{n=1}^{\infty}{\frac{1}{\sqrt{1}+\sqrt{2}+...+\sqrt{n}}}$

hint

For $k=1,2,...n $,

$$\int_{k-1}^k\sqrt {t}dt <\sqrt {k}<\int_k^{k+1}\sqrt {t}dt $$

by sum,

$$\int_0^n\sqrt {t}dt <\sum_{k=1}^n\sqrt {k}. $$

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 $s>1$

Compare it with harmonic series. As $\sum_{k=1}^n\sqrt{k} = \Omega(n^{1.5})$, you can conclude that this series is convergent. And for the claim you can prove it by the following:

$$\sum_{k=1}^n\sqrt{k} \geq \sqrt{\frac{n}{2}} + \cdots+\sqrt{n}\geq \frac{n}{2}\sqrt{\frac{n}{2}}$$