Prove :$\sum_{k＝1}^{n}\frac {1}{(k＋1)\sqrt {k}}<2$

Question

My attemp $$\sum_{k＝1}^{n}\frac {1}{(k＋1)\sqrt {k}}＝\sum_{k＝1}^{n}\frac {\sqrt {k}}{k(k＋1)}＝\sum_{k＝1}^{n}\left({\frac {\sqrt {k}}{k}-\frac {\sqrt {k}}{k＋1}}\right)\\~\\＝1－\frac {\sqrt {n}}{n＋1}＋\sum_{k＝2}^{n}\frac {\sqrt {k}}{k}-\sum_{k＝2}^{n}\frac {\sqrt {k-1}}{k}\\~\\＝1－\frac {\sqrt {n}}{n＋1}＋\sum_{k＝2}^{n}\left({\frac {\sqrt {k}-\sqrt {k-1}}{k}}\right)＝1－\frac {\sqrt {n}}{n＋1}＋\sum_{k＝2}^{n}\left({\frac {1}{k(\sqrt {k}＋\sqrt {k-1})}}\right)$$ I need help proving this : $$\sum_{k＝2}^{n}\left({\frac {1}{k(\sqrt {k}＋\sqrt {k-1})}}\right)<1$$ Thank you

Answer 1

Since the function $f(x) = \frac {1}{x(\sqrt {x}＋\sqrt{x-1})}$ is monotonic decreasing,

$ \sum_{k=2}^{\infty} f(k) = f(2)+f(3)+f(4)+\sum_{k=5}^{n} f(k) \\ \leq f(2)+f(3)+f(4)+\int_{x=4}^{\infty}\left({\frac {1}{x(\sqrt {x}＋\sqrt {x-1})}}\right)dx \\ \leq f(2)+f(3)+f(4)+\int_{x=4}^{\infty}\left({\frac {1}{x(\sqrt {x-1}＋\sqrt {x-1})}}\right)dx \\ = f(2)+f(3)+f(4) + \lbrack \arctan(\sqrt{x-1})\rbrack_4^{\infty} \\ = \frac{1}{2+2\sqrt2}+\frac{1}{3\sqrt{2}+3\sqrt3}+\frac{1}{4\sqrt3+8} + \frac{\pi}{6} \\ < \frac{1}{2+2}+\frac{1}{4+5}+\frac{1}{6+8} + \frac{22/7}{6} = \frac{242}{252} < 1 $,

(according to computational calculation.)