TITLE: $\left\lfloor \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{1024}}\right\rfloor =?$
QUESTION [9 upvotes]: $$\left\lfloor \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{1024}}\right\rfloor =?$$ 
I can't find $$? \leq  \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{n}} \leq ?$$ I remeber there was a sharp bound , Am i right  ?
I am thankful if someone help me to find the bound  , or give another idea
Last question : Is it possible to find $\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{1024}}$  by integral  ?

REPLY [5 votes]: Use
$$\sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n+1}+\sqrt{n}}\in\left(\frac{1}{2\sqrt{n+1}}, \frac{1}{2\sqrt{n}}\right)$$
or, to put it differently:
$$2(\sqrt{n+1}-\sqrt{n})\lt\frac{1}{\sqrt{n}}\lt2(\sqrt{n}-\sqrt{n-1})$$
so you can squeeze your sum between two telescopic sums.