TITLE: Finding summation of inverse of square roots.
QUESTION [11 upvotes]: Problem:  Find the integer part of
$$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} +...+\frac{1}{\sqrt{1000000}}$$
How should I go about approaching this question?  I have never seen such a question before, where you cannot multiply any conjugate or whatsoever.  What is the method to go about approaching such problems?

REPLY [13 votes]: For an precalculus (telescoping) approach.
Notice that $$\frac{1}{\sqrt n}>\frac{2}{\sqrt{n}+\sqrt{n+1}}=2(\sqrt{n+1}-\sqrt{n})$$
From this $$\frac{1}{\sqrt{1}}+\frac1{\sqrt2}+\cdots+\frac1{\sqrt{1000^2}}\geq 2(\sqrt 2-1+\sqrt 3-\sqrt 2+\cdots+\sqrt{1000^2+1}-\sqrt{1000^2})\geq 2\sqrt{1000^2+1}-2>2000-2=1998$$
And also $$\frac{1}{\sqrt n}\leq\frac{2}{\sqrt n+\sqrt{n-1}}=2(\sqrt n-\sqrt{n-1})$$
Leaving the $n=1$ case alone we get
$$\frac{1}{\sqrt{1}}+\frac1{\sqrt2}+\cdots+\frac1{\sqrt{1000^2}}<1+2(\sqrt 2-1+\sqrt 3-\sqrt 2+\cdots \sqrt{1000^2}-\sqrt{1000^2-1})=1+2\sqrt{1000^2}-2=1999$$
And it follows that the integer part is $1998$