2015/043) Find the integer part of $1 + \dfrac{1}{\sqrt{2}}+ \dfrac{1}{\sqrt{3}}+ \dfrac{1}{\sqrt{4}}+ \cdots+ \dfrac{1}{\sqrt{1000000}}$

note that $\dfrac{1}{\sqrt{k}} = \dfrac{2}{2 \sqrt{k}}= \dfrac{2}{\sqrt{k}+ \sqrt{k}}$

now

 $\dfrac{2}{\sqrt{k}+ \sqrt{k}}\lt \dfrac{2}{\sqrt{k}+ \sqrt{k-1}}\lt 2(\sqrt{k}- \sqrt{k-1})$
adding from k = 2to n we get sum $\lt 2(\sqrt{n}-1)$ or $\lt 1998$ (when $n= 10^6$)

adding 1 we get sum = 1999

again
$\dfrac{2}{\sqrt{k}+ \sqrt{k}}\gt \dfrac{2}{\sqrt{k}+ \sqrt{k+1}}\gt 2(\sqrt{k+1}- \sqrt{k})$

 adding sum from 2 to n we get $\gt 2(\sqrt{n+1} - \sqrt{2}) \gt 1997$
so sum from 1 > 1998

so integer part of sum = 1998