How to calculate the integer part of the value of the following equation?
$$y=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+\ldots+\frac{1}{\sqrt{1000000}}$$
It should be calculated in a special way, after all the equation is so long.
We have from the decreasing of the function $x\mapsto\frac{1}{\sqrt x}$ $$\sum_{k=1}^n\int_{k}^{k+1}\frac{dx}{\sqrt x}\le\sum_{k=1}^n\frac{1}{\sqrt k}\leq\sum_{k=2}^n\int_{k-1}^k\frac{dx}{\sqrt x}+1$$ hence
$$ 2(\sqrt{n+1}-1)=\int_1^{n+1}\frac{dx}{\sqrt x}\le\sum_{k=1}^n\frac{1}{\sqrt k}\leq\int_{1}^n\frac{dx}{\sqrt x}+1= 2(\sqrt n-1)+1$$
Numerically for $n=1000000$ the integer part is $1998$