Show that $1998 < \sum_1^{10^6}\frac{1}{\sqrt{n}}<1999$ Show that $1998 < \sum_1^{10^6}\frac{1}{\sqrt{n}}<1999$ by first recalling the inequality:
$$2(\sqrt{n+1}-\sqrt{n})<\frac{1}{\sqrt{n}}<2(\sqrt{n}-\sqrt{n-1})$$ when $n=1, 2, 3...$
What I have tried to show using the inequality:
By the first step of induction to see if the inequality holds  :
$2(\sqrt{1+1}-\sqrt{1})<\frac{1}{\sqrt{1}}<2(\sqrt{1}-\sqrt{1-1})$ is true
Then by induction we have:
$2(\sqrt{n+1+1}-\sqrt{n+1})<\frac{1}{\sqrt{n+1}}<2(\sqrt{n+1}-\sqrt{n+1-1})$
Then by using the fact that if $a>b$ and $c>0$ then $ac>bc$
The following should be true if the induction is right.
$\frac{2(\sqrt{n+1}-\sqrt{n})}{\sqrt{n+1}}>2(\sqrt{n+1+1}-\sqrt{n+1})$ by taking the square of both sides we have after expanding:
$$\frac{8n-8\sqrt{n^2+n}+4}{n+1}>8n+12-8\sqrt{n^2+3n+2}$$
Then placing everything onto one side:
$$\frac{-8n^2+8n\sqrt{n^2+3n+2}-12n+8\sqrt{n^2+3n+2}-8\sqrt{n^2+n}-8}{n+1}>0$$
Thouigh I'm not sure whether I'm going down the right direction. I had assumed that the LHS would cancel to provide something more exact. Any detailed hint son how to proceed?

You don't need induction here. We have
$$n(n+1)=n^2+n<n^2+n+\frac14=\left(n+\frac12\right)^2$$
Take square roots:
$$\sqrt n\sqrt{n+1}<n+\frac12$$
Multiply by $2/\sqrt n$ and rearrange:
$$2(\sqrt{n+1}-\sqrt n)<\frac{1}{\sqrt n}$$
Do the same with $n(n-1)$ to get the right-hand side of the inequality.
