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?

More directly, multiply and divide by the conjugate:
$2 ( \sqrt{n+1} - \sqrt{n} ) = 2 \frac{  ( \sqrt{n+1} - \sqrt{n} )  ( \sqrt{n+1} + \sqrt{n} ) }{ ( \sqrt{n+1} + \sqrt{n} )} = \frac{2}{ ( \sqrt{n+1} + \sqrt{n} )} < \frac{2}{2\sqrt{n}} = \frac{1}{ \sqrt{n}}. $
Same for the RHS, can you complete it?
