Prove Inequality: $18<\sum\limits_{i=2}^{99}\dfrac 1{\sqrt i} <19$

[Albert1]

prove :
$18<1+\dfrac {1}{\sqrt 2}+\dfrac {1}{\sqrt 3}+----+\dfrac{1}{\sqrt {99}}<19$

 

[kaliprasad]

Spoiler

we have
$n+ 1 \gt n\gt n-1$

hence$\sqrt{n+1} + \sqrt{n}> 2 \sqrt{n} > \sqrt{n-1}+ \sqrt{n}$

hence taking reciprocal
$\dfrac{1}{\sqrt{n+1} + \sqrt{n}}<\dfrac{1}{ 2 \sqrt{n}} < \dfrac{1}{\sqrt{n-1}+ \sqrt{n}} $

or
$\sqrt{n+1} - \sqrt{n}< \dfrac{1}{ 2 \sqrt{n}} < \sqrt{n}- \sqrt{n-1}$

using the above we get the telescopic sum we get
the result

 

[Albert1]

hence$\sqrt{n+1} + \sqrt{n}> 2 \sqrt{n} > \sqrt{n-1}+ \sqrt{n}$

taking reciprocal it should be:
$\dfrac{1}{\sqrt{n+1} + \sqrt{n}}<\dfrac{1}{ 2 \sqrt{n}} < \dfrac{1}{\sqrt{n-1}+ \sqrt{n}} $

or
$\sqrt{n+1} - \sqrt{n}< \dfrac{1}{ 2 \sqrt{n}} < \sqrt{n}- \sqrt{n-1}$

a very good solution !