LMVT Application: $\sqrt {n + 1} - \sqrt n < \frac{1}{{2N}}$

Problem

Prove that $\sqrt {n + 1}  - \sqrt n <  \frac{1}{{2N}}$ if $n > N^2$, n being a natural number.

Solution

${\rm{Let }}f(x) = \sqrt x ,x \in [n,n + 1]$

Since n is a natural number, clearly f(x) satisfies the continuity and differentiability requirements of LMVT.

$f'(c) = \frac{1}{{2\sqrt c }} = \frac{{f(n + 1) - f(n)}}{{(n + 1) - n}},n < c < n + 1$

$\frac{1}{{2\sqrt c }} = f(n + 1) - f(n) = \sqrt {n + 1}  - \sqrt n .........(*)$

Since, $n < c < n + 1$

$\sqrt n  < \sqrt c  < \sqrt {n + 1}$

$\frac{1}{{\sqrt n }} > \frac{1}{{\sqrt c }} > \frac{1}{{\sqrt {n + 1} }}$..........(#)

Given, $n > {N^2}$

$\sqrt n  > N$

$\frac{1}{{\sqrt n }} < \frac{1}{N},\frac{1}{{\sqrt n }} > \frac{1}{{\sqrt c }}$      from (#)

Hence, $\frac{1}{N} > \frac{1}{{\sqrt c }}$

$\frac{1}{{2N}} > \frac{1}{{2\sqrt c }}$ 

$\frac{1}{{2N}} > \sqrt {n + 1}  - \sqrt n$      from (*)