Prove the sum is less than 2016

[anemone]

Prove the inequality

\(\displaystyle \sqrt{\frac{1\cdot 2}{3^2}}+\sqrt{\frac{2\cdot 3}{5^2}}+\sqrt{\frac{3\cdot 4}{7^2}}+\cdots+\sqrt{\frac{4032\cdot 4033}{8065^2}}<2016\)

 

[kaliprasad]

Prove the inequality

\(\displaystyle \sqrt{\frac{1\cdot 2}{3^2}}+\sqrt{\frac{2\cdot 3}{5^2}}+\sqrt{\frac{3\cdot 4}{7^2}}+\cdots+\sqrt{\frac{4032\cdot 4033}{8065^2}}<2016\)

Spoiler

we have n^{th} term = $\frac{\sqrt{n\cdot (n+1)}}{2n+1}$
$= \frac{\sqrt{n^2+n}}{2n+1}$
$= \frac{\sqrt{n^2+n+\frac{1}{4}-\frac{1}{4}}}{2n+1}$
$= \frac{\sqrt{(n+\frac{1}{2})^2-\frac{1}{4}}}{2n+1}$
$ < \frac{n+\frac{1}{2}}{2n+1}$
$ < \frac{1}{2}$
each term is $ < \frac{1}{2}$ and there are 4032 terms so sum is less than 2016

 

[anemone]

Very well done, kaliprasad!