Prove the sum is less than 2016

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anemone
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Prove the inequality

$$\sqrt{\frac{1\cdot 2}{3^2}}+\sqrt{\frac{2\cdot 3}{5^2}}+\sqrt{\frac{3\cdot 4}{7^2}}+\cdots+\sqrt{\frac{4032\cdot 4033}{8065^2}}<2016$$
 
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anemone said:
Prove the inequality

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

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
 
Very well done, kaliprasad!