MHB Prove the sum is less than 2016

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The discussion focuses on proving the inequality that the sum of the series $$\sqrt{\frac{1\cdot 2}{3^2}}+\sqrt{\frac{2\cdot 3}{5^2}}+\sqrt{\frac{3\cdot 4}{7^2}}+\cdots+\sqrt{\frac{4032\cdot 4033}{8065^2}}$$ is less than 2016. Participants explore various mathematical approaches and techniques to demonstrate this inequality. The series involves terms that combine factorial-like products divided by squares of odd integers. The conversation highlights the importance of bounding the terms effectively to establish the overall sum's limit. The conclusion emphasizes the successful proof of the inequality.
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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!
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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