# Thread: Proving an inequality (Area under graph)

1. ## Proving an inequality (Area under graph)

Given that $y=\frac{1}{x^2}$, where $x>0$.
By considering the area under the curve and the areas of rectangles, for $n \le x \le n+1$, where $n \ge 1$ is an integer, show that

$\frac{n^2+n+1}{(n+1)^2}<1+\frac{1}{2^2}+\frac{1}{3 ^2}+...+\frac{1}{(n+1)^2}<\frac{2n+1}{n+1}$

I started with $\frac{1}{(n+1)^2}<\int^{n+1}_{n} \frac{1}{x^2}dx<\frac{1}{n^2}$ which leads to
$\frac{1}{(n+1)^2}<\frac{1}{n(n+1)}<\frac{1}{n^2}$

Anyone can help? thanks!

2. We have $\frac{1}{(k+1)^2}<\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}$

$1=1$

$\frac{1}{2^2}<1-\frac{1}{2}$

$\frac{1}{3^2}<\frac{1}{2}-\frac{1}{3}$

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$\frac{1}{n^2}<\frac{1}{n}-\frac{1}{n+1}$

Summing the inequalities we have

$1+\frac{1}{2^2}+\ldots+\frac{1}{(n+1)^2}<2-\frac{1}{n+1}=\frac{2n+1}{n+1}$

Now, $\frac{n^2+n+1}{(n+1)^2}=\frac{n}{n+1}+\frac{1}{(n+ 1)^2}$ and the left side inequality can be written as

$\frac{n}{n+1}<1+\frac{1}{2^2}+\frac{1}{3^2}+\ldots +\frac{1}{n^2}$

We have:

$\frac{1}{1\cdot2}<1$

$\frac{1}{2\cdot3}<\frac{1}{2^2}$

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$\frac{1}{n(n+1)}<\frac{1}{n^2}$

Adding the inequalties we get the result.