If $x=\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{2015^2}$, show that $\frac{201}{403}
If $$x=\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{2015^2}$$
show that
$$\frac{201}{403}<x<\frac{2014}{2015}$$
So, I manage to do the RH inequality using that $$x<\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\cdots+\frac{1}{2014\cdot2015}=1-\frac{1}{2015}=\frac{2014}{2015}$$ Unfortunately I can't find a way to do the LH inequality. I saw that
$$\frac{1}{2015}+\frac{1}{2015}+\cdots+\frac{1}{2015}$$ for 1005 times is equal to $\frac{201}{403}$ but I don't see why
$$x> \frac{1}{2015}+\frac{1}{2015}+\cdots+\frac{1}{2015}$$ thx!

We have the bound
$$\sum_{n=2}^{2015}\frac{1}{n^2}>\int_2^{2016}\frac{1}{x^2}dx=\frac{1}{2}-\frac{1}{2016}>\frac{1}{2}-\frac{1}{806}=\frac{402}{806}=\frac{201}{403}$$
by thinking about the graph of the function $y=1/x^2$ and noticing that if you draw rectangles of base length $1$ with four vertices $(n,0),(n+1,0),(n,1/n^2),(n+1,1/n^2)$ for $n=2,\dots,2015$ then the union of the resulting rectangles have an area bounded below by the integral given above.
By the way, this also works to establish the upper bound $2014/2015$.
