If $x=\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{2015^2}$, show that $\frac{201}{403}<x<\frac{2014}{2015}$

Question

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!

Answer 1

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 $2014/2015$.

Answer 2

We have $$\frac{1}{k}-\frac{1}{k+1} < \frac{1}{k^2} < \frac{1}{k-1} - \frac{1}{k}$$

and so (telescoping sum)

$$\frac{1}{2} - \frac{1}{2016} < \sum_{k=2}^{2015} \frac{1}{k^2} < \frac{1}{1} - \frac{1}{2015}$$ that is $$\frac{1007}{2016}< \sum_{k=2}^{2015} \frac{1}{k^2} < \frac{2014}{2015}$$

Now, LHS $= \frac{2014}{4032}= \frac{201.4}{403.2}> \frac{201}{403}$ (mediant)