The sum of reciprocal squares: estimating the remainder

Question

Let $a_n$ denote the $n$th remainder of the series $$ 1+\frac{1}{2^2}+\frac{1}{3^2}+\ldots $$ In other words, $$ a_n = \frac{\pi^2}{6}-\left(1+\frac{1}{2^2}+\ldots +\frac{1}{n^2}\right). $$ I noticed that for small $n$ the following is true $$ \frac{1}{n+1}<a_n<\frac{1}{n}\tag{$*$} $$ and tried to prove it for all $n$. Using induction on $n$, I ended up having to prove the estimates $$\frac{1}{n+1}\cdot \frac{n^2+3n+3}{(n+1)(n+2)}<a_n<\frac{1}{n}\cdot \frac{n(n+2)}{(n+1)^2}$$ which are even stronger than $(*)$.

My question is whether $(*)$ is true for all $n$ and, if so, how could one prove it?

Answer 1

The easiest way to show $(*)$ is to note that for $k>1$ we have $$ \frac{1}{k}-\frac{1}{k+1}< \frac{1}{k^2}<\frac{1}{k-1}-\frac{1}{k} $$ then add these inequalities starting from $k=n+1$, using the telescoping aspect.

Answer 2

It seems to be true from computing large $a_i$'s by Mathematica but induction does not work here. However, induction does work on the bounds $\frac{1}{(n+1)^2} < a_n < \frac{1}{n^2}$.