# An Inequality by Uncommon Induction

Prove that for every $n\gt 1$,

$\displaystyle 1+\frac{1}{2^2}+\frac{1}{3^2}+\ldots +\frac{1}{n^2}\gt \frac{3n}{2n+1}.$

Proof

Prove that for every $n\gt 1$,

$\displaystyle 1+\frac{1}{2^2}+\frac{1}{3^2}+\ldots +\frac{1}{n^2}\gt \frac{3n}{2n+1}.$

The first idea that comes to mind is that the method of mathematical induction ought to be of use for the proof. This is indeed so, but not without a workaround. For $n=1$, the two expressions are equal: $\displaystyle 1=\frac{3\cdot 1}{2\cdot 1+1}$, and this is why $n=1$ is excluded. From then on, the two sides grow. The left-hand side grows by $\displaystyle \frac{1}{n^2}$, the right-hand side grows by

$\displaystyle \frac{3n}{2n+1} - \frac{3(n-1)}{2(n-1)+1} = \frac{3}{4n^{2}-1}.$

Now, it is easy to verify that, for $n\gt 1$, $\displaystyle \frac{1}{n^2}\gt \frac{3}{4n^{2}-1}$. This exactly means that the left-hand side grows faster than the right-hand side which, thus, proves the inequality.

The two sides monotone increasing as $n\rightarrow\infty$; the left-hand side is known as Euler series, with the famous value:

$\displaystyle \sum_{k=1}^{\infty}\frac{1}{k^2}=\frac{\pi ^2}{6}\approx 1.645.$

This is certainly greater than the limit $\displaystyle\frac{3}{2}$ of the right-hand side. In itself, though, this is not yet sufficient to prove the inequality for all $n\gt 1$!

Jack D'Aurizio came up with another solution. He starts with

$\displaystyle \sum_{k=2}^{n}\frac{1}{k^2} \lt \sum_{k=2}^{n}\frac{1}{k^2-1/4} = 2\sum_{k=2}^{n}\bigg(\frac{1}{2k-1} - \frac{1}{2k+1}\bigg) = 2\bigg(\frac{1}{3} - \frac{1}{2n+1}\bigg),$

which holds for $n\ge 2$. By adding $1$ to both sides we get:

$\displaystyle \sum_{k=1}^{n}\frac{1}{k^2} > \frac{10n-1}{6n+3},$

which is stronger than the inequality we set out to prove, because $\displaystyle\frac{10n-1}{6n+3}\gt \frac{3n}{2n+1}$, for $n\gt 1.$

If we apply the "telescoping estimation" technique later, we get even stronger inequalities. For example, starting from

$\displaystyle \sum_{k=3}^{n}\frac{1}{k^2} \lt \sum_{k=3}^{n}\frac{1}{k^2-1/4} = 2\sum_{k=3}^{n}\bigg(\frac{1}{2k-1} - \frac{1}{2k+1}\bigg) = 2\bigg(\frac{1}{5} - \frac{1}{2n+1}\bigg)$

and by adding $\displaystyle\frac{5}{4}$ to both sides, we get

$\displaystyle \sum_{k=1}^{n} \frac{1}{k^2} \gt \frac{66n-7}{20(2n+1)},$

which also holds for $n\gt 1.$ Starting with $k=4$ and adding $\displaystyle\frac{5}{4}$ gives

$\displaystyle \sum_{k=1}^{n} \frac{1}{k^2} \gt \frac{830 n - 89}{252 (2n+1)},$

### Reference

1. R. Honsberger, More Mathematical Morsels, MAA, New Math Library, 1991, 33-35