# An Inequality by Uncommon Induction

Prove that for every \(n\gt 1\),

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

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Copyright © 1996-2018 Alexander Bogomolny

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

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

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Copyright © 1996-2018 Alexander Bogomolny