Inequality with Harmonic Differences

The following problem comes from the 2007 China Mathematical Competition [Xu Jiagu, 5-6]:

For each positive integer \(n\gt 1\) prove that

\(\displaystyle \frac{2n}{3n+1} \lt \frac{1}{n+1}+\frac{1}{n+2}+\ldots +\frac{1}{n+n}\lt \frac{25}{36} \)

If we introduce Harmonic numbers as \(\displaystyle H_{n}=\sum_{k=1}^{n}\frac{1}{k}\), then the problem defines the low and upper bounds for the difference \(H_{2n}-H_{n}\).



  1. Xu Jiagu, Lecture Notes on Mathematical Olympiad Courses, v 8, (For senior section, v 2), World Scientific, 2012

|Contact| |Front page| |Contents| |Up|

Copyright © 1996-2018 Alexander Bogomolny

For each positive integer \(n\gt 1\) prove that

\(\displaystyle \frac{2n}{3n+1} \lt H_{2n}-H_{n} \lt \frac{25}{36} \)


Both the left and right inequalities succumb to the mathematical induction. The left one is more or less straightforward. The right one is much less so.

For the left inequality, define

\(\displaystyle f(n)= H_{2n}-H_{n}-\frac{2n}{3n+1} \)

Since \(\displaystyle f(2) = \frac{1}{3}+\frac{1}{4}-\frac{4}{7}=\frac{1}{84}\gt 0\), we have a basis for induction. So assume that \(f(n)\gt 0\) for some \(n\) and prove that \(f(n+1)\gt 0\). To this end suffice it to show that \(f(n+1)\gt f(n)\). This is indeed so:

\(\displaystyle \begin{align} f(n+1)-f(n) &= \frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n+1}-\frac{2n}{3n+1}+\frac{2n+2}{3n+4} \\ &= \frac{1}{2n+1}-\frac{1}{2n+2}-\frac{2n}{3n+1}+\frac{2n+2}{3n+4} \\ &= \frac{1}{(2n+1)(2n+2)}-\frac{2}{(3n+1)(3n+4)} \\ &= \frac{n(n+3)}{(2n+1)(2n+2)(3n+1)(3n+4)} \gt 0. \end{align} \)

Trying the same approach for the right inequality leads to a dead end: it would have been nice if function \(\displaystyle g(n)=H_{2n}-H_{n}-\frac{25}{36}\) was decreasing, but - as can be easily verified - it's not. It comes out that a stronger inequality is more amenable to a proof by the mathematical induction. So let's define

\(\displaystyle g(n)=H_{2n}-H_{n}-\frac{25}{36}+\frac{1}{4n+1}. \)

First of all,

\(\displaystyle g(2) = \frac{1}{3}+\frac{1}{4}-\frac{25}{36}+\frac{1}{9} = \frac{12 + 9 - 25 + 4}{36} = 0. \)


\(\displaystyle \begin{align} g(n+1) - g(n) &= \frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n+1}+\frac{1}{4n+5}-\frac{1}{4n+1} \\ &= \frac{1}{2n+1}-\frac{1}{2n+2}+\frac{1}{4n+5}-\frac{1}{4n+1} \\ &= \frac{1}{(2n+1)(2n+2)}-\frac{4}{(4n+5)(4n+1)} \\ &= -\frac{3}{(2n+1)(2n+2)(4n+5)(4n+1)} \lt 0. \end{align} \)

It follows that \(g(n)\) is decreasing, \(g(n)\le 0\) for \(n\gt 1\) and, therefore,

\(\displaystyle \begin{align} H_{2n}-H_{n}-\frac{25}{36} \lt g(n) \le 0, \end{align} \)

as required.

|Contact| |Front page| |Contents| |Up|

Copyright © 1996-2018 Alexander Bogomolny