A Problem for the 2015 New Year's Eve

Here's a slight modification of a problem from a 2009 Chinese math olympiad (see the reference below, p 19.) The problem has been modified to be suitable for a consideration on the eve of the 2015 New Year.

Find the least positive integer $a$ such that the inequality

$\displaystyle\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n+1}\lt a-2014\frac{1}{7}.$

holds for every positive integer $n.$

Denote $f(n)=\displaystyle\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n+1}$ and observe that $f(n)\gt f(n+1)$ for all positive $n.$ Indeed,

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

This shows that $f$ attends its maximum at $n=1.$ Therefore, suffice it to consider the inequality $\displaystyle f(1)\lt a-2014\frac{1}{7},$ i.e.,

$\displaystyle \frac{1}{2}+\frac{1}{3}\lt a-2014\frac{1}{7}.$

From here $a$ satisfies $\displaystyle a\gt 2014+\frac{5}{6}+\frac{1}{7}=2014\frac{41}{42}.$ Obviously $a=2015$ is the least integer that satisfies this inequality.


  1. Xiong Bin, Lee Peng Yee, Mathematical Olympiads in China (2009-2010). Problems and Solutions, World Scientific, 2013

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