Competition problem — Algebra — Maths Olympiad Prep

Library / /47 of 100

Algebra Difficulty 4.8 AIME Prove it China

The inequality
$\frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{2n+1} < a - 2007 \frac{1}{3}$
holds for every positive integer $n$ . Then the least positive integer of $a$ is $\underline{\hspace{2cm}}$ .

Solution

Obviously,
$f(n) = \frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{2n+1}$
is monotonically decreasing. Therefore, $f(1)$ reaches the maximum of $f(n)$ . From
$f(1) = \frac{1}{2} + \frac{1}{3} < a - 2007 \frac{1}{3},$
we have $a > 2008$ . Therefore, the least positive integer of $a$ is $2009$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.