Track / Stage 5 / 189 of 400 #789 of 1964

Problem 789

AIME late

Algebra Difficulty 5.3 Prove it

Prove that for every natural number $n$ the following holds

$\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{3 n+1}>1$

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Solution. Let $a_{n}=\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{3 n+1}$ . Then $a_{n+1}=\frac{1}{n+2}+\frac{1}{n+3}+\ldots+\frac{1}{3 n+4}$ , so

$a_{n+1}-a_{n}=\frac{1}{3 n+2}+\frac{1}{3 n+3}+\frac{1}{3 n+4}-\frac{1}{n+1}=\frac{2}{(3 n+2)(3 n+3)(3 n+4)}>0$

Thus, the sequence $a_{n}$ is increasing. Moreover, $a_{1}=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}>1$ , so $a_{n}>1$ for every natural number.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.