Track / Stage 5 / 42 of 400 #1282 of 2604

Problem 1282

AIME late

Algebra Difficulty 5.0 Prove it Bulgarian Mathematical Competitions · Bulgaria

Let $a_{1}>0$ and $a_{n+1}=a_{n}+\frac{n}{a_{n}}$ for $n \geq 1$ . Prove that:

a) $a_{n} \geq n$ for $n \geq 2$ ;

b) the sequence $\left\{\frac{a_{n}}{n}\right\}_{n \geq 1}$ converges and find its limit.

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:

a) We have $a_{2}=a_{1}+\frac{1}{a_{1}} \geq 2$ . If $a_{n} \geq n$ , then
$a_{n+1}-n-1=a_{n}+\frac{n}{a_{n}}-n-1=\frac{\left(a_{n}-1\right)\left(a_{n}-n\right)}{a_{n}} \geq 0$
and the assertion follows by induction.

b) Let $n \geq 2$ . It follows from a) that $a_{n+1} \leq a_{n}+1$ . Then $a_{n} \leq a_{2}+n-2$ , whence $1 \leq \frac{a_{n}}{n} \leq 1+\frac{a_{2}-2}{n}$ . Therefore the sequence $\left(\frac{a_{n}}{n}\right)_{n \geq 1}$ is convergent and its limit equals $1$ .

Remark. One can prove the stronger statement that $\lim _{n \rightarrow \infty}\left(a_{n}-n\right)=0$ .

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.