Problem 37: Turkmenistan National MO 2016, Day 2, Problem 1

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Joined: Mar 10, 2015

by henderson, Jul 10, 2016, 2:55 PM

$\color{red}\bf{Problem \ 37}$ The sequence $\{a_n \}_{n\geq 1}$ is defined by
$a_1 =3 \ \text{and} \ 2a_{n+1} =a^2_n+1.$ Prove that
$\sum_{k=1}^{n}\frac{1}{a_k +1} < \frac{1}{2}.$ $\color{red}\bf{My \ solution}$ $\color{green}\textbf{Claim.}$ $\sum_{k=1}^{n}\frac{1}{a_k +1}=\frac{1}{2}-\frac{1}{a_{n+1}-1}$ for all $n \in \mathbb{N}.$ $($ it's obvious that $a_n>1$ for all $n \in \mathbb{N},$ so $\frac{1}{a_{n+1}-1}$ is always positive, thus the proof of the claim will be sufficient to finish the solution. $)$
$\color{green}\textbf{Proof.}$ We will apply induction.

$n=1$ is trivial. Let us assume that
$\frac{1}{a_1+1}+\frac{1}{a_2+1}+...+\frac{1}{a_n+1}=\frac{1}{2}-\frac{1}{a_{n+1}-1}$ holds and try to prove it for $n+1.$ Or we just need to prove the following equation:
$\frac{1}{a_1+1}+\frac{1}{a_2+1}+...+\frac{1}{a_{n+1}+1}=\frac{1}{2}-\frac{1}{a_{n+2}-1}.$

$\frac{1}{a_1+1}+\frac{1}{a_2+1}+...+\frac{1}{a_{n+1}+1}=\frac{1}{2}-\frac{1}{a_{n+2}-1}.$ $\iff$
$\frac{1}{2}-\frac{1}{a_{n+1}-1}+\frac{1}{a_{n+1}+1}=\frac{1}{2}-\frac{1}{a_{n+2}-1}.$ $\iff$
$\frac{1}{a_{n+1}+1}-\frac{1}{a_{n+1}-1}=-\frac{1}{a_{n+2}-1}.$ $\iff$
$\frac{2}{a^2_{n+1}-1}=\frac{1}{a_{n+2}-1}$ $\iff$
$2a_{n+2}-2=a^2_{n+1}-1$ $\iff$
$2a_{n+2}=a^2_{n+1}+1,$ which is true according to the given condition
$2a_{n+1} =a^2_n+1$ for all $n \in \mathbb{N}.$
We are done!