Define a sequence $\{a_n\}_{n \geq 1}$ of real number by $a_1 = 2, a_{n+1} = \frac {a_n^2 +1 } {2} , \text{for}\ n \geq 1 $

Question

Define a sequence $\{a_n\}_{n \geq 1}$ of real number by $$a_1 = 2, a_{n+1} = \frac {a_n^2 +1 } {2} , \text{for}\ n \geq 1 $$ Then prove that $$\displaystyle {\sum_{j=1}^N \frac {1}{a_j +1} < 1 }$$for every natural number N.

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I don't know how to solve this problem, please give me some ideas to start in a right way to reach the solution.

This question is from R.M.O 2018 Tamilnadu region.

Answer 1

I'm pretty sure I have seen this many years ago (back in the days in the discussion group sci.math). In any event, notice $$\frac{1}{x-1} - \frac{1}{x+1} = \frac{2}{x^2-1} = \frac{1}{\frac{x^2+1}{2}-1}$$ we have $$\frac{1}{a_j-1} - \frac{1}{a_j+1} = \frac{1}{a_{j+1}+1}$$ This leads to $$\begin{align}\sum_{j=1}^N \frac{1}{a_j+1} &= \sum_{j=1}^N \left(\frac{1}{a_j-1} - \frac{1}{a_{j+1}-1}\right) = \frac{1}{a_1-1} - \frac{1}{a_{N+1}-1}\\ &< \frac{1}{a_1-1} = \frac{1}{2-1} = 1\end{align}$$ because $\displaystyle\;\frac{1}{a_{N+1}-1}$ is clearly positive.

Answer 2

Let $S_n=\displaystyle {\sum_{j=1}^n \frac {1}{a_j +1}}$

I'm going to show by induction that $\displaystyle {1-S_n=\frac{1}{a_{n+1}-1} \Rightarrow 1-S_{n+1}=\frac{1}{a_{n+2}-1}}$

If this is proven to be true then $\displaystyle {S_n=1-\frac{1}{a_{n+1}-1}<1}$

Base case: $\displaystyle {S_1=\frac{1}{3},a_2=\frac{5}{2}}$

$\displaystyle {1-\frac{1}{3}=\frac{1}{\frac{5}{2}-1}\Rightarrow\frac{2}{3}=\frac{1}{\frac{3}{2}}\Rightarrow\frac{2}{3}=\frac{2}{3}}$

Inductive Step:

$\displaystyle {1-S_n=\frac{1}{a_{n+1}-1}}$

$\displaystyle {1-S_n-\frac{1}{a_{n+1}+1}=\frac{1}{a_{n+1}-1}-\frac{1}{a_{n+1}+1}}$

$\displaystyle {1-(S_n+\frac{1}{a_{n+1}+1})=\frac{1}{a_{n+1}-1}-\frac{1}{a_{n+1}+1}}$

$\displaystyle {1-(S_{n+1})=\frac{a_{n+1}+1}{(a_{n+1}-1)(a_{n+1}+1)}-\frac{a_{n+1}-1}{(a_{n+1}-1)(a_{n+1}+1)}}$

$\displaystyle {1-S_{n+1}=\frac{2}{(a_{n+1}-1)(a_{n+1}+1)}}$

$\displaystyle {1-S_{n+1}=\frac{2}{a_{n+1}^2-1}}$

By definition $\displaystyle {a_{n+2}=\frac{a_{n+1}^2+1}{2}\Rightarrow a_{n+2}-1=\frac{a_{n+1}^2-1}{2}\Rightarrow \frac{1}{a_{n+2}-1}=\frac{2}{a_{n+1}^2-1}}$

$\displaystyle {1-S_{n+1}=\frac{2}{a_{n+1}^2-1}=\frac{1}{a_{n+2}-1}}$

Thus completing the proof.