<s>[INST] 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 $ 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.

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.
 [/INST] 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.
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