Let $\{a_{n}\}$ be the sequence of real numbers defined by $a_{1}=3$ and for all $n\ge 1$, $$a_{n+1}=\dfrac{1}{2}(a^2_{n}+1)$$ Evaluate $$\sum_{k=1}^{\infty}\dfrac{1}{1+a_{k}}$$
My idea 1: since $$2a_{n+1}=a^2_{n}+1$$ so we have $$2(a_{n+1}-1)=(a_{n}+1)(a_{n}-1)\Longrightarrow \dfrac{1}{1+a_{n}}=2\cdot\dfrac{a_{n}-1}{a_{n+1}-1}$$ so we must find this sum $$\sum_{n=1}^{\infty}\dfrac{1}{1+a_{n}}=\sum_{n=1}^{\infty}\dfrac{2(a_{n}-1)}{a_{n+1}-1}$$ then I can't find this sum
other idea: maybe we can find this $a_{n}$ closed form? $$2a_{n+1}-1=(a_{n})^2$$ I want let $a_{n}=\cos^2{b_{n}}$,so $$\cos{2b_{n+1}}=(\cos{b_{n}})^4$$ then I can't follow works.
First, observe that $$ a_{n+1} -1= \frac{a_n^2-1}{2} = \frac{(a_n+1)(a_n-1)}{2}, $$ and consequently, (one can easily show that $a_n >1$ for all $n\ge 1$) $$ \frac{1}{a_{n+1}-1}=\frac{1}{a_n-1} - \frac{1}{a_n+1}. $$ Rewriting this as $$ \frac{1}{a_n+1} = \frac{1}{a_n-1}-\frac{1}{a_{n+1}-1} $$ Summing up this identity from $n=1$ to $N$ we get $$ \sum_{n=1}^N \frac{1}{a_n+1} = \frac{1}{a_1-1} - \frac{1}{a_{N+1}-1} $$ Finally, observe that $a_n \ge n$ for all $n\ge 1$ (since $\frac{n^2+1}{2}\ge n+1$ as soon as $n\ge 2$) and conclude that the sum is $\frac{1}{2}$.