Define a sequence {$a_n$} of real numbers by $a_1 = 2$ and $a_{n+1} = \dfrac{a_n^2+1}{2}$, for $n\ge 1$,
Prove that for every natural number $N, \sum_{j=1}^{N} \frac{1}{1+a_j} \lt 1$
I tried mathematical induction after coming to the step where
$Sum_n = \frac{1}{2}\left(\frac{a_1-1}{a_2-1} + \frac{a_2-1}{a_3-1} +\cdots \frac{a_n-1}{a_{n+1}-1}\right)$.
Having gotten this how would go about proving it using induction?
Best Answer
To get $\dfrac{1}{a_n+1}$, Firse subtract $1$ from both sides $$a_{n+1}-1=\frac{a_{n}^2-1}{2}=\frac{(a_n+1)(a_n-1)}{2}.$$ Then take the multiplicative inverse ( assume $a_n\neq \pm1$, as we'll see later) $$\frac{1}{a_{n+1}-1}=\frac{2}{(a_n+1)(a_n-1)}$$ where the right side can be written as $$\frac{1}{a_n-1}-\frac{1}{a_n+1}.$$ Thus $$\frac{1}{a_n+1} = \frac{1}{a_n-1}-\frac{1}{a_{n+1}-1}$$ Sum up both sides from $1$ to $N$ we get $$\sum_{k=1}^{N}\frac{1}{a_k+1}=\frac{1}{a_1-1}-\frac{1}{a_{N+1}-1}=1-\frac{1}{a_{N+1}-1}.$$ Now we just need to prove $a_n>1$ for all $n\geqslant2$. Notice that $$a_{n+1}-a_{n}=\frac{a_{n}^2+1}{2}-a_n=\frac{(a_n-1)^2}{2}>0.$$ Then $\{a_n\}$ is incresing with $a_1=2$. Hence the proof is done.