Proving a recursion is less than two

Question

The sequence $(a_n)$ is defined by $a_1 = \frac{1}{2}$ and $a_n = a_{n - 1}^2 + a_{n - 1}$ for $n \ge 2.$

Prove that $\frac{1}{a_1 + 1} + \frac{1}{a_2 + 1} + \dots + \frac{1}{a_n + 1} < 2$ for all $n \ge 1.$

So far, I have manipulated the given recursion into $a_n = a_{n - 1} (a_{n - 1} + 1)$. But now I am stuck.

Answer 1

After all we have by @hdighfan solution: $$\frac{1}{a_n~+1}=\frac{1}{a_n}-\frac{1}{a_{n+1}}$$ By telescopic summing we have $$S_n=\sum_{k=1}^n \frac{1}{a_k~+1}=\frac{1}{a_1}-\frac{1}{a_{n+1}}<2,$$ as $a_1=1/2$.

Answer 2

We claim a stronger result; namely that $\frac{1}{a_1+1} + \cdots + \frac{1}{a_{n-1}+1} + \frac{1}{a_n}<2$, for all $n$. This obviously holds for small $n$; you can check this yourself.

Observe that $$\frac{1}{a_n} = \frac{1}{(a_{n+1}+1)a_{n+1}} = \frac{1}{a_n+1} + \frac{1}{a_{n+1}}.$$

The claim follows by induction.