Sum of given telescopic series

Question

A sequence is defined as follows: $a_1=\frac{1}{2}$ and $a_{n+1}=a_n^2+a_n$

If $$S=\frac{1}{a_1+1}+\frac{1}{a_2+1}+\frac{1}{a_3+1}+.....+\frac{1}{a_{100}+1}$$

Find $\lfloor S \rfloor$.

Now using $a_{n+1}=a_n^2+a_n$, I generated telescopic series to get $S=2-\frac{1}{a_{101}}$ but I am not able to find $\frac{1}{a_{101}}$

Could someone help me with this?

Answer 1

You don't need actually ${1\over a_{101}}$. Since $a_{n+1}-a_n = a_n^2>0$ and $a_3>1$ we have $$0<{1\over a_{101}}<1$$ so

$$ 1 <2-{1\over a_{101}}<2 \Longrightarrow \lfloor S \rfloor = 1$$