Calculate floor of $\frac{1}{x_1 + 1} + \frac{1}{x_2 + 1} + ... + \frac{1}{x_{100} + 1}$ There is a recurrence sequence $x_1 = \frac{1}{2}$, $x_{n+1} = x_n^2 + x_n$. How much is floor of $\frac{1}{x_1 + 1} + \frac{1}{x_2 + 1} + ... + \frac{1}{x_{100} + 1}$? Floor is an integer part of a real number.
 A: The instructive hints of @AchilleHui deserve an answer by its own.

Given the recurrence relation 
  \begin{align*}
x_{n+1}&=x_n(x_n+1)\tag{1}\\
x_1&=\frac{1}{2}\\
\end{align*}
we derive from (1) a telescoping representation of the reciprocal values via
  \begin{align*}
\frac{1}{x_{n+1}}&=\frac{1}{x_n(x_n+1)}
=\frac{1}{x_n}-\frac{1}{x_n+1}
\end{align*}
We obtain
  \begin{align*}
\sum_{j=1}^{100}\frac{1}{x_j+1}&=\sum_{j=1}^{100}\left(\frac{1}{x_j}-\frac{1}{x_{j+1}}\right)\\
&=\sum_{j=1}^{100}\frac{1}{x_j}-\sum_{j=2}^{101}\frac{1}{x_j}\tag{2}\\
&=\frac{1}{x_1}-\frac{1}{x_{101}}\\
&=2-\frac{1}{x_{101}}\tag{2}
\end{align*}
From $x_1=\frac{1}{2}$ and (1) we get $x_2=\frac{3}{4}, x_3=\frac{21}{16}>1$ and it follows again from (1) $x_n>1$ with $n\geq 3$.
We finally get from (2) 
  \begin{align*}
\color{blue}{\left\lfloor 2-\frac{1}{x_{101}} \right\rfloor = 1}
\end{align*} 

Note: Since $x_1=\frac{1}{2}$ and $x_{n+1}\approx x_n^2$ we have $x_{101}\approx \frac{1}{2^{2^{100}}}$. So, the difference of $2-\frac{1}{x_{101}}$ to the value $2$ is very small. 
