Calculate floor of $\frac{1}{x_1 + 1} + \frac{1}{x_2 + 1} + ... + \frac{1}{x_{100} + 1}$

Question

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.

Answer 1

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.