# How to find Find $\left\lfloor\sum_1^{100}\frac{1}{x_n+1}\right\rfloor$ with $x_1 =\frac{1}{2}, x_{k+1} =x_k^2+x_k$.

The sequence $\{x_n\}$ is defined by $x_1 =\frac{1}{2}, x_{k+1} =x_k^2+x_k$. Find $$\left\lfloor\frac{1}{x_1+1}+\frac{1}{x_2+1}+\cdots+\frac{1}{x_{100}+1}\right\rfloor$$ where $\left\lfloor\dots\right\rfloor$ is greatest integer function.

If we put the values of $k$ then we get the numerator part of the series =2 and then taking 2 as common from the series how can we solve the rest of the series..

How do we proceed in this case ... please suggest thanks..

## 1 Answer

Hint The fact that $$x_{k+1} =x_k^2+x_k$$ implies that $$\frac{{{x_k}}}{{{x_{k + 1}}}} = \frac{1}{{{x_k} + 1}}$$

But note that by the recursion yet again we obtain $$\frac{{{x_k}}}{{{x_{k + 1}}}} = \frac{{x_k^2}}{{{x_{k + 1}}{x_k}}} = \frac{{x_k^2 + {x_k} - {x_k}}}{{{x_{k + 1}}{x_k}}} = \frac{{{x_{k + 1}} - {x_k}}}{{{x_{k + 1}}{x_k}}} = \frac{1}{{{x_k}}} - \frac{1}{{{x_{k + 1}}}}$$

• Nice. What motivated you finding this telescoping? Was it from trying small values, or based on experience? – Calvin Lin Aug 8 '13 at 15:31
• @CalvinLin I was trying something more far-fetched and bumped into the telescopy. – Pedro Tamaroff Aug 8 '13 at 15:34