Olympiad Problem 4

We have : $S=\sum _{ r=1 }^{ \infty }{ \frac { 1 }{ { x }_{ r }+1 } }$

We will add and subtract $\frac { 1 }{ { x }_{ 1 } }$ to S. So we will write :

S= $(\frac { 1 }{ { x }_{ 1 } } )+(-\frac { 1 }{ { x }_{ 1 } } +\frac { 1 }{ { x }_{ 1 }+1 } +........)$ Step 1

$=\frac { 1 }{ { x }_{ 1 } } +(\frac { -1 }{ { x }_{ 1 }(1+{ x }_{ 1 }) } +\frac { 1 }{ { (x }_{ 2 }+1) } +.......)$ Step 2

Using ${x}_{k+1}={x}_{k}^{2}+{x}_{k}$ we get :

$S=\frac { 1 }{ { x }_{ 1 } } +(\frac { -1 }{ { x }_{ 2 } } +\frac { 1 }{ { (x }_{ 2 }+1) } +.......)$ Step 3

Repeating the same pattern we get

$S=\frac { 1 }{ { x }_{ 1 } } +(\frac { -1 }{ { x }_{ 3 } } +\frac { 1 }{ { (x }_{ 3 }+1) } +.......)$ Step 4

Repeating it till infinity we get :

$S=\frac { 1 }{ { x }_{ 1 } } -\lim _{ n\rightarrow \infty }{ \frac { 1 }{ { x }_{ n } } } =\frac { 1 }{ { x }_{ 1 } }$ Step 5

$\Rightarrow S=\frac { 1 }{ { x }_{ 1 } }$ Step 6

Put the value of ${x}_{1}$ to get :

$S=2$ Hence it is clear that the sum of finite terms of this series is less than 2. Hence the integral part of the series is 1.

$x_{k+1} = x_{k}(x_{k}+1)$

$\Rightarrow \dfrac{1}{x_{k}+1} = \dfrac{x_{k}}{x_{k+1}} = \dfrac{x_{k}^2}{x_{k}x_{k+1}}$

= $\dfrac{x_{k+1} - x_{k}}{x_{k}x_{k+1} }= \dfrac{1}{x_{k}} - \dfrac{1}{x_{k+1}}$

$\displaystyle \sum_{k=1}^{n} \dfrac{1}{x_{k}+1}$

= $\displaystyle \sum_{k=1}^{n} \bigg( \dfrac{1}{x_{k}} - \dfrac{1}{x_{k+1}} \bigg)$

= $\dfrac{1}{x_{1}} - \dfrac{1}{x_{n}}$

Clearly, as the sequence goes to $\infty$ at $k = \infty$ ,

S = $\displaystyle \sum_{k=1}^{n} \dfrac{1}{x_{k}+1} \leq \displaystyle \sum_{k=1}^{\infty} \dfrac{1}{x_{k}+1} = 2 - 0 = 2$

Hence the answer is 2.

Note : This is an increasing sequence, and $x_{3} = 9/16 + 3/4 > 1$ , hence, $x_{100} > 1, \Rightarrow \frac{1}{x_{100}} < 1$ ,and the sum is greater than 1.