A sequence of negative real numbers $x_1,x_2,x_3,\dots$ satisfies the relation $x_{n+1} = -\sqrt{1-2x_n},\text{ for } n\ge 1.$ Given that the sequence converges to $l$, find the exact value of $l$. By expressing $\dfrac{2}{r(r+2)}$ in the form $\dfrac{A}{r}+\dfrac{B}{r+2}$, where $A$ and $B$ are real constants to be determined, show that $\sum_{r=1}^n \dfrac{1}{r(r+2)} = \dfrac{3}{4} - \dfrac{2n+3}{2(n+1)(n+2)}.$ Prove the result in (i) using mathematical induction. Deduce the value of $\sum_{r=2}^{\infty} \dfrac{1}{r(r+2)}.$

1. Set up the equation for the limit by taking $n \to \infty$.

Since the sequence converges to $l$, as $n \to \infty$, both $x_n \to l$ and $x_{n+1} \to l$.
Substituting $l$ into the given recurrence relation:
$l = -\sqrt{1-2l}$

2. Transform the radical equation into a quadratic equation and solve for $l$.

Squaring both sides of the equation:
$l^2 = 1 - 2l$
$l^2 + 2l - 1 = 0$
Using the quadratic formula:
$l = \frac{-2 \pm \sqrt{2^2 - 4(1)(-1)}}{2}$
$l = \frac{-2 \pm \sqrt{8}}{2}$
$l = -1 \pm \sqrt{2}$

3. Reject the invalid root to find the exact final answer.

Since $x_n$ is a sequence of negative real numbers, its limit must be non-positive ($l \le 0$). Moreover, from the equation $l = -\sqrt{1-2l}$, it is clear that $l$ must be strictly less than or equal to $0$.
Since $-1 + \sqrt{2} \approx 0.414$ is positive, we reject it.
Thus, the exact value of $l$ is:
$l = -1 - \sqrt{2}$

1. Find the constants $A$ and $B$ using partial fractions.

$\frac{2}{r(r+2)} = \frac{A}{r} + \frac{B}{r+2}$
$2 = A(r+2) + Br$
Substituting $r = 0$, we get $2 = 2A \implies A = 1$.
Substituting $r = -2$, we get $2 = -2B \implies B = -1$.
So, $\frac{2}{r(r+2)} = \frac{1}{r} - \frac{1}{r+2}$.

2. Set up the summation expression using the partial fractions.

$\sum_{r=1}^n \frac{1}{r(r+2)} = \frac{1}{2} \sum_{r=1}^n \frac{2}{r(r+2)}$
$= \frac{1}{2} \sum_{r=1}^n \left( \frac{1}{r} - \frac{1}{r+2} \right)$

3. Expand the sum to apply the method of differences.

Listing the terms to observe the telescoping cancellations:
$\sum_{r=1}^n \left( \frac{1}{r} - \frac{1}{r+2} \right) = \left( \frac{1}{1} - \frac{1}{3} \right) + \left( \frac{1}{2} - \frac{1}{4} \right) + \left( \frac{1}{3} - \frac{1}{5} \right) + \dots$
$\dots + \left( \frac{1}{n-1} - \frac{1}{n+1} \right) + \left( \frac{1}{n} - \frac{1}{n+2} \right)$
All terms cancel except the first two positive terms and the last two negative terms:
$= 1 + \frac{1}{2} - \frac{1}{n+1} - \frac{1}{n+2}$

4. Simplify the remaining terms to obtain the required expression.

$= \frac{3}{2} - \left( \frac{(n+2) + (n+1)}{(n+1)(n+2)} \right)$
$= \frac{3}{2} - \frac{2n+3}{(n+1)(n+2)}$
Multiplying by the $\frac{1}{2}$ from earlier:
$\sum_{r=1}^n \frac{1}{r(r+2)} = \frac{1}{2} \left[ \frac{3}{2} - \frac{2n+3}{(n+1)(n+2)} \right]$
$= \frac{3}{4} - \frac{2n+3}{2(n+1)(n+2)} \quad \text{(shown)}$

1. State the proposition and verify the base case for $n=1$.

Let $P_n$ be the proposition that $\sum_{r=1}^n \frac{1}{r(r+2)} = \frac{3}{4} - \frac{2n+3}{2(n+1)(n+2)}$ for all $n \in \mathbb{Z}^+$.
When $n=1$,
$\text{LHS} = \sum_{r=1}^1 \frac{1}{r(r+2)} = \frac{1}{1(1+2)} = \frac{1}{3}$
$\text{RHS} = \frac{3}{4} - \frac{2(1)+3}{2(1+1)(1+2)} = \frac{3}{4} - \frac{5}{12} = \frac{9-5}{12} = \frac{1}{3}$
Since $\text{LHS} = \text{RHS}$, $P_1$ is true.

2. State the induction hypothesis.

Assume $P_k$ is true for some $k \in \mathbb{Z}^+$, that is:
$\sum_{r=1}^k \frac{1}{r(r+2)} = \frac{3}{4} - \frac{2k+3}{2(k+1)(k+2)}$

3. Express the sum for $n=k+1$ using the sum up to $k$.

We want to show that $P_{k+1}$ is true. Consider the sum for $n=k+1$:
$\sum_{r=1}^{k+1} \frac{1}{r(r+2)} = \sum_{r=1}^k \frac{1}{r(r+2)} + \frac{1}{(k+1)(k+3)}$

4. Substitute the induction hypothesis and simplify the expression.

$= \frac{3}{4} - \frac{2k+3}{2(k+1)(k+2)} + \frac{1}{(k+1)(k+3)}$
$= \frac{3}{4} - \left[ \frac{(2k+3)(k+3) - 2(k+2)}{2(k+1)(k+2)(k+3)} \right]$
$= \frac{3}{4} - \frac{2k^2 + 9k + 9 - 2k - 4}{2(k+1)(k+2)(k+3)}$
$= \frac{3}{4} - \frac{2k^2 + 7k + 5}{2(k+1)(k+2)(k+3)}$
$= \frac{3}{4} - \frac{(2k+5)(k+1)}{2(k+1)(k+2)(k+3)}$
$= \frac{3}{4} - \frac{2k+5}{2(k+2)(k+3)}$

5. Demonstrate the expression matches the required form for $k+1$ and conclude the proof.

Observe that the simplified expression can be rewritten as:
$= \frac{3}{4} - \frac{2(k+1)+3}{2((k+1)+1)((k+1)+2)}$
This is exactly $P_{k+1}$.
Since $P_1$ is true, and if $P_k$ is true then $P_{k+1}$ is true, by the Principle of Mathematical Induction, $P_n$ is true for all $n \in \mathbb{Z}^+$.

1. Evaluate the limit of the sum to $n$ terms to find the sum to infinity from $r=1$.

$\sum_{r=1}^{\infty} \frac{1}{r(r+2)} = \lim_{n \to \infty} \left( \frac{3}{4} - \frac{2n+3}{2(n+1)(n+2)} \right)$
As $n \to \infty$, $\dfrac{2n+3}{2(n+1)(n+2)} \to 0$, so:
$\sum_{r=1}^{\infty} \frac{1}{r(r+2)} = \frac{3}{4} - 0 = \frac{3}{4}$

2. Subtract the $r=1$ term to deduce the sum starting from $r=2$.

$\sum_{r=2}^{\infty} \frac{1}{r(r+2)} = \sum_{r=1}^{\infty} \frac{1}{r(r+2)} - \text{term where } r=1$
$= \frac{3}{4} - \frac{1}{1(1+2)}$
$= \frac{3}{4} - \frac{1}{3} = \frac{5}{12}$