负实数数列 $x_1, x_2, x_3, \dots$ 满足关系式 $x_{n+1} = -\sqrt{1-2x_n}, \text{ 对于 } n \ge 1.$ 已知该数列收敛于 $l$，求 $l$ 的精确值。 通过将 $\dfrac{2}{r(r+2)}$ 表示为 $\dfrac{A}{r}+\dfrac{B}{r+2}$ 的形式（其中 $A$ 和 $B$ 是待测定的实常数），证明 $\sum_{r=1}^n \dfrac{1}{r(r+2)} = \dfrac{3}{4} - \dfrac{2n+3}{2(n+1)(n+2)}.$ 使用数学归纳法证明 (i) 中的结论。 推导 $\sum_{r=2}^{\infty} \dfrac{1}{r(r+2)}$ 的值。

1. Set up the equation for the limit $l$.

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 recurrence relation gives:
$l = -\sqrt{1-2l}$

2. Form and solve the resulting quadratic equation.

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

3. Select the correct valid root.

Since the terms $x_n$ are negative real numbers, the limit $l$ must be less than or equal to 0. (Furthermore, the definition $l = -\sqrt{1-2l}$ mandates $l \le 0$).
Because $-1 + \sqrt{2} \approx 0.414 > 0$, we reject this root.
Therefore, $l = -1 - \sqrt{2}$.

1. Find the values of constants $A$ and $B$ for the partial fraction decomposition.

$\dfrac{2}{r(r+2)} = \dfrac{A}{r} + \dfrac{B}{r+2}$

Multiplying by $r(r+2)$:
$2 = A(r+2) + Br$

Let $r=0$:
$2 = 2A \implies A=1$

Let $r=-2$:
$2 = -2B \implies B=-1$

So, $\dfrac{2}{r(r+2)} = \dfrac{1}{r} - \dfrac{1}{r+2}$

2. Express the required summation using the partial fractions form.

Since $\dfrac{1}{r(r+2)} = \dfrac{1}{2} \left( \dfrac{1}{r} - \dfrac{1}{r+2} \right)$,

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

3. Expand the series to reveal the telescoping pattern and cancel terms.

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

$= \left( \dfrac{1}{1} - \dfrac{1}{3} \right)$
$+ \left( \dfrac{1}{2} - \dfrac{1}{4} \right)$
$+ \left( \dfrac{1}{3} - \dfrac{1}{5} \right)$
$+ \dots$
$+ \left( \dfrac{1}{n-1} - \dfrac{1}{n+1} \right)$
$+ \left( \dfrac{1}{n} - \dfrac{1}{n+2} \right)$

Notice that the terms cancel each other out diagonally (method of differences). The remaining uncancelled terms are the first two positive parts and the last two negative parts:

$= 1 + \dfrac{1}{2} - \dfrac{1}{n+1} - \dfrac{1}{n+2}$

4. Combine the remaining terms and multiply by $\frac{1}{2}$ to show the required result.

$= \dfrac{3}{2} - \left( \dfrac{(n+2) + (n+1)}{(n+1)(n+2)} \right)$
$= \dfrac{3}{2} - \dfrac{2n+3}{(n+1)(n+2)}$

Returning to the main expression:
$\sum_{r=1}^n \dfrac{1}{r(r+2)} = \dfrac{1}{2} \left( \dfrac{3}{2} - \dfrac{2n+3}{(n+1)(n+2)} \right)$

$= \dfrac{3}{4} - \dfrac{2n+3}{2(n+1)(n+2)}$ (shown)

1. Establish the base case for $n=1$.

Let $P_n$ be the statement: $\sum_{r=1}^n \frac{1}{r(r+2)} = \frac{3}{4} - \frac{2n+3}{2(n+1)(n+2)}$.
When $n=1$,
$\text{LHS} = \sum_{r=1}^1 \frac{1}{r(r+2)} = \frac{1}{1(1+2)} = \frac{1}{3}$
$\text{RHS} = \frac{3}{4} - \frac{2(1)+3}{2(1+1)(1+2)} = \frac{3}{4} - \frac{5}{12} = \frac{4}{12} = \frac{1}{3}$
Since $\text{LHS} = \text{RHS}$, $P_1$ is true.

2. State the inductive hypothesis and set up the expression for $n=k+1$.

Assume $P_k$ is true for some positive integer $k$. That is,
$\sum_{r=1}^k \frac{1}{r(r+2)} = \frac{3}{4} - \frac{2k+3}{2(k+1)(k+2)}$
For $n = k+1$,
$\sum_{r=1}^{k+1} \frac{1}{r(r+2)} = \sum_{r=1}^k \frac{1}{r(r+2)} + \frac{1}{(k+1)(k+3)}$

3. Perform algebraic manipulation to show that $P_{k+1}$ matches the required formula.

$\sum_{r=1}^{k+1} \frac{1}{r(r+2)} = \left[ \frac{3}{4} - \frac{2k+3}{2(k+1)(k+2)} \right] + \frac{1}{(k+1)(k+3)}$
$= \frac{3}{4} - \left[ \frac{2k+3}{2(k+1)(k+2)} - \frac{1}{(k+1)(k+3)} \right]$
$= \frac{3}{4} - \left[ \frac{(2k+3)(k+3) - 2(k+2)}{2(k+1)(k+2)(k+3)} \right]$
$= \frac{3}{4} - \left[ \frac{2k^2 + 9k + 9 - 2k - 4}{2(k+1)(k+2)(k+3)} \right]$
$= \frac{3}{4} - \frac{2k^2 + 7k + 5}{2(k+1)(k+2)(k+3)}$
$= \frac{3}{4} - \frac{(2k+5)(k+1)}{2(k+1)(k+2)(k+3)}$
$= \frac{3}{4} - \frac{2k+5}{2(k+2)(k+3)}$
$= \frac{3}{4} - \frac{2(k+1)+3}{2((k+1)+1)((k+1)+2)}$
Hence, $P_{k+1}$ is true if $P_k$ is true.

4. Conclude the proof using the Principle of Mathematical Induction.

Since $P_1$ is true, and $P_k$ being true implies $P_{k+1}$ is true, by the Principle of Mathematical Induction, $P_n$ is true for all $n \in \mathbb{Z}^+$.

1. Find the infinite sum starting from $r=1$ by taking the limit of $S_n$ as $n \to \infty$.

$\sum_{r=1}^{\infty} \frac{1}{r(r+2)} = \lim_{n \to \infty} \left( \frac{3}{4} - \frac{2n+3}{2(n+1)(n+2)} \right)$
As $n \to \infty$, the denominator of the second term is a quadratic in $n$ while the numerator is linear, so $\frac{2n+3}{2(n+1)(n+2)} \to 0$.
$\sum_{r=1}^{\infty} \frac{1}{r(r+2)} = \frac{3}{4}$

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

$\sum_{r=2}^{\infty} \frac{1}{r(r+2)} = \sum_{r=1}^{\infty} \frac{1}{r(r+2)} - \frac{1}{1(1+2)}$
$= \frac{3}{4} - \frac{1}{3}$
$= \frac{9}{12} - \frac{4}{12}$
$= \frac{5}{12}$