通过将 $\frac{2}{(2r+1)(2r+3)}$ 化为部分分式，求出用 $n$ 表示 $\sum_{r=2}^{n} \frac{2}{(2r+1)(2r+3)}$ 的表达式。 由此求出 $\frac{2}{19} \times \frac{1}{21} + \frac{2}{21} \times \frac{1}{23} + \frac{2}{23} \times \frac{1}{25} + \dots$ 的精确值。

1. Express the given algebraic fraction in partial fractions.

Let $\frac{2}{(2r+1)(2r+3)} = \frac{A}{2r+1} + \frac{B}{2r+3}$

$2 = A(2r+3) + B(2r+1)$

Substitute $r = -\frac{1}{2}$:
$2 = A(2(-\frac{1}{2})+3) \implies 2 = 2A \implies A = 1$

Substitute $r = -\frac{3}{2}$:
$2 = B(2(-\frac{3}{2})+1) \implies 2 = -2B \implies B = -1$

$\therefore \frac{2}{(2r+1)(2r+3)} = \frac{1}{2r+1} - \frac{1}{2r+3}$

2. List the terms of the sum to identify the cancellation pattern (method of differences).

$\sum_{r=2}^{n} \frac{2}{(2r+1)(2r+3)} = \sum_{r=2}^{n} \left( \frac{1}{2r+1} - \frac{1}{2r+3} \right)$

Writing out the first few and last few terms:
$= \left( \frac{1}{5} - \frac{1}{7} \right)$
$+ \left( \frac{1}{7} - \frac{1}{9} \right)$
$+ \dots$
$+ \left( \frac{1}{2n-1} - \frac{1}{2n+1} \right)$
$+ \left( \frac{1}{2n+1} - \frac{1}{2n+3} \right)$

3. Cancel the intermediate terms to find the simplified expression.

Notice that $-\frac{1}{7}$ cancels with $+\frac{1}{7}$, $-\frac{1}{9}$ cancels with the next term, and so on. This leaves only the first and last terms:

$= \frac{1}{5} - \frac{1}{2n+3}$

(Note: This can also be written as a single fraction $\frac{2n-2}{5(2n+3)}$)

1. Establish the general expression for the sum from part (i) using the method of differences.

From part (i), resolving the general term into partial fractions:
$\frac{2}{(2r+1)(2r+3)} = \frac{1}{2r+1} - \frac{1}{2r+3}$
Using the method of differences:
$\sum_{r=2}^{n} \frac{2}{(2r+1)(2r+3)} = \left( \frac{1}{5} - \frac{1}{7} \right) + \left( \frac{1}{7} - \frac{1}{9} \right) + \dots + \left( \frac{1}{2n+1} - \frac{1}{2n+3} \right)$
$= \frac{1}{5} - \frac{1}{2n+3}$

2. Express the given infinite series in sigma notation and relate it to the sum from part (i).

The given series can be written as:
$\frac{2}{19 \times 21} + \frac{2}{21 \times 23} + \dots = \sum_{r=9}^{\infty} \frac{2}{(2r+1)(2r+3)}$
Relating this to the sum from part (i), we subtract the sum of the first 7 terms (from $r=2$ to $r=8$):
$\sum_{r=9}^{\infty} \frac{2}{(2r+1)(2r+3)} = \lim_{n \to \infty} \left( \sum_{r=2}^{n} \frac{2}{(2r+1)(2r+3)} - \sum_{r=2}^{8} \frac{2}{(2r+1)(2r+3)} \right)$

3. Substitute the algebraic expression from part (i) and evaluate the limit to infinity.

$= \lim_{n \to \infty} \left( \left[ \frac{1}{5} - \frac{1}{2n+3} \right] - \left[ \frac{1}{5} - \frac{1}{2(8)+3} \right] \right)$
$= \lim_{n \to \infty} \left( \frac{1}{19} - \frac{1}{2n+3} \right)$
As $n \to \infty$, the term $\frac{1}{2n+3} \to 0$.
$\therefore \text{Exact value} = \frac{1}{19}$