证明 $\frac{3}{2r-1} - \frac{4}{2r+1} + \frac{1}{2r+3} = \frac{8r+20}{(2r-1)(2r+1)(2r+3)}$。 由此证明 $\sum_{r=1}^{n} \frac{2r+5}{(2r-1)(2r+1)(2r+3)} = \frac{2}{3} - \frac{n+2}{(2n+1)(2n+3)}$。 求最小的可能整数 $k$，使得 $\sum_{r=1}^{k} \frac{2r+5}{(2r-1)(2r+1)(2r+3)}$ 与无穷级数之和的差在 $0.004$ 以内。

1. Combine the fractions over a common denominator.

$\frac{3}{2r-1} - \frac{4}{2r+1} + \frac{1}{2r+3}$
$= \frac{3(2r+1)(2r+3) - 4(2r-1)(2r+3) + (2r-1)(2r+1)}{(2r-1)(2r+1)(2r+3)}$

2. Expand the terms in the numerator.

Numerator $= 3(4r^2 + 8r + 3) - 4(4r^2 + 4r - 3) + (4r^2 - 1)$
$= (12r^2 + 24r + 9) - (16r^2 + 16r - 12) + (4r^2 - 1)$

3. Simplify the numerator to obtain the final expression.

$= (12 - 16 + 4)r^2 + (24 - 16)r + (9 + 12 - 1)$
$= 8r + 20$

Hence, the expression simplifies to:
$\frac{8r+20}{(2r-1)(2r+1)(2r+3)}$

1. Relate the summand to the identity given in part (a).

$\frac{2r+5}{(2r-1)(2r+1)(2r+3)} = \frac{1}{4} \left( \frac{8r+20}{(2r-1)(2r+1)(2r+3)} \right)$
Using the result from (a),
$= \frac{1}{4} \left( \frac{3}{2r-1} - \frac{4}{2r+1} + \frac{1}{2r+3} \right)$

2. List the terms of the sum to identify the cancellation pattern.

$\begin{aligned} &\sum_{r=1}^{n} \frac{2r+5}{(2r-1)(2r+1)(2r+3)} \\ &= \frac{1}{4} \sum_{r=1}^{n} \left( \frac{3}{2r-1} - \frac{4}{2r+1} + \frac{1}{2r+3} \right) \\ &= \frac{1}{4} \left[ \begin{array}{l} \left( \frac{3}{1} - \frac{4}{3} + \frac{1}{5} \right) \\ + \left( \frac{3}{3} - \frac{4}{5} + \frac{1}{7} \right) \\ + \left( \frac{3}{5} - \frac{4}{7} + \frac{1}{9} \right) \\ + \dots \\ + \left( \frac{3}{2n-5} - \frac{4}{2n-3} + \frac{1}{2n-1} \right) \\ + \left( \frac{3}{2n-3} - \frac{4}{2n-1} + \frac{1}{2n+1} \right) \\ + \left( \frac{3}{2n-1} - \frac{4}{2n+1} + \frac{1}{2n+3} \right) \end{array} \right] \end{aligned}$

3. Cancel intermediate terms across the columns and sum the remaining terms.

Observe that terms offset by two rows cancel completely (e.g., $\frac{1}{5} - \frac{4}{5} + \frac{3}{5} = 0$). The surviving uncancelled terms are found at the very beginning and the very end of the series:
$\begin{aligned} &= \frac{1}{4} \left[ 3 - \frac{4}{3} + \frac{3}{3} + \frac{1}{2n+1} - \frac{4}{2n+1} + \frac{1}{2n+3} \right] \\ &= \frac{1}{4} \left[ \frac{8}{3} - \frac{3}{2n+1} + \frac{1}{2n+3} \right] \end{aligned}$

4. Simplify the remaining fractions to reach the final required expression.

$\begin{aligned} &= \frac{1}{4} \left[ \frac{8}{3} - \left( \frac{3(2n+3) - 1(2n+1)}{(2n+1)(2n+3)} \right) \right] \\ &= \frac{1}{4} \left[ \frac{8}{3} - \frac{6n+9 - 2n - 1}{(2n+1)(2n+3)} \right] \\ &= \frac{1}{4} \left[ \frac{8}{3} - \frac{4n+8}{(2n+1)(2n+3)} \right] \\ &= \frac{1}{4} \left[ \frac{8}{3} - \frac{4(n+2)}{(2n+1)(2n+3)} \right] \\ &= \frac{2}{3} - \frac{n+2}{(2n+1)(2n+3)} \quad \text{(Shown)} \end{aligned}$

1. Find the sum to infinity by taking the limit as $n \to \infty$.

$S_\infty = \lim_{n \to \infty} \left( \frac{2}{3} - \frac{n+2}{(2n+1)(2n+3)} \right)$
As $n \to \infty$, the fraction $\frac{n+2}{(2n+1)(2n+3)} \to 0$.
$S_\infty = \frac{2}{3}$

2. Set up an inequality for the difference between the sum to infinity and the sum to $k$ terms.

We require $|S_\infty - S_k| \le 0.004$.
$\left| \frac{2}{3} - \left( \frac{2}{3} - \frac{k+2}{(2k+1)(2k+3)} \right) \right| \le 0.004$
$\left| \frac{k+2}{(2k+1)(2k+3)} \right| \le 0.004$
Since $k \ge 1$, all terms in the fraction are positive, so we can remove the modulus sign:
$\frac{k+2}{(2k+1)(2k+3)} \le \frac{1}{250}$

3. Solve the inequality to find the smallest integer $k$.

Since the denominator $(2k+1)(2k+3) = 4k^2+8k+3$ is strictly positive for $k \ge 1$, we can cross-multiply:
$250(k+2) \le 4k^2+8k+3$
$250k + 500 \le 4k^2+8k+3$
$4k^2 - 242k - 497 \ge 0$
Solving the quadratic equation $4k^2 - 242k - 497 = 0$ using the quadratic formula:
$k = \frac{242 \pm \sqrt{(-242)^2 - 4(4)(-497)}}{2(4)}$
$k \approx -1.988 \text{ or } k \approx 62.488$
The inequality holds for $k \le -1.988$ or $k \ge 62.488$.
Since $k$ is a positive integer, $k \ge 63$.
Therefore, the smallest possible integer $k$ is $63$.