通过将 $\frac{1}{(2r+1)(2r+3)}$ 写为部分分式，求出用 $n$ 表示 $\sum_{r=1}^{n} \frac{1}{(2r+1)(2r+3)}$ 的表达式。 解释为什么 $\sum_{r=1}^{n} \left[ \left(-\frac{1}{3}\right)^r + \frac{1}{(2r+1)(2r+3)} \right]$ 是一个收敛级数，并写出其无穷级数之和。

1. Express the given term in partial fractions.

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

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

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

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

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

2. Expand the sum to observe telescoping cancellation.

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

Expanding the terms:
$= \frac{1}{2} \left[ \left(\frac{1}{3} - \frac{1}{5}\right) + \left(\frac{1}{5} - \frac{1}{7}\right) + \dots + \left(\frac{1}{2n-1} - \frac{1}{2n+1}\right) + \left(\frac{1}{2n+1} - \frac{1}{2n+3}\right) \right]$

3. Cancel intermediate terms and simplify to a single fraction.

Most terms cancel out, leaving only the first and last parts:
$= \frac{1}{2} \left( \frac{1}{3} - \frac{1}{2n+3} \right)$

Combining into a single fraction:
$= \frac{1}{2} \left( \frac{2n+3-3}{3(2n+3)} \right)$
$= \frac{1}{2} \left( \frac{2n}{3(2n+3)} \right)$
$= \frac{n}{3(2n+3)}$

1. Concisely find the expression for the sum of the rational terms from part (a).

Let $\\frac{1}{(2r+1)(2r+3)} = \\frac{A}{2r+1} + \\frac{B}{2r+3}$.
$1 = A(2r+3) + B(2r+1) \\implies A = \\frac{1}{2}, B = -\\frac{1}{2}$
Thus, $\\sum_{r=1}^{n} \\frac{1}{(2r+1)(2r+3)} = \\frac{1}{2} \\sum_{r=1}^{n} \\left( \\frac{1}{2r+1} - \\frac{1}{2r+3} \\right)$
By the method of differences:
$= \\frac{1}{2} \\left[ \\left(\\frac{1}{3} - \\frac{1}{5}\\right) + \\left(\\frac{1}{5} - \\frac{1}{7}\\right) + \\dots + \\left(\\frac{1}{2n+1} - \\frac{1}{2n+3}\\right) \\right]$
$= \\frac{1}{2} \\left( \\frac{1}{3} - \\frac{1}{2n+3} \\right)$

2. Explain the convergence of both components of the series.

The overall series is the sum of two components.
The first component, $\\sum_{r=1}^{n} \\left(-\\frac{1}{3}\\right)^r$, is a geometric series with common ratio $r = -\\frac{1}{3}$. Since $|r| = \\left|-\\frac{1}{3}\\right| < 1$, it is a convergent series.
For the second component, as $n \\to \\infty$, the term $\\frac{1}{2n+3} \\to 0$. Therefore, its sum approaches a finite limit of $\\frac{1}{2}\\left(\\frac{1}{3} - 0\\right) = \\frac{1}{6}$, meaning it also converges.
Since both individual series converge to finite limits, their sum forms a convergent series.

3. Calculate the sum to infinity of the geometric series component.

For the geometric series, the first term is $a = -\\frac{1}{3}$ and the common ratio is $r = -\\frac{1}{3}$.
$S_{\\infty, \\text{GP}} = \\frac{a}{1 - r} = \\frac{-\\frac{1}{3}}{1 - \\left(-\\frac{1}{3}\\right)} = \\frac{-\\frac{1}{3}}{\\frac{4}{3}} = -\\frac{1}{4}$

4. Combine the limits to find the total sum to infinity.

The total sum to infinity is the sum of the infinite limits of both components:
$S_\infty = -\\frac{1}{4} + \\frac{1}{6} = -\\frac{3}{12} + \\frac{2}{12} = -\\frac{1}{12}$