证明 $\frac{3}{2r+1}-\frac{4}{2r+3}+\frac{1}{2r+5}=\frac{8r+k}{(2r+1)(2r+3)(2r+5)}$，其中 $k$ 是待求常数。 由此求 $\sum_{r=0}^n \frac{2r+7}{(2r+1)(2r+3)(2r+5)}$。 （不需要将答案表示为单一的代数分数。） 解释为什么 $\displaystyle\sum_{r=0}^{\infty} \frac{2r+7}{(2r+1)(2r+3)(2r+5)}$ 是一个收敛级数，并写出该无穷级数的和。

1. Combine the algebraic fractions over a single common denominator

$\frac{3}{2r+1} - \frac{4}{2r+3} + \frac{1}{2r+5} = \frac{3(2r+3)(2r+5) - 4(2r+1)(2r+5) + (2r+1)(2r+3)}{(2r+1)(2r+3)(2r+5)}$

2. Expand and simplify the numerator to deduce the value of $k$

Comparing with the given expression $\frac{8r+k}{(2r+1)(2r+3)(2r+5)}$, we find $k = 28$.

1. Find the constant $k$ from part (i) to establish the relationship between the required sum and the given partial fractions.

$\frac{3}{2r+1} - \frac{4}{2r+3} + \frac{1}{2r+5} = \frac{3(2r+3)(2r+5) - 4(2r+1)(2r+5) + (2r+1)(2r+3)}{(2r+1)(2r+3)(2r+5)}$
$= \frac{12r^2+48r+45 - (16r^2+48r+20) + 4r^2+8r+3}{(2r+1)(2r+3)(2r+5)}$
$= \frac{8r+28}{(2r+1)(2r+3)(2r+5)} = \frac{4(2r+7)}{(2r+1)(2r+3)(2r+5)}$
Thus, $k = 28$ and we can rewrite the summand as:
$\frac{2r+7}{(2r+1)(2r+3)(2r+5)} = \frac{1}{4} \left( \frac{3}{2r+1} - \frac{4}{2r+3} + \frac{1}{2r+5} \right)$

2. Expand the sum to identify the cancelling terms using the Method of Differences.

$\sum_{r=0}^n \frac{2r+7}{(2r+1)(2r+3)(2r+5)} = \frac{1}{4} \sum_{r=0}^n \left( \frac{3}{2r+1} - \frac{4}{2r+3} + \frac{1}{2r+5} \right)$
$= \frac{1}{4} \left[ \begin{array}{l} \left( \frac{3}{1} - \frac{4}{3} + \frac{1}{5} \right) \\ + \left( \frac{3}{3} - \frac{4}{5} + \frac{1}{7} \right) \\ + \left( \frac{3}{5} - \frac{4}{7} + \frac{1}{9} \right) \\ + \dots \\ + \left( \frac{3}{2n-1} - \frac{4}{2n+1} + \frac{1}{2n+3} \right) \\ + \left( \frac{3}{2n+1} - \frac{4}{2n+3} + \frac{1}{2n+5} \right) \end{array} \right]$

3. Simplify the remaining uncancelled terms to find the final expression for the sum.

Summing the terms by columns, all intermediate terms cancel out.
The uncancelled terms at the start are:
$\frac{3}{1} - \frac{4}{3} + \frac{3}{3} = 3 - \frac{1}{3} = \frac{8}{3}$
The uncancelled terms at the end are:
$\frac{1}{2n+3} - \frac{4}{2n+3} + \frac{1}{2n+5} = -\frac{3}{2n+3} + \frac{1}{2n+5}$
Adding these together and multiplying by the overall factor of $\frac{1}{4}$:
$\text{Sum} = \frac{1}{4} \left( \frac{8}{3} - \frac{3}{2n+3} + \frac{1}{2n+5} \right)$

1. Determine the expression for the partial sum $S_n$ (from part ii).

From part (i), solving for $k$ gives $k = 28$, so:
$\frac{2r+7}{(2r+1)(2r+3)(2r+5)} = \frac{1}{4} \left( \frac{3}{2r+1} - \frac{4}{2r+3} + \frac{1}{2r+5} \right)$

Using the method of differences to evaluate the sum of the first $n+1$ terms (from $r=0$ to $n$) as required in part (ii):

2. Explain the convergence of the series and state the sum to infinity.

As $n \to \infty$, the terms $\frac{3}{2n+3} \to 0$ and $\frac{1}{2n+5} \to 0$.
Since the sequence of partial sums $S_n$ approaches a finite limit as $n$ tends to infinity, the series is convergent.

The sum to infinity is given by:
$\lim_{n \to \infty} S_n = \frac{1}{4} \left( \frac{8}{3} - 0 + 0 \right) = \frac{2}{3}$