将 $\dfrac{4}{4r^{2}+12r+5}$ 表示为部分分式。 求 $\displaystyle\sum_{r=2}^{n} \dfrac{4}{4r^{2}+12r+5}$。 数列 $u_{1},u_{2},u_{3},\dots$ 满足 $u_{1}=3$，且对于 $r\ge 2$，有 $u_{r}=u_{r-1}+\dfrac{4}{4r^{2}+12r+5}$。 通过考虑 $\displaystyle\sum_{r=2}^{n}(u_{r}-u_{r-1})$，用 $n$ 表示 $u_{n}$。 根值审敛法是判断形如 $\displaystyle\sum_{r=0}^{\infty}a_{r}$ 的无穷级数敛散性的一种方法。该审敛法指出，当 $\displaystyle\lim_{n\to\infty}\sqrt[n]{|a_{n}|}<1$ 时级数收敛，而当 $\displaystyle\lim_{n\to\infty}\sqrt[n]{|a_{n}|}>1$ 时级数发散。 当 $\displaystyle\lim_{n\to\infty}\sqrt[n]{|a_{n}|}=1$ 时，该审敛法失效。使用该审敛法解释为什么级数 $\displaystyle\sum_{r=0}^{\infty} \left(\dfrac{-3r^{3}+rx}{5r^{3}+7}\right)^{r}$ 对所有的 $x$ 值都收敛。

1. Factorise the quadratic expression in the denominator.

$4r^2+12r+5 = (2r+1)(2r+5)$

2. Set up the partial fractions identity and solve for the constants by substituting suitable values of $r$.

Let $\dfrac{4}{(2r+1)(2r+5)} \equiv \dfrac{A}{2r+1} + \dfrac{B}{2r+5}$.

Multiplying throughout by $(2r+1)(2r+5)$ gives:
$4 = A(2r+5) + B(2r+1)$

Substitute $r = -\frac{1}{2}$:
$4 = A\left(2\left(-\frac{1}{2}\right)+5\right)$
$4 = 4A \implies A = 1$

Substitute $r = -\frac{5}{2}$:
$4 = B\left(2\left(-\frac{5}{2}\right)+1\right)$
$4 = -4B \implies B = -1$

3. Write the final decomposed expression.

$\dfrac{4}{4r^2+12r+5} = \dfrac{1}{2r+1} - \dfrac{1}{2r+5}$

1. Obtain the partial fraction decomposition of the summand (from part a(i)).

Factorise the denominator: $4r^2+12r+5 = (2r+1)(2r+5)$.
Let $\dfrac{4}{(2r+1)(2r+5)} \equiv \dfrac{A}{2r+1} + \dfrac{B}{2r+5}$.
$4 \equiv A(2r+5) + B(2r+1)$.
Substitute $r = -\dfrac{1}{2}$: $4 = A(4) \implies A = 1$.
Substitute $r = -\dfrac{5}{2}$: $4 = B(-4) \implies B = -1$.
Thus, $\dfrac{4}{4r^2+12r+5} = \dfrac{1}{2r+1} - \dfrac{1}{2r+5}$.

2. List the first few and last few terms of the series to apply the method of differences.

$\sum_{r=2}^{n} \dfrac{4}{4r^{2}+12r+5} = \sum_{r=2}^{n} \left( \dfrac{1}{2r+1} - \dfrac{1}{2r+5} \right)$
Writing out the terms systematically:

3. Cancel out intermediate terms and simplify the remaining terms.

Observe the diagonal cancellation: the second term of each line cancels with the first term two lines down (e.g., $-\dfrac{1}{9}$ from $r=2$ cancels with $+\dfrac{1}{9}$ from $r=4$).
The surviving terms are the first terms from $r=2$ and $r=3$, and the second terms from $r=n-1$ and $r=n$.

1. Evaluate the sum from part (ii) using the method of differences.

From part (i), we have $\dfrac{4}{4r^2+12r+5} = \dfrac{1}{2r+1} - \dfrac{1}{2r+5}$.
Evaluating the sum for part (ii):
$\sum_{r=2}^{n} \dfrac{4}{4r^{2}+12r+5} = \sum_{r=2}^{n} \left( \dfrac{1}{2r+1} - \dfrac{1}{2r+5} \right)$
$= \left(\dfrac{1}{5} - \dfrac{1}{9}\right) + \left(\dfrac{1}{7} - \dfrac{1}{11}\right) + \left(\dfrac{1}{9} - \dfrac{1}{13}\right) + \dots + \left(\dfrac{1}{2n-1} - \dfrac{1}{2n+3}\right) + \left(\dfrac{1}{2n+1} - \dfrac{1}{2n+5}\right)$
Most terms cancel out, leaving the first two positive terms and the last two negative terms:
$= \dfrac{1}{5} + \dfrac{1}{7} - \dfrac{1}{2n+3} - \dfrac{1}{2n+5} = \dfrac{12}{35} - \dfrac{1}{2n+3} - \dfrac{1}{2n+5}$

2. Expand the summation $\displaystyle\sum_{r=2}^{n}(u_{r}-u_{r-1})$.

By writing out the terms of the sum, we get a telescoping series:
$\sum_{r=2}^{n}(u_{r}-u_{r-1}) = (u_2 - u_1) + (u_3 - u_2) + (u_4 - u_3) + \dots + (u_n - u_{n-1})$
Notice that all intermediate terms cancel out:
$= u_n - u_1$

3. Equate the sums and substitute the initial condition to find $u_n$.

From the given recurrence relation, $u_r - u_{r-1} = \dfrac{4}{4r^{2}+12r+5}$.
Therefore, the sums evaluated in the previous steps are equal:
$\sum_{r=2}^{n}(u_r - u_{r-1}) = \sum_{r=2}^{n}\dfrac{4}{4r^{2}+12r+5}$
$u_n - u_1 = \dfrac{12}{35} - \dfrac{1}{2n+3} - \dfrac{1}{2n+5}$
Given $u_1 = 3$:
$u_n - 3 = \dfrac{12}{35} - \dfrac{1}{2n+3} - \dfrac{1}{2n+5}$
$u_n = \dfrac{117}{35} - \dfrac{1}{2n+3} - \dfrac{1}{2n+5}$

1. Identify the general term $a_n$ and formulate an expression for $\sqrt[n]{|a_n|}$.

Let $a_n = \left(\dfrac{-3n^3+nx}{5n^3+7}\right)^n$.

Then, $\sqrt[n]{|a_n|} = \left| \dfrac{-3n^3+nx}{5n^3+7} \right|$.

2. Evaluate the limit of $\sqrt[n]{|a_n|}$ as $n \to \infty$.

$\begin{aligned} \lim_{n\to\infty}\sqrt[n]{|a_{n}|} &= \lim_{n\to\infty} \left| \dfrac{-3n^3+nx}{5n^3+7} \right| \\ &= \lim_{n\to\infty} \left| \dfrac{-3 + \frac{x}{n^2}}{5 + \frac{7}{n^3}} \right| \\ &= \left| \dfrac{-3 + 0}{5 + 0} \right| \\ &= \dfrac{3}{5} \end{aligned}$

3. Conclude the convergence of the series using the condition provided by the Root Test.

Since $\displaystyle\lim_{n\to\infty}\sqrt[n]{|a_{n}|} = \dfrac{3}{5}$, which is strictly less than $1$, and this limit evaluates to a constant completely independent of $x$, the test states that the series converges for all values of $x$.