Show that $\frac{7}{n-2} - \frac{5}{n-1} - \frac{2}{n} = \frac{9n-4}{(n-2)(n-1)n}$. Hence find $\sum_{n=3}^{N} \frac{9n-4}{(n-2)(n-1)n}$, giving your answer in the form $k - \text{f}(N)$, where $k$ is a constant. Show that $\sum_{n=3}^{N} \frac{9n-4}{(n-2)(n-1)n}$ is convergent and state the sum to infinity of this series. Use your answer in (ii) to find $\sum_{n=2}^{N} \frac{9n+14}{n(n+1)(n+2)}$.

1. Express the fractions with a common denominator.

$\frac{7}{n-2} - \frac{5}{n-1} - \frac{2}{n} = \frac{7n(n-1) - 5n(n-2) - 2(n-1)(n-2)}{(n-2)(n-1)n}$

2. Expand and simplify the numerator to obtain the final expression.

$= \frac{7n^2 - 7n - (5n^2 - 10n) - 2(n^2 - 3n + 2)}{(n-2)(n-1)n}$
$= \frac{7n^2 - 7n - 5n^2 + 10n - 2n^2 + 6n - 4}{(n-2)(n-1)n}$
$= \frac{9n-4}{(n-2)(n-1)n} \text{ (shown)}$

1. Express the sum using the partial fractions from part (i) and list the terms to identify the pattern.

$\sum_{n=3}^{N} \frac{9n-4}{(n-2)(n-1)n} = \sum_{n=3}^{N} \left( \frac{7}{n-2} - \frac{5}{n-1} - \frac{2}{n} \right)$
Listing the first few and last few terms of the series:

2. Group the terms by common denominators to show the cancellation (method of differences).

Summing vertically by grouping terms with the same denominator:
Denominator $1$: $\frac{7}{1} = 7$
Denominator $2$: $-\frac{5}{2} + \frac{7}{2} = \frac{2}{2} = 1$
Denominator $3$: $-\frac{2}{3} - \frac{5}{3} + \frac{7}{3} = 0$
$\vdots$
Denominator $N-2$: $-\frac{2}{N-2} - \frac{5}{N-2} + \frac{7}{N-2} = 0$
Denominator $N-1$: $-\frac{2}{N-1} - \frac{5}{N-1} = -\frac{7}{N-1}$
Denominator $N$: $-\frac{2}{N}$

All intermediate terms with denominators from $3$ to $N-2$ cancel out completely to sum to $0$.

3. Collect the non-cancelled terms and format the final expression.

The sum $S_N$ of the remaining terms is:
$S_N = 7 + 1 - \frac{7}{N-1} - \frac{2}{N}$
$S_N = 8 - \left( \frac{7}{N-1} + \frac{2}{N} \right)$

Alternatively, expressing the function of $N$ as a single fraction:
$S_N = 8 - \frac{7N + 2(N-1)}{N(N-1)} = 8 - \frac{9N-2}{N(N-1)}$

This is in the required form $k - \text{f}(N)$, with constant $k=8$ and $\text{f}(N) = \frac{9N-2}{N(N-1)}$.

1. Concisely evaluate the sum $S_N$ from part (ii) as it is required for this part.

$\begin{aligned} S_N &= \sum_{n=3}^{N} \left( \frac{7}{n-2} - \frac{5}{n-1} - \frac{2}{n} \right) \\ &= \left(\frac{7}{1} - \frac{5}{2} - \frac{2}{3}\right) + \left(\frac{7}{2} - \frac{5}{3} - \frac{2}{4}\right) + \dots + \left(\frac{7}{N-2} - \frac{5}{N-1} - \frac{2}{N}\right) \\ &= 7 + \left(\frac{7}{2} - \frac{5}{2}\right) - \frac{5}{N-1} - \frac{2}{N-1} - \frac{2}{N} \\ &= 8 - \frac{7}{N-1} - \frac{2}{N} \end{aligned}$

2. Determine the limit of $S_N$ as $N \to \infty$ to demonstrate convergence.

As $N \to \infty$, $\frac{7}{N-1} \to 0$ and $\frac{2}{N} \to 0$.

$S_\infty = \lim_{N \to \infty} S_N = 8 - 0 - 0 = 8$.

Since the limit evaluates to a finite constant, the series is convergent.

3. State the sum to infinity.

Sum to infinity = $8$

1. Calculate the result for part (ii), as it is required to solve this target

Since the result of (ii) is required, we first derive it. Using the partial fractions from (i), the sum is:
$\sum_{n=3}^{N} \left( \frac{7}{n-2} - \frac{5}{n-1} - \frac{2}{n} \right)$
Writing out the terms:
$\begin{aligned} n=3 &: \frac{7}{1} - \frac{5}{2} - \frac{2}{3} \\ n=4 &: \frac{7}{2} - \frac{5}{3} - \frac{2}{4} \\ n=5 &: \frac{7}{3} - \frac{5}{4} - \frac{2}{5} \\ &\vdots \\ n=N-1 &: \frac{7}{N-3} - \frac{5}{N-2} - \frac{2}{N-1} \\ n=N &: \frac{7}{N-2} - \frac{5}{N-1} - \frac{2}{N} \end{aligned}$
Summing vertically, most intermediate terms cancel out:
$\text{Sum} = 7 + \left(-\frac{5}{2} + \frac{7}{2}\right) - \left(\frac{2}{N-1} + \frac{5}{N-1}\right) - \frac{2}{N}$
$= 8 - \frac{7}{N-1} - \frac{2}{N}$

2. Relate the given series in (iv) to the series in (ii) using a substitution

Let $r = n+2$. The summation limits change from $n=2 \dots N$ to $r=4 \dots N+2$.
Substitute $n = r-2$ into the general term:
$\frac{9n+14}{n(n+1)(n+2)} = \frac{9(r-2)+14}{(r-2)(r-1)r} = \frac{9r-4}{(r-2)(r-1)r}$
Hence, the required sum can be rewritten using the series structure from (ii):
$\sum_{n=2}^{N} \frac{9n+14}{n(n+1)(n+2)} = \sum_{r=4}^{N+2} \frac{9r-4}{(r-2)(r-1)r}$

3. Evaluate the sum using the result from (ii) by adjusting the limits

We evaluate this by taking the sum from $r=3$ to $N+2$ and subtracting the unneeded $r=3$ term:
$\sum_{r=4}^{N+2} \frac{9r-4}{(r-2)(r-1)r} = \sum_{r=3}^{N+2} \frac{9r-4}{(r-2)(r-1)r} - \left[ \frac{9(3)-4}{(3-2)(3-1)3} \right]$
Using the answer from (ii), substitute the upper limit as $N+2$:
$\sum_{r=3}^{N+2} \frac{9r-4}{(r-2)(r-1)r} = 8 - \frac{7}{(N+2)-1} - \frac{2}{N+2} = 8 - \frac{7}{N+1} - \frac{2}{N+2}$
The term for $r=3$ is $\frac{23}{6}$.
Therefore, the required sum is:
$\left( 8 - \frac{7}{N+1} - \frac{2}{N+2} \right) - \frac{23}{6}$
$= \frac{25}{6} - \frac{7}{N+1} - \frac{2}{N+2}$