It is given that $u_n = \frac{2n-1}{n(n-1)}$, where $n \in \mathbb{Z}^+$. Show that $u_n - u_{n+1} = \frac{2}{(n+1)(n-1)}$. It is given that $u_n = \frac{2n-1}{n(n-1)}$, where $n \in \mathbb{Z}^+$. Show that $u_n - u_{n+1} = \frac{2}{(n+1)(n-1)}$.

1. Relate the general term of the series to the identity from the previous part.

From the previous part, $\frac{2}{(r+1)(r-1)} = u_r - u_{r+1}$.

Therefore, $\frac{1}{(r+1)(r-1)} = \frac{1}{2}(u_r - u_{r+1})$.

2. Apply the method of differences to evaluate the sum.

$\sum_{r=2}^{N} \frac{1}{(r+1)(r-1)} = \frac{1}{2} \sum_{r=2}^{N} (u_r - u_{r+1})$

$= \frac{1}{2} [ (u_2 - u_3) + (u_3 - u_4) + \dots + (u_N - u_{N+1}) ]$

$= \frac{1}{2} (u_2 - u_{N+1})$

3. Substitute the expression for $u_n$ to obtain the final result in terms of $N$.

Since $u_n = \frac{2n-1}{n(n-1)}$, we evaluate $u_2$ and $u_{N+1}$:

$u_2 = \frac{2(2)-1}{2(1)} = \frac{3}{2}$

$u_{N+1} = \frac{2(N+1)-1}{(N+1)(N)} = \frac{2N+1}{N(N+1)}$

$\text{Sum} = \frac{1}{2} \left( \frac{3}{2} - \frac{2N+1}{N(N+1)} \right)$

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

1. Establish the expression for the sum of the first N terms (from part i) to analyse its limit.

$\sum_{r=2}^{N} \frac{1}{(r+1)(r-1)} = \frac{1}{2} \sum_{r=2}^{N} (u_r - u_{r+1}) = \frac{1}{2} (u_2 - u_{N+1})$

Since $u_n = \frac{2n-1}{n(n-1)}$, $u_2 = \frac{3}{2}$ and $u_{N+1} = \frac{2N+1}{(N+1)N}$.

$S_N = \frac{1}{2} \left( \frac{3}{2} - \frac{2N+1}{N(N+1)} \right) = \frac{3}{4} - \frac{2N+1}{2N(N+1)}$

2. Provide a valid reason for the convergence of the series.

As $N \to \infty$, the denominator of the fractional term $\frac{2N+1}{2N(N+1)}$ is a quadratic in $N$ while the numerator is linear, so $\frac{2N+1}{2N(N+1)} \to 0$.

Since the sequence of partial sums $S_N$ approaches a finite limit, the series is convergent.

3. Calculate the sum to infinity.

$S_\infty = \lim_{N \to \infty} \left( \frac{3}{4} - \frac{2N+1}{2N(N+1)} \right) = \frac{3}{4} - 0 = \frac{3}{4}$

1. Evaluate the sum in part (i) as it is required for this part.

$\sum_{r=2}^{N} \frac{1}{(r+1)(r-1)} = \frac{1}{2} \sum_{r=2}^{N} (u_r - u_{r+1})$
$= \frac{1}{2} (u_2 - u_{N+1})$
$= \frac{1}{2} \left( \frac{2(2)-1}{2(1)} - \frac{2(N+1)-1}{(N+1)N} \right)$
$= \frac{1}{2} \left( \frac{3}{2} - \frac{2N+1}{N(N+1)} \right) = \frac{3}{4} - \frac{2N+1}{2N(N+1)}$

2. Relate the target series to the series from part (i) by substituting the index.

Let $k = r+1$. When $r=2$, $k=3$. When $r=N-1$, $k=N$.
$\sum_{r=2}^{N-1} \frac{1}{(r+2)(r)} = \sum_{k=3}^{N} \frac{1}{(k+1)(k-1)}$

3. Express the shifted series using the result from part (i) and subtract the missing first term.

$\sum_{k=3}^{N} \frac{1}{(k+1)(k-1)} = \sum_{k=2}^{N} \frac{1}{(k+1)(k-1)} - \frac{1}{(2+1)(2-1)}$
$= \left( \frac{3}{4} - \frac{2N+1}{2N(N+1)} \right) - \frac{1}{3}$

4. Simplify the constants to obtain the final expression.

$= \left( \frac{3}{4} - \frac{1}{3} \right) - \frac{2N+1}{2N(N+1)}$
$= \frac{9-4}{12} - \frac{2N+1}{2N(N+1)}$
$= \frac{5}{12} - \frac{2N+1}{2N(N+1)} \quad \text{(Shown)}$