A sequence $u_1, u_2, u_3, \dots$ is such that $u_1 = 0$ and $u_{n+1} = u_n + \frac{4 + n - n^2}{[(n+1)(n+2)]^2}, \text{ for all } n \in \mathbb{Z}^+.$

1. Establish the base case for $n=1$

Let $P_n$ be the statement that $u_n = \frac{n - 1}{(n + 1)^2}$ for $n \in \mathbb{Z}^+$.

When $n = 1$,
$\text{LHS} = u_1 = 0$ (given)
$\text{RHS} = \frac{1 - 1}{(1 + 1)^2} = 0$

Since $\text{LHS} = \text{RHS}$, $P_1$ is true.

2. State the inductive hypothesis

Assume $P_k$ is true for some positive integer $k$, i.e., $u_k = \frac{k - 1}{(k + 1)^2}$.

We want to show that $P_{k+1}$ is true, i.e., $u_{k+1} = \frac{(k+1) - 1}{((k+1) + 1)^2} = \frac{k}{(k + 2)^2}$.

3. Substitute the inductive hypothesis into the recurrence relation

From the given recurrence relation:
$u_{k+1} = u_k + \frac{4 + k - k^2}{[(k+1)(k+2)]^2}$

Substituting $u_k$ from the inductive hypothesis:
$u_{k+1} = \frac{k - 1}{(k + 1)^2} + \frac{4 + k - k^2}{(k+1)^2(k+2)^2}$

4. Simplify algebraically to show $P_{k+1}$ holds

Combine into a single fraction with a common denominator:
$u_{k+1} = \frac{(k - 1)(k + 2)^2 + (4 + k - k^2)}{(k+1)^2(k+2)^2}$

Expanding the terms in the numerator:
$u_{k+1} = \frac{(k - 1)(k^2 + 4k + 4) + 4 + k - k^2}{(k+1)^2(k+2)^2}$
$u_{k+1} = \frac{(k^3 + 3k^2 - 4) + 4 + k - k^2}{(k+1)^2(k+2)^2}$

Simplifying the numerator:
$u_{k+1} = \frac{k^3 + 2k^2 + k}{(k+1)^2(k+2)^2}$
$u_{k+1} = \frac{k(k^2 + 2k + 1)}{(k+1)^2(k+2)^2}$
$u_{k+1} = \frac{k(k+1)^2}{(k+1)^2(k+2)^2}$
$u_{k+1} = \frac{k}{(k+2)^2}$

Thus, $P_{k+1}$ is true.

5. State the concluding statement for mathematical induction

Since $P_1$ is true, and if $P_k$ is true then $P_{k+1}$ is true, by the Principle of Mathematical Induction, $P_n$ is true for all positive integers $n$.

1. Recognise that the summand can be expressed as a difference of consecutive terms using the given recurrence relation.

From the recurrence relation given in the question,
$u_{n+1} = u_n + \frac{4 + n - n^2}{[(n+1)(n+2)]^2}$
Rearranging gives:
$u_{n+1} - u_n = \frac{4 + n - n^2}{[(n+1)(n+2)]^2}$

2. Apply the method of differences to evaluate the finite sum and substitute the general formula for $u_n$.

$\sum_{n=2}^{N} \frac{4 + n - n^2}{[(n+1)(n+2)]^2} = \sum_{n=2}^{N} (u_{n+1} - u_n)$
$= (u_3 - u_2) + (u_4 - u_3) + \dots + (u_{N+1} - u_N)$
$= u_{N+1} - u_2$
Using $u_n = \frac{n - 1}{(n + 1)^2}$ from part (i):
$u_{N+1} = \frac{(N+1) - 1}{((N+1) + 1)^2} = \frac{N}{(N+2)^2}$
$u_2 = \frac{2 - 1}{(2 + 1)^2} = \frac{1}{9}$
Thus, the finite sum is $\frac{N}{(N+2)^2} - \frac{1}{9}$.

3. Take the limit of the finite sum as $N \to \infty$ to find the sum to infinity.

Sum to infinity $= \lim_{N \to \infty} \left( \frac{N}{(N+2)^2} - \frac{1}{9} \right)$
$= \lim_{N \to \infty} \left( \frac{\frac{1}{N}}{\left(1+\frac{2}{N}\right)^2} - \frac{1}{9} \right)$
As $N \to \infty$, $\frac{1}{N} \to 0$, so $\frac{N}{(N+2)^2} \to 0$.
Therefore, the sum to infinity is $0 - \frac{1}{9} = -\frac{1}{9}$.

1. Relate the given series' general term to the one from part (ii) by shifting the index.

Let $n = m + 1$.
Substituting this into the general term:
$\frac{2 + 3n - n^2}{[n(n+1)]^2} = \frac{2 + 3(m+1) - (m+1)^2}{[(m+1)(m+2)]^2}$
$= \frac{2 + 3m + 3 - (m^2 + 2m + 1)}{[(m+1)(m+2)]^2}$
$= \frac{4 + m - m^2}{[(m+1)(m+2)]^2}$

2. Rewrite the required infinite series using the shifted index and split off the first term to match part (ii).

Changing the limits of summation by substituting $n = m + 1$ (when $n = 2$, $m = 1$):
$\sum_{n=2}^{\infty} \frac{2 + 3n - n^2}{[n(n+1)]^2} = \sum_{m=1}^{\infty} \frac{4 + m - m^2}{[(m+1)(m+2)]^2}$
Separating the $m=1$ term from the rest of the series to directly use the series evaluated in part (ii):
$= \left( \frac{4 + 1 - 1^2}{[(1+1)(1+2)]^2} \right) + \sum_{m=2}^{\infty} \frac{4 + m - m^2}{[(m+1)(m+2)]^2}$
$= \frac{1}{9} + \sum_{m=2}^{\infty} (u_{m+1} - u_m)$

3. Evaluate the remaining infinite sum using the method of differences and limit to infinity.

First, evaluate the sum to infinity from part (ii):
$\sum_{m=2}^{N} (u_{m+1} - u_m) = u_{N+1} - u_2$
From part (i), $u_n = \frac{n-1}{(n+1)^2}$. Thus, $u_2 = \frac{2-1}{(2+1)^2} = \frac{1}{9}$.
Taking the limit as $N \to \infty$, $u_{N+1} = \frac{N}{(N+2)^2} \to 0$.
The sum to infinity for $m \ge 2$ is $0 - \frac{1}{9} = -\frac{1}{9}$.
Therefore, the required total sum is:
$= \frac{1}{9} - \frac{1}{9} = 0$