数列 $u_1, u_2, u_3, \dots$ 满足 $u_1 = \frac{1}{2}$ 且
$u_{n+1} = u_n - \frac{n^2 + n - 1}{(n+2)!}, \quad \text{对于所有 } n \ge 1.$

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

Let $P_n$ be the statement that $u_n = \frac{n}{(n+1)!}$ for $n \in \mathbb{Z}^+$.
For $n = 1$:
$\text{LHS} = u_1 = \frac{1}{2}$ (given)
$\text{RHS} = \frac{1}{(1+1)!} = \frac{1}{2}$
Since $\text{LHS} = \text{RHS}$, $P_1$ is true.

2. State the induction hypothesis.

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

3. Express $u_{k+1}$ using the recurrence relation and substitute the induction hypothesis.

We want to show that $P_{k+1}$ is true, i.e., $u_{k+1} = \frac{k+1}{(k+2)!}$.
Using the given recurrence relation for $n=k$:
$u_{k+1} = u_k - \frac{k^2 + k - 1}{(k+2)!}$
Substitute $u_k = \frac{k}{(k+1)!}$:
$u_{k+1} = \frac{k}{(k+1)!} - \frac{k^2 + k - 1}{(k+2)!}$

4. Algebraically manipulate the expression to obtain the form for $P_{k+1}$.

Make the denominators common by multiplying the first term by $\frac{k+2}{k+2}$:
$u_{k+1} = \frac{k(k+2)}{(k+2)!} - \frac{k^2 + k - 1}{(k+2)!}$
$u_{k+1} = \frac{k^2 + 2k - (k^2 + k - 1)}{(k+2)!}$
$u_{k+1} = \frac{k^2 + 2k - k^2 - k + 1}{(k+2)!}$
$u_{k+1} = \frac{k+1}{(k+2)!}$
Hence, $P_{k+1}$ is true.

5. State the standard conclusion 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. Express the term inside the summation as a difference of terms in the sequence $u_n$ by rearranging the given recurrence relation.

From the given recurrence relation $u_{n+1} = u_n - \frac{n^2 + n - 1}{(n+2)!}$, we rearrange the terms:
$\frac{n^2 + n - 1}{(n+2)!} = u_n - u_{n+1}$

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

$\sum_{n=1}^{N} \frac{n^2 + n - 1}{(n+2)!} = \sum_{n=1}^{N} (u_n - u_{n+1})$
$= (u_1 - u_2) + (u_2 - u_3) + \dots + (u_N - u_{N+1})$
Cancelling the intermediate terms yields:
$= u_1 - u_{N+1}$

3. Substitute the explicit formula for $u_n$ from part (i) to obtain the final expression in terms of $N$.

From part (i), we proved that $u_n = \frac{n}{(n+1)!}$.
Thus, $u_1 = \frac{1}{2}$ and $u_{N+1} = \frac{N+1}{(N+2)!}$.
Substituting these into the sum expression gives:
$\sum_{n=1}^{N} \frac{n^2 + n - 1}{(n+2)!} = \frac{1}{2} - \frac{N+1}{(N+2)!}$

1. Find the expression for the partial sum of $N$ terms (evaluating the result of part (ii)).

From the given recurrence relation, we have $\frac{n^2 + n - 1}{(n+2)!} = u_n - u_{n+1}$.
Using the method of differences to sum from $n=1$ to $N$:
$\sum_{n=1}^{N} \frac{n^2 + n - 1}{(n+2)!} = (u_1 - u_2) + (u_2 - u_3) + \dots + (u_N - u_{N+1})$
$= u_1 - u_{N+1}$
Using $u_1 = \frac{1}{2}$ and the formula from part (i) $u_{N+1} = \frac{N+1}{(N+2)!}$, the sum to $N$ terms is:
$S_N = \frac{1}{2} - \frac{N+1}{(N+2)!}$

2. Explain the convergence of the series by considering the limit of $S_N$ as $N \to \infty$.

As $N \to \infty$, $(N+2)!$ grows much faster than $N+1$. We can observe this by simplifying the fraction:
$\frac{N+1}{(N+2)!} = \frac{1}{(N+2)N!} \to 0$
Taking the limit of the partial sum:
$\lim_{N \to \infty} S_N = \frac{1}{2} - 0 = \frac{1}{2}$
Because the limit of the partial sums as $N \to \infty$ is a finite constant, the infinite series is convergent.

3. State the sum to infinity.

The value of the sum to infinity is exactly the limit of the partial sums as $N \to \infty$.
$\sum_{n=1}^{\infty} \frac{n^2 + n - 1}{(n+2)!} = \frac{1}{2}$