A sequence $u_1, u_2, u_3, \dots$ is such that $u_1 = \frac{1}{3}$ and $u_{r+1} = u_r - \frac{1}{(2r-1)(2r+3)}, \quad \text{for all } r \ge 1.$

1. State the proposition and prove the base case for $n=1$.

Let $P_n$ be the statement that $u_n = \frac{n}{4n^2 - 1}$, for $n \in \mathbb{Z}^+$.
When $n = 1$:
$\text{LHS} = u_1 = \frac{1}{3}$ (given)
$\text{RHS} = \frac{1}{4(1)^2 - 1} = \frac{1}{4 - 1} = \frac{1}{3}$
Since $\text{LHS} = \text{RHS}$, $P_1$ is true.

2. Assume the proposition holds for $n=k$ and substitute into the recurrence relation for $n=k+1$.

Assume $P_k$ is true for some $k \in \mathbb{Z}^+$, i.e., $u_k = \frac{k}{4k^2 - 1} = \frac{k}{(2k-1)(2k+1)}$.
We want to show $P_{k+1}$ is true, meaning $u_{k+1} = \frac{k+1}{4(k+1)^2 - 1}$.
Using the given recurrence relation $u_{r+1} = u_r - \frac{1}{(2r-1)(2r+3)}$ for $r = k$:
$u_{k+1} = u_k - \frac{1}{(2k-1)(2k+3)}$
$u_{k+1} = \frac{k}{(2k-1)(2k+1)} - \frac{1}{(2k-1)(2k+3)}$

3. Combine the expression into a single algebraic fraction over a common denominator.

$u_{k+1} = \frac{k(2k+3) - (2k+1)}{(2k-1)(2k+1)(2k+3)}$
$u_{k+1} = \frac{2k^2 + 3k - 2k - 1}{(2k-1)(2k+1)(2k+3)}$
$u_{k+1} = \frac{2k^2 + k - 1}{(2k-1)(2k+1)(2k+3)}$

4. Factorise the numerator, simplify, and match it to the proposition $P_{k+1}$.

Factorising the quadratic numerator $2k^2 + k - 1 = (2k-1)(k+1)$:
$u_{k+1} = \frac{(2k-1)(k+1)}{(2k-1)(2k+1)(2k+3)}$
$u_{k+1} = \frac{k+1}{(2k+1)(2k+3)}$
Expanding the denominator to show the required form:
$u_{k+1} = \frac{k+1}{4k^2 + 8k + 3}$
$u_{k+1} = \frac{k+1}{4(k^2 + 2k + 1) - 1}$
$u_{k+1} = \frac{k+1}{4(k+1)^2 - 1}$
Hence, $P_{k+1}$ is true if $P_k$ is true.

5. State the final conclusion of the induction.

Since $P_1$ is true, and $P_k$ being true implies $P_{k+1}$ is true, by the Principle of Mathematical Induction, $P_n$ is true for all $n \in \mathbb{Z}^+$.

1. Express the sum of the first $n$ terms using sigma notation by relating its terms to the given recurrence equation.

Let $S_n$ be the sum of the first $n$ terms of the series.
The terms are of the form $\frac{1}{(2r-1)(2r+3)}$.
For the first term $\frac{1}{5 \times 9}$, we have $2r-1=5 \implies r=3$.
Therefore, the $n$ terms correspond to $r = 3, 4, \dots, n+2$.
$S_n = \sum_{r=3}^{n+2} \frac{1}{(2r-1)(2r+3)}$
Rearranging the given equation $u_{r+1} = u_r - \frac{1}{(2r-1)(2r+3)}$, we obtain:
$\frac{1}{(2r-1)(2r+3)} = u_r - u_{r+1}$

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

$S_n = \sum_{r=3}^{n+2} (u_r - u_{r+1})$
Expanding the sum:
$S_n = (u_3 - u_4) + (u_4 - u_5) + \dots + (u_{n+1} - u_{n+2}) + (u_{n+2} - u_{n+3})$
By cancellation (telescoping sum), all intermediate terms cancel out:
$S_n = u_3 - u_{n+3}$

3. Substitute the formula for $u_n$ from part (i) to establish the final result.

From part (i), we know $u_n = \frac{n}{4n^2 - 1}$.
Evaluating $u_3$:
$u_3 = \frac{3}{4(3)^2 - 1} = \frac{3}{36 - 1} = \frac{3}{35}$
Evaluating $u_{n+3}$:
$u_{n+3} = \frac{n+3}{4(n+3)^2 - 1}$
Substituting these into our expression for $S_n$:
$S_n = \frac{3}{35} - \frac{n+3}{4(n+3)^2 - 1}$
(Shown)

1. Show that the limit of the sequence of partial sums exists and is finite

For the sum to $n$ terms $S_n = \frac{3}{35} - \frac{n+3}{4(n+3)^2 - 1}$, we evaluate the limit as $n \to \infty$.
As $n \to \infty$, the term $\frac{n+3}{4(n+3)^2 - 1} \to 0$ because the degree of the denominator (quadratic in $n$) is strictly greater than the degree of the numerator (linear in $n$).
Thus, $\lim_{n\to\infty} S_n = \frac{3}{35}$.
Since the sequence of partial sums converges to a finite limit, the series is convergent.

2. State the sum to infinity

The sum to infinity is the limit of $S_n$ as $n \to \infty$.
$S_\infty = \frac{3}{35}$