Given that $\sum_{r=1}^{n} \frac{1}{r(r+1)} = \frac{n}{n+1}$.

1. Write out the first few terms of the sum to recognise it as the given series with $n-1$ terms.

$\sum_{r=2}^{n} \frac{1}{r(r-1)} = \frac{1}{2(1)} + \frac{1}{3(2)} + \frac{1}{4(3)} + \dots + \frac{1}{n(n-1)}$
$= \frac{1}{1(2)} + \frac{1}{2(3)} + \dots + \frac{1}{(n-1)n}$
$= \sum_{r=1}^{n-1} \frac{1}{r(r+1)}$

2. Apply the given identity and simplify to obtain the final expression.

Since we are given $\sum_{r=1}^{n} \frac{1}{r(r+1)} = \frac{n}{n+1}$, we replace $n$ with $n-1$:
$\sum_{r=1}^{n-1} \frac{1}{r(r+1)} = \frac{n-1}{(n-1)+1}$
$= \frac{n-1}{n}$
$= \frac{n}{n} - \frac{1}{n}$
$= 1 - \frac{1}{n} \text{ (shown)}$

1. Split the infinite series into a geometric series and the limit of the series from part (a).

$\sum_{r=2}^{\infty} \left( \frac{1}{2^r} + \frac{1}{r(r-1)} \right) = \sum_{r=2}^{\infty} \left(\frac{1}{2}\right)^r + \lim_{n \to \infty} \sum_{r=2}^{n} \frac{1}{r(r-1)}$

2. Evaluate the sum to infinity of the geometric series.

For the geometric series $\sum_{r=2}^{\infty} \left(\frac{1}{2}\right)^r$, the first term is $a = \frac{1}{2^2} = \frac{1}{4}$ and the common ratio is $r = \frac{1}{2}$.
$S_\infty = \frac{\frac{1}{4}}{1 - \frac{1}{2}} = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2}$

3. Evaluate the limit of the remaining series using the result from (a) and find the final sum.

Using the result from part (a):
$\lim_{n \to \infty} \sum_{r=2}^{n} \frac{1}{r(r-1)} = \lim_{n \to \infty} \left( 1 - \frac{1}{n} \right) = 1$
Total sum:
$\frac{1}{2} + 1 = \frac{3}{2}$