Summation and series: ACJC 2025 Promo Q3

(a)

Replacing $r$ with $r-1$ (let $s = r-1$, so $s$ runs from $1$ to $n-1$): $$\begin{aligned} \sum_{r=2}^{n}\frac{1}{r(r-1)} &= \sum_{s=1}^{n-1}\frac{1}{(s+1)s}\\ &= \frac{n-1}{(n-1)+1}\\ &= \frac{n-1}{n}\\ &= 1 - \frac{1}{n}. \quad\text{((shown))} \end{aligned}$$

(b)

Splitting the sum: $$\begin{aligned} \sum_{r=2}^{\infty}\left(\frac{1}{2^r} + \frac{1}{r(r-1)}\right) &= \sum_{r=2}^{\infty}\frac{1}{2^r} + \lim_{n\to\infty}\left(1 - \frac{1}{n}\right). \end{aligned}$$ The first sum is a geometric series with first term $\tfrac{1}{4}$ and common ratio $\tfrac{1}{2}$: $$\begin{aligned} \sum_{r=2}^{\infty}\frac{1}{2^r} &= \frac{\tfrac{1}{4}}{1-\tfrac{1}{2}} = \frac{1}{2}. \end{aligned}$$ Therefore $$\begin{aligned} \sum_{r=2}^{\infty}\left(\frac{1}{2^r} + \frac{1}{r(r-1)}\right) &= \frac{1}{2} + 1 = \frac{3}{2}. \end{aligned}$$