Summation of Series: Telescoping; equation — RVHS 2025 H2 Math Prelim Paper 1

(a) Find \(\displaystyle\sum_{r=1}^{N} \frac{1}{(r+1)(r+2)}\) in terms of \(N\).

Let \(k - 1 = r\), i.e. \(k = r + 1\). When \(r = 1\), \(k = 2\); when \(r = N\), \(k = N+1\). \[\begin{aligned} \sum_{r=1}^{N} \frac{1}{(r+1)(r+2)} &= \sum_{k=2}^{N+1} \frac{1}{k(k+1)}\\[4pt] &= \left(1 - \frac{1}{(N+1)+1}\right) - \frac{1}{(1)(1+1)}\\[4pt] &= 1 - \frac{1}{N+2} - \frac{1}{2} \end{aligned}\]

(b) Find the positive integer \(k\) such that \(\displaystyle 8\sum_{r=k+1}^{\infty} \frac{1}{r(r+1)} = \sum_{r=1}^{k} \frac{1}{r(r+1)}\).

\[\begin{aligned} 8\left(\sum_{r=1}^{\infty} \frac{1}{r(r+1)} - \sum_{r=1}^{k} \frac{1}{r(r+1)}\right) &= \sum_{r=1}^{k} \frac{1}{r(r+1)}\\[4pt] 8\sum_{r=1}^{\infty} \frac{1}{r(r+1)} - 8\sum_{r=1}^{k} \frac{1}{r(r+1)} &= \sum_{r=1}^{k} \frac{1}{r(r+1)}\\[4pt] 8\sum_{r=1}^{\infty} \frac{1}{r(r+1)} &= 9\sum_{r=1}^{k} \frac{1}{r(r+1)} \end{aligned}\]

As \(n \to \infty\), \(\dfrac{1}{n+1} \to 0\). Hence \(\displaystyle\sum_{r=1}^{\infty} \frac{1}{r(r+1)} = 1\).

\[ 8(1) = 9\sum_{r=1}^{k} \frac{1}{r(r+1)} \implies \sum_{r=1}^{k} \frac{1}{r(r+1)} = \frac{8}{9} \]

Using part (a) with \(N = k\): \[ 1 - \frac{1}{k+1} = \frac{8}{9} \implies \frac{k}{k+1} = \frac{8}{9} \implies 9k = 8(k+1) \implies k = 8 \]