2009 H2 Mathematics Paper 1 Question 3

Sigma Notation

Answers

(i)

1n−1−2n+1n+1=2n3−n{\frac{ 1 }{ n - 1 } - \frac{ 2 }{ n } + \frac{ 1 }{ n + 1 } = \frac{ 2 }{ n^3 - n }}

(ii)

12(12−1n+1n+1){\frac{1}{2} \left( \frac{1}{2} - \frac{1}{n} + \frac{1}{n + 1} \right) }

(iii)

As n→∞,{ n \to \infty, } 1n,1n+1→0{\frac{1}{n},\frac{1}{n + 1} \to 0} so 12(12−1n+1n+1)→14{\frac{1}{2} \left( \frac{1}{2} - \frac{1}{n} + \frac{1}{n + 1} \right) \to \frac{1}{4}}
Hence the series converges
∑r=2∞1r3−r=14{\displaystyle \sum_{r=2}^\infty \frac{1}{r^3-r} = \frac{1}{4}}

Full solutions

(i)

1n−1−2n+1n+1=n(n+1)(n−1)n(n+1)−2(n−1)(n+1)(n−1)n(n+1)+(n−1)(n)(n−1)n(n+1)=n2+n−2(n2−1)+n2−nn(n2−1)=2n3−n  ■\begin{align*} & \frac{ 1 }{ n - 1 } - \frac{ 2 }{ n } + \frac{ 1 }{ n + 1 } \\ & = \frac{n(n+1)}{(n-1)n(n+1)} - \frac{2(n-1)(n+1)}{(n-1)n(n+1)} + \frac{(n-1)(n)}{(n-1)n(n+1)} \\ & = \frac{n^2+n - 2(n^2-1) + n^2-n}{n(n^2-1)} \\ & =\frac{ 2 }{ n^3 - n } \; \blacksquare \end{align*}

(ii)

∑r=2n1r3−r=12∑r=2n(1r−1−2r+1r+1)=12(1−1+13+12−23+14+13−12+15+⋯+1n−3−2n−2+1n−1+1n−2−2n−1+1n+1n−1−2n+1n+1)=12(12−1n+1n+1)  ■\begin{align*} & \sum_{r=2}^n \frac{1}{r^3-r} \\ &= \frac{1}{2} \sum_{r=2}^n \left(\frac{1}{r - 1} - \frac{2}{r} + \frac{1}{r + 1} \right) \\ & = \frac{1}{2} \left(\def\arraystretch{1.5} \begin{array}{lclclc} & 1 &-& 1 &+& \cancel{\frac{1}{3}} \\ + & \frac{1}{2} &-& \cancel{\frac{2}{3}} &+& \cancel{\frac{1}{4}} \\ + & \cancel{\frac{1}{3}} &-& \cancel{\frac{1}{2}} &+& \cancel{\frac{1}{5}} \\ + &&& \cdots && \\ + & \cancel{\frac{1}{n - 3}} &-& \cancel{\frac{2}{n - 2}} &+& \cancel{\frac{1}{n - 1}} \\ + & \cancel{\frac{1}{n - 2}} &-& \cancel{\frac{2}{n - 1}} &+& \frac{1}{n} \\ + & \cancel{\frac{1}{n - 1}} &-& \frac{2}{n} &+& \frac{1}{n + 1} \end{array}\right) \\ &= \frac{1}{2} \left( \frac{1}{2} - \frac{1}{n} + \frac{1}{n + 1} \right) \; \blacksquare \end{align*}

(iii)

As n→∞,{ n \to \infty, } 1n,1n+1→0{\displaystyle \frac{1}{n},\frac{1}{n + 1} \to 0} so
∑r=2n1r3−r=12(12−1n+1n+1)→14\begin{align*} & \sum_{r=2}^n \frac{1}{r^3-r} \\ &= \frac{1}{2} \left( \frac{1}{2} - \frac{1}{n} + \frac{1}{n + 1} \right) \\ &\to \frac{1}{4} \end{align*}
Hence ∑r=2∞1r3−r{\displaystyle \sum_{r=2}^\infty \frac{1}{r^3-r}} converges ■{\blacksquare}
∑r=1∞1r3−r=14  ■\sum_{r=1}^\infty \frac{1}{r^3-r} = \frac{1}{4} \; \blacksquare