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Summation and series: CJC 2024 Promo Q3

Catholic Junior College2024 PromoPromo Q3●●○ Standard5 marks

What this question tests

Summation and series.

Question

It is given that ∑r=1n1r(r+1)=1−1n+1\displaystyle\sum_{r=1}^{n} \frac{1}{r(r+1)} = 1 - \frac{1}{n+1}.

  1. Find ∑r=5n+11r(r+1)\displaystyle\sum_{r=5}^{n+1} \frac{1}{r(r+1)}. [3]

  2. Give a reason why the series in part (a) is convergent and state the value of ∑r=5∞1r(r+1)\displaystyle\sum_{r=5}^{\infty} \frac{1}{r(r+1)}. [2]

Show full worked solution▾

(a)

∑r=5n+11r(r+1)=∑r=1n+11r(r+1)−∑r=141r(r+1)=(1−1(n+1)+1)−(1−14+1)=15−1n+2.\begin{aligned} \sum_{r=5}^{n+1} \frac{1}{r(r+1)} &= \sum_{r=1}^{n+1} \frac{1}{r(r+1)} - \sum_{r=1}^{4} \frac{1}{r(r+1)}\\ &= \left(1 - \frac{1}{(n+1)+1}\right) - \left(1 - \frac{1}{4+1}\right)\\ &= \frac{1}{5} - \frac{1}{n+2}. \end{aligned}

(b)

As n→∞n\to\infty, 1n+2→0\dfrac{1}{n+2}\to0. ∴∑r=5∞1r(r+1)=15\begin{aligned} \therefore\sum_{r=5}^{\infty} \frac{1}{r(r+1)} &= \frac{1}{5} \end{aligned}

Answers

(a)15−1n+2\dfrac{1}{5} - \dfrac{1}{n+2}(b)15\dfrac{1}{5}

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