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Summation and series: CJC 2024 Promo Q3

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Catholic Junior College2024 PromoPromo Q3●●○ Standard5 marks

What this question tests

Summation and series.

Question

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

Find $\displaystyle\sum_{r=5}^{n+1} \frac{1}{r(r+1)}$ . [3]
Give a reason why the series in part (a) is convergent and state the value of $\displaystyle\sum_{r=5}^{\infty} \frac{1}{r(r+1)}$ . [2]

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(a)

$\begin{aligned} \sum_{r=5}^{n+1} \frac{1}{r(r+1)} &= \sum_{r=1}^{n+1} \frac{1}{r(r+1)} - \sum_{r=1}^{4} \frac{1}{r(r+1)}\\ &= \left(1 - \frac{1}{(n+1)+1}\right) - \left(1 - \frac{1}{4+1}\right)\\ &= \frac{1}{5} - \frac{1}{n+2}. \end{aligned}$

(b)

As $n\to\infty$ , $\dfrac{1}{n+2}\to0$ . $\begin{aligned} \therefore\sum_{r=5}^{\infty} \frac{1}{r(r+1)} &= \frac{1}{5} \end{aligned}$

Answers

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

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