Using the method of differences, find $\sum\limits_{r=1}^{n}{\frac{1}{r\left( r+1 \right)}}$.[3] H

Question

Using the method of differences, find $\sum\limits_{r=1}^{n}{\frac{1}{r\left( r+1 \right)}}$.[3] Hence find $\sum\limits_{r=1}^{n}{\left[ {{3}^{-r}}-\frac{1}{r\left( r+1 \right)} \right]}$.[3] Use your results in part (i) to show $\sum\limits_{r=3}^{2N}{\left[ {{3}^{1-r}}-\frac{1}{r\left( r-1 \right)} \right]}=-\frac{1}{3}+\frac{1}{2}\left( \frac{1}{N}-\frac{1}{{{3}^{2N-1}}} \right)$[3] Hence find $\sum\limits_{r=3}^{\infty }{\left[ {{3}^{1-r}}-\frac{1}{r\left( r-1 \right)} \right]}$.[1]

Accepted Answer

(i) $\frac{1}{n+1} - \frac{1}{2(3^n)} - \frac{1}{2}$; (ii) $-\frac{1}{3}$

2018 JJC P2 Q3

2018 JJC P2 Q3

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