N2023 P1 Q5
N2023 P1 Q5
Junior College 2
7 marks
Free
- Use the method of differences to show that \(\displaystyle\sum_{r=2}^{n}\ln\left[\frac{(r-1)(r+1)}{r^2}\right]=\ln\left(\frac{n+1}{n}\right)-\ln 2\).[3]
- Show that the corresponding infinite series is convergent and state the sum to infinity.[2]
- Show that \(\displaystyle\sum_{r=10}^{20}\ln\left[\frac{(r-1)(r+1)}{r^2}\right]=\ln\left(\frac{a}{b}\right)\), where \(a\) and \(b\) are integers to be found.[2]
Default solution
- For \(r\ge2\), all logarithm arguments below are positive.\(\ln\left[\dfrac{(r-1)(r+1)}{r^2}\right]\)\({}=\ln\left(\dfrac{r+1}r\right)-\ln\left(\dfrac r{r-1}\right)\)Let \(a_r=\ln\left(\dfrac{r+1}r\right)\). Method of differences:\(S_n\)\({}=(a_2-a_1)+(a_3-a_2)+\cdots+(a_n-a_{n-1})\)\(S_n\)\({}=a_n-a_1\)\(\therefore S_n\)\({}=\ln\left(\dfrac{n+1}n\right)-\ln2\)Intermediate logarithms cancel. (shown)
- \(\lim_{n\to\infty}S_n\)\({}=\lim_{n\to\infty}\ln\left(1+\dfrac1n\right)-\ln2\)\(\lim_{n\to\infty}S_n\)\({}=-\ln2\)The partial sums have a finite limit, so the infinite series converges, with sum \(-\ln2\). (shown)
- \(\sum_{r=10}^{20}\ln\left[\dfrac{(r-1)(r+1)}{r^2}\right]\)\({}=S_{20}-S_9\)\(S_{20}-S_9\)\({}=\left[\ln\left(\dfrac{21}{20}\right)-\ln2\right]-\left[\ln\left(\dfrac{10}9\right)-\ln2\right]\)\(S_{20}-S_9\)\({}=\ln\left(\dfrac{21}{20}\times\dfrac9{10}\right)=\ln\left(\dfrac{189}{200}\right)\)\(\therefore a=189\) and \(b=200\). (shown)
Answer:(a) \(S_n=\ln\left(\dfrac{n+1}n\right)-\ln2\) (b) Convergent; sum \(-\ln2\) (c) \(a=189, b=200\)
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