1 problem found
Show that the sum \(S_N\) of the first \(N\) terms of the series $$\frac{1}{1\cdot2\cdot3}+\frac{3}{\cdot3\cdot4}+\frac{5}{3\cdot4\cdot5}+\cdots +\frac{2n-1}{n(n+1)(n+2)}+\cdots$$ is $${1\over2}\left({3\over2}+{1\over N+1}-{5\over N+2}\right).$$ What is the limit of \(S_N\) as \(N\to\infty\)? The numbers \(a_n\) are such that $$\frac{a_n}{a_{n-1}}=\frac{(n-1)(2n-1)}{(n+2)(2n-3)}.$$ Find an expression for \(a_n/a_1\) and hence, or otherwise, evaluate \(\sum\limits_{n=1}^\infty a_n\) when \(\displaystyle a_1=\frac{2}{9}\;\).
Solution: First notice by partial fractions: \begin{align*} \frac{2n-1}{n(n+1)(n+2)} &= \frac{-1/2}{n} + \frac{3}{n+1} + \frac{-5/2}{n+2} \\ &= \frac{-1}{2n} + \frac{3}{n+1} - \frac{5}{2(n+2)} \end{align*} And therefore: \begin{align*} \sum_{n = 1}^N \frac{2n-1}{n(n+1)(n+2)} &= -\frac12 \sum_{n=1}^N \frac1n +3\sum_{n=1}^N \frac1{n+1} -\frac52 \sum_{n=1}^N \frac1{n+2} \\ &= -\frac12-\frac14 + \frac{3}{2}+ \sum_{n=3}^N (3-\frac12 -\frac52)\frac1n + \frac{3}{N+1} - \frac{5}{2(N+1)} - \frac{5}{2(N+2)} \\ &= \frac12 \l \frac32+\frac1{N+1}-\frac{5}{N+2} \r \end{align*} As \(N \to \infty, S_N \to \frac{3}{4}\). \begin{align*} && \frac{a_n}{a_{n-1}}&=\frac{(n-1)(2n-1)}{(n+2)(2n-3)} \\ \Rightarrow && \frac{a_n}{a_1} &= \frac{a_n}{a_{n-1}} \cdot \frac{a_{n-1}}{a_{n-2}} \cdots \frac{a_2}{a_1} \\ &&&= \frac{(n-1)(2n-1)}{(n+2)(2n-3)} \cdot \frac{(n-2)(2n-3)}{(n+1)(2n-5)} \cdots \frac{(1)(3)}{(4)(1)} \\ &&&= \frac{(2n-1)3\cdot 2\cdot 1}{(n+2)(n+1)n} \\ &&& = \frac{6(2n-1)}{n(n+1)(n+2)} \end{align*} Therefore \(a_n = \frac{4}{3} \frac{2n-1}{n(n+1)(n+2)}\) and so our sequence is \(\frac43\) the earlier sum, ie \(1\)