The sum of the first \(n\) terms of a sequence, \(u_r\), is given by \(\sum_{r=1}^{n} u_r = 3 - \frac{n-1}{(n+1)!}\) where \(n \ge 1\).
Explain why \(\sum_{r=1}^{\infty} u_r\) is a convergent series.[1]
Find \(u_n, n \ge 2\) in terms of \(n\), expressing your answer as a single fraction.[2]
Determine whether \(\sum_{r=4}^{n} u_r < \frac{1}{12}\) is true for all \(n \ge 4\).\n\nShow your working clearly.[2]
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Answer:(a) convergent, with sum (3) (b) (u_n=\dfrac{n^2-2n-1}{(n+1)!}) (c) (\displaystyle\sum_{r=4}^{n}u_r=\dfrac1{12}-\dfrac{n-1}{(n+1)!}<\dfrac1{12})