Question 0507

Sigma Notation
2007 Paper 2 Question 2 Variant

Question

A sequence u1,u2,u3,…u_1, \allowbreak u_2, \allowbreak u_3, \ldots is such that
un+1=un−1n(n+1),for all n≥1,u_{n+1} = u_n - \frac{1}{n(n+1)}, \quad \textrm{for all } n \geq 1,
and un=1n.{\displaystyle u_n = \frac{1}{n}.}
(i)
Find ∑n=1N1n(n+1).{\displaystyle \sum_{n=1}^N \frac{1}{n(n+1)}.}
[2]
(ii)
Give a reason why the series in part (i) is convergent and state the sum to infinity.
[2]
(iii)
Use your answer to part (i) to find ∑n=2N1n(n−1).{\displaystyle \sum_{n=2}^N \frac{1}{n(n - 1)}.}
[2]

Answer

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