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−3n2+3n+1n3(n+1)3,for all n≥1,u_{n+1} = u_n - \frac{3 n^2 + 3 n + 1}{n^3(n + 1)^3}, \quad \textrm{for all } n \geq 1,
and un=1n3.{\displaystyle u_n = \frac{1}{n^3}.}
(i)
Find ∑n=2N3n2+3n+1n3(n+1)3.{\displaystyle \sum_{n=2}^N \frac{3 n^2 + 3 n + 1}{n^3(n + 1)^3}.}
[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=0N3n2+15n+19(n+2)3(n+3)3.{\displaystyle \sum_{n=0}^N \frac{3 n^2 + 15 n + 19}{(n + 2)^3(n + 3)^3}.}
[2]

Answer

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