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=un2n1(n1)2n2,for all n1,u_{n+1} = u_n - \frac{2 n - 1}{(n - 1)^2n^2}, \quad \textrm{for all } n \geq 1,
and un=1(n1)2.{\displaystyle u_n = \frac{1}{(n - 1)^2}.}
(i)
Find n=2N2n1(n1)2n2.{\displaystyle \sum_{n=2}^N \frac{2 n - 1}{(n - 1)^2n^2}.}
[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=0N2n+3(n+1)2(n+2)2.{\displaystyle \sum_{n=0}^N \frac{2 n + 3}{(n + 1)^2(n + 2)^2}.}
[2]

Answer