Sigma notation and summations

Question code: 12132

Question 0504

A sequence u1,u2,u3,…{u_1, u_2, u_3, \ldots} is such that un=1n{\displaystyle u_n = \frac{1}{n}}.

(i)

Show that un−1−un+1=2(n−1)(n+1){\displaystyle u_{ n - 1 } - u_{ n + 1 } = \frac{ 2 }{ ( n - 1 ) ( n + 1 ) }}.
[2]

(ii)

Hence find ∑n=4N2(n−1)(n+1){\displaystyle \sum_{ n = 4 }^N \frac{ 2 }{ ( n - 1 ) ( n + 1 ) }}.
[4]

(iii)

Give a reason why the series in part (ii) is convergent and state the sum to infinity.
[2]

(iv)

Use your answer to part (ii) to find ∑n=2N2(n+3)(n+1){\displaystyle \sum_{ n = 2 }^N \frac{ 2 }{ ( n + 3 ) ( n + 1 ) }}.
[2]
Answer

(ii)

712−1N−1N+1{\frac{7}{12} - \frac{1}{ N } - \frac{1}{ N + 1 }}.

(iii)

As N→∞,{N \to \infty,} 1N,1N+1→0{\frac{1}{ N }, \frac{1}{ N + 1 } \to 0} so 712−1N−1N+1→712.{\frac{7}{12} - \frac{1}{ N } - \frac{1}{ N + 1 }\to\frac{7}{12}.}
Hence the series is convergent and the sum to infinity is 712.{\frac{7}{12}.}

(iv)

712−1N+2−1N+3{\frac{7}{12} - \frac{1}{ N + 2 } - \frac{1}{ N + 3 }}.
Question code: 12132