N2007 P2 Q2 [Modified]

N2007 P2 Q2 [Modified]

6 marks
Past Year National Exams
Free

A sequence \({{u}_{1}}\), \({{u}_{2}}\), \({{u}_{3}}\), … is such that \({{u}_{1}}=1\) and

\({{u}_{n+1}}={{u}_{n}}-\frac{2n+1}{{{n}^{2}}{{\left( n+1 \right)}^{2}}}\), for all \(n\ge 1\).

  1. Given that \({{u}_{n}}=\frac{1}{{{n}^{2}}}\), find \(\sum\limits_{n=1}^{N}{\frac{2n+1}{{{n}^{2}}{{\left( n+1 \right)}^{2}}}}\).[2]
  2. Give a reason why the series in part (i) is convergent and state the sum to infinity.[2]
  3. Use your answer to part (i) to find \(\sum\limits_{n=2}^{N}{\frac{2n-1}{{{n}^{2}}{{\left( n-1 \right)}^{2}}}}\).[2]

Video Solution:

Default solution

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Answer:(i) \(1-\frac1{(N+1)^2}\); (ii) convergent with sum \(1\); (iii) \(1-\frac1{N^2}\)

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