A sequence ${{u}_{1}},{{u}_{2}},{{u}_{3}},...$ is such that ${{u}_{n}}=\frac{1}{{{\left( 2n+1 \right

Question

A sequence ${{u}_{1}},{{u}_{2}},{{u}_{3}},...$ is such that ${{u}_{n}}=\frac{1}{{{\left( 2n+1 \right)}^{2}}}$. By considering ${{u}_{n-1}}-{{u}_{n}}$, find $\sum\limits_{n=2}^{N}{\frac{n}{{{\left( 2n-1 \right)}^{2}}{{\left( 2n+1 \right)}^{2}}}}$.[4] Give a reason why the series in part (i) is convergent and state the sum to infinity.[2] Use your answer to part (i) to find $\sum\limits_{n=3}^{N+2}{\frac{n-1}{{{\left( 2n-1 \right)}^{2}}{{\left( 2n-3 \right)}^{2}}}}$[3]

Accepted Answer

(i) $\frac18\left(\frac19-\frac1{(2N+1)^2}\right)$ (ii) $\frac1{72}$ (iii) $\frac18\left(\frac19-\frac1{(2N+3)^2}\right)$

2017 PJC MYCT P1 Q2

2017 PJC MYCT P1 Q2

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