A sequence ${{u}_{1}}$, ${{u}_{2}}$,${{u}_{3}}$, … is defined by \({{u}_{n}}=\sum\limits_{r=1}

Question

A sequence ${{u}_{1}}$, ${{u}_{2}}$,${{u}_{3}}$, … is defined by ${{u}_{n}}=\sum\limits_{r=1}^{n}{\left( 2r+n+1 \right)}$. Another sequence ${{v}_{1}}$, ${{v}_{2}}$, ${{v}_{3}}$, … is given by ${{v}_{n}}=\frac{2}{{{u}_{n}}}$, where $n\in {{\mathbb{Z}}^{+}}$. Find ${{u}_{n}}$ in terms of $n$.[2] Show that ${{v}_{n}}=\frac{1}{n}-\frac{1}{n+1}$.[1] Describe the behaviour of the sequence ${{v}_{1}}$, ${{v}_{2}}$, ${{v}_{3}}$… .[1] Find the sum, ${{S}_{N}}$, of the first $N$ terms of the sequence ${{v}_{1}}$, ${{v}_{2}}$, ${{v}_{3}}$… .[2] Give a reason why the series ${{S}_{N}}$ converges, and write down the value of the sum to infinity.[1]

Accepted Answer

(i) $u_n = 2n(n+1)$, (iii) The sequence ${{v}_{1}}$, ${{v}_{2}}$, ${{v}_{3}}$… decreases and converges to zero as $\frac{1}{n}\to 0$, and $\frac{1}{n+1}\to 0$ (iv) $S_N = 1 - \frac{1}{N+1}$, (v) $S_\infty = 1$

2021 RI Promo Q4

2021 RI Promo Q4

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