RI/2021/JC1/Promo/P1/Q04

A sequence $u_1, u_2, u_3, \ldots$ is defined by

$u_n=\sum_{r=1}^n(2 r+n+1)$

Another sequence $v_1, v_2, v_3, \ldots$ is given by $v_n=\frac{2}{u_n}$, where $n \in \mathbb{Z}^{+}$.
(i) Find $u_n$ in terms of $n$.

[2]

(ii) Show that $v_n=\frac{1}{n}-\frac{1}{n+1}$.

[1]

(iii) Describe the behaviour of the sequence $v_1, v_2, v_3, \ldots$

[1]

(iv) Find the sum, $S_N$, of the first $N$ terms of the sequence $v_1, v_2, v_3, \ldots$

[2]

(v) Give a reason why the series $S_N$ converges, and write down the value of the sum to infinity.

[2]

(i) $2 n(n+1)$

(iii) The sequence $v_1, v_2, v_3, \ldots$ decreases and converges to zero as $\frac{1}{n} \rightarrow 0$, and $\frac{1}{n+1} \rightarrow 0$

(iv) $1-\frac{1}{N+1}$

(v)

As $N \rightarrow \infty, S_N=\sum_{n=1}^N v_n=1-\frac{1}{N+1} \rightarrow 1$, since $\frac{1}{N+1} \rightarrow 0$. Thus, the series $S_N$ converges.
Sum to infinity of the series $=1$

(v)

As $N \rightarrow \infty, S_N=\sum_{n=1}^N v_n=1-\frac{1}{N+1} \rightarrow 1$, since $\frac{1}{N+1} \rightarrow 0$. Thus, the series $S_N$ converges.
Sum to infinity of the series $=1$