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 $=1$
(i) $u_n=\sum_{r=1}^n(2 r+n+1)=\frac{n}{2}[n+3+3 n+1]=2 n(n+1)$
(ii) $v_n=\frac{2}{u_n}=\frac{1}{n(n+1)}=\frac{(n+1)-n}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$ (shown)
(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) $S_N=\sum_{n=1}^N\left(\frac{1}{n}-\frac{1}{n+1}\right)$

$=1-\frac{1}{N+1}$