All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(a)
(i)
Let P(n) be the preposition, $p_{n}=\frac{1}{3}(7-4^{n})$ for $n\in {\mathbb{Z}}^{+}$
When $n=1, \mathrm{LHS}=p_{1}=1$ and $\mathrm{RHS}=\frac{1}{3}(7-4)=1$
Since $\mathrm{LHS}=\mathrm{RHS}$, P(1) is true.
Assume P(k) is true for some $k \in {\mathbb{Z}}^{+}$.
To prove that P(K+1) is true, i.e., $p_{k+1}=\frac{1}{3}(7-4^{k+1})$
$\mathrm{LHS}$
$=p_{k+1}$

$=4p_{k}-7$

$=4(\frac{1}{3}(7-4^{k}))-7$

$= \frac{1}{3}(4\times{7}-4\times{4^{k}}-21)$

$= \frac{1}{3}(7-4^{k+1})$
$= \mathrm{RHS}$
Therefore, P(k) is true $\Rightarrow$ P(k+1) is also true.
Since P(1) is true and P(k) is true $\Rightarrow$ P(k+1) is true, by principle of mathematical induction, P(n) is true for all $n\in {\mathbb{Z}}^{+}$.

(ii)
$\sum_{r=1}^{n} p_{r}$

$= \sum_{r=1}^{n} {\frac{1}{3}}(7-4^{r})$

$= \frac{1}{3}[\sum_{r=1}^{n}7 - \sum_{r=1}^{n} 4^{r}]$

$= \frac{1}{3}\times{7n} - \frac{1}{3}[\frac{4(4^{n}-1)}{4-1}]$

$= \frac{7n}{3} - \frac{4}{9}(4^{n}-1)$

(b)
(i)
As $n\rightarrow \infty, \frac{1}{(n+1)!} \rightarrow 0$
Therefore, the series converges and the $\mathrm{Sum~to~Infinity}=1$.

(ii)
$u_{n}=S_{n}-S_{n-1}$

$u_{n}=1-\frac{1}{(n+1)!}-(1-\frac{1}{n!})$

$u_{n}=\frac{-1}{(n+1)!}+\frac{1}{n!}$

$u_{n}=\frac{-1+n+1}{(n+1)!}$

$u_{n}=\frac{n}{(n+1)!}$

Convergence of a series can be proved in a few ways, the given is the most direct, which is to show that the sum to infinity is a finite value. Question wants students to write down, so please write it explicitly. It is surprising that some students confuse the formulas and write $u_{n}=S_{n+1}-S_{n}$, this is totally wrong.