Step 1

Take sequence \(\displaystyle{x}_{{{n}}}={\left(-{1}\right)}^{{{n}}}\) where 1 occurs infinitely many times but does not converge to 1.

This is because if we choose \(\displaystyle\epsilon\le{2}\), given any \(\displaystyle{N}\in{\mathbb{{{N}}}}\), where exists \(\displaystyle{n}={2}{N}+{1}\) such that

\(\displaystyle{\left|{\left({1}\right)}^{{{n}}}-{1}\right|}={2}\geq\epsilon\)

Step 2

b) Suppose \(\displaystyle{x}_{{{n}}}\) is a convergent sequence with infinitely many 1 but converges to \(\displaystyle{x}\ne{1}\). Let \(\displaystyle\epsilon={\left|{1}-{x}\right|},\) then for any \(\displaystyle{N}\in{\mathbb{{{N}}}}\), there exist \(\displaystyle{n}\geq{N}\) such that \(\displaystyle{x}_{{{n}}}={1}\). So

\(\displaystyle{\left|{x}_{{{n}}}-{x}\right|}={\left|{1}-{x}\right|}\geq\epsilon\)

So this is not possible

Step 3

c) Consider a sequence \(\displaystyle{\left({x}_{{{n}}}\right)}\) given by

\(x_{n}=\begin{cases}r & if\ n=\frac{r(r+1)}{2}-1\ \text{for some}\ r\in\mathbb{N}\\ 1 & \text{otherwise}\end{cases}\)

(or put any thing in place of r other than 1)

There n consecutive 1's occur in the sequence. Some of the terms in sequence are

\(\displaystyle{\left({x}_{{{n}}}\right)}_{{{n}\in{\mathbb{{{N}}}}}}={\left({1},\ {2},\ {1},\ {1},\ {5},\ {1},\ {1},\ {1},\ {9},\ {1},\ {1},\ {1},\ {1},\ {14},\ \cdots\right)}\)

This is constructed this way

\(\begin{matrix} 1 & 2 \\ 1 & 1 & 5 \\ 1 & 1 & 1 & 9 \\ \cdots\end{matrix}\)