# How do you do this?

Jul 17, 2018

See below:

#### Explanation:

The sequence is

$1 , 0 , - 1 , 0 , 1 , 0 , - 1 , 0 , \ldots$

Although it is rather clear from inspection that this sequence does not converge, a more formal proof can be given by several different approaches.

A more advanced approach would be to notice that the series has sub-sequences that converge to different limits (namely $1$, $- 1$, and $0$) - and this shows that the sequence proper has no limit.

A direct approach from the definition uses a proof by contradiction. We assume that a limit $L$ exists. This implies that

$\setminus \forall \epsilon > 0 , \setminus \exists N \in \mathbb{N} \setminus : \setminus n > N \implies | {a}_{n} - L | < \epsilon$

Since $1 \in \left\{{a}_{n} | n > N\right\}$, we have $| 1 - L | < \epsilon$
Similarly $- 1 \in \left\{{a}_{n} | n > N\right\} \implies | - 1 - L | < \epsilon$

Thus

$2 = | 1 - \left(- 1\right) | = | \left(1 - L\right) - \left(- 1 - L\right) |$
$q \quad \le | 1 - L | + | - 1 - L | < 2 \epsilon$

Thus, we end up with $\forall \epsilon > 0 , \epsilon > 1$ - which is obviously a contradiction. Hence the limit does not exist.