My proof of divergence of $(-1)^n$

Question

Should I shorten my proof? (Also, should I try to prove without contradiction?)

We consider the sequence $(x_n)_{n \in \mathbb{N}}$, where $x_n = (-1)^n$.

$\textbf{Lemma.}$ For every element $x_n$ of the sequence $(x_n)$, we have $|x_n| = 1$. (We could prove this by induction on $n$.)

$\textbf{Theorem.}$ $(x_n)$ diverges.

$\textit{Proof.}$ We prove the theorem by contradiction. To that end, we assume that $(x_n)$ is not divergent, i.e. we assume that it is convergent. With that said, we are done as soon as a contradiction is deduced. By assumption, there is an $x \in \mathbb{R}$ such that \begin{equation*} \forall \varepsilon \in \mathbb{R}, \varepsilon > 0 : \exists N \in \mathbb{N} : \forall n \in \mathbb{N}, n > N : |x_n - x| < \varepsilon . \end{equation*} We choose $\varepsilon = 1$. By assumption, there is an $N \in \mathbb{N}$ such that \begin{equation*} \forall n \in \mathbb{N}, n > N : |x_n - x| < 1 . \end{equation*} We choose $n = N + 1$. Hence, both $|x_n - x| < 1$ and $|x_{n + 1} - x| < 1$. Thus, \begin{equation*} |x_{n + 1} - x| + |x_n - x| < 2 . \end{equation*} Moreover, \begin{equation*} \begin{split} 2 & = |2| \\ & = |2| \cdot 1 \\ & = |2| \cdot |x_{n + 1}| && | \text{ by Lemma} \\ & = |2 x_{n + 1}| && | \text{ by multiplicativeness of abs. val.} \\ & = |x_{n + 1} + x_{n + 1}| \\ & = |x_{n + 1} + (-1)x_{n}| \\ & = |x_{n + 1} - x_{n}| \\ & = |x_{n + 1} + 0 - x_{n}| \\ & = |x_{n + 1} + (-x + x) - x_{n}| \\ & = |(x_{n + 1} - x) + (x - x_{n})| \\ & \le |x_{n + 1} - x| + |x - x_{n}| && | \text{ by subadditivity of abs. val.} \\ & = |x_{n + 1} - x| + |x_{n} - x| \qquad && | \text{ by evenness of abs. val.} \\ \end{split} \end{equation*} Hence, by transitivity, we have $2 < 2$. Obviously, we deduced a contradiction. QED

Answer 1

The shortest proof of this would be to show that the sequence is not Cauchy, therefore cannot converge. Let $\epsilon =1$. Assume for contradiction sake that the sequence is cauchy, therefore there exists an $N$ such that for all $m\ge N,n\ge N,|a_n-a_m|<1$. Thus, taking $n=N,m=N+1$, we have $|(-1)^N-(-1)^{N+1}|<1$, but this is false, since this number is actually 2 (Since $N$ and $N+1$ have opposite parity). Therefore it's not cauchy, hence it does not converge

Answer 2

You can use the following theorem: "If a sequence converges, then every subsequence converges to the same limit".

Answer 3

I do not know how much you know about sequences but the shortest proof of the divergence of this sequence is that the two subsequences $x_{2n}$ and $x_{2n+1}$ that are constant therefore convergent do not converge to the same limit

Answer 4

I think there are better ways to prove it.

Proposition A: If $x_n$ converges, then the sequence of differences $x_{n+1}{-}x_n$ converges to $0$.

Proposition B: If $x_n$ has subsequences $a_n$, $b_n$ which converge to different limits, then $x_n$ doesn't converge.

Either of these two immediately proves it and their proofs are no harder than what you are trying to do.

Take e.g. the first: For all $\varepsilon>0$ there's $N$ such that for all $n\ge N$, $|x_n-L|<\varepsilon/2$, so for all $n\ge N$, $$|x_{n+1}-x_n|=|(x_{n+1}-L)-(x_n-L)|\le |x_{n+1}-L|+|x_n-L|\le\varepsilon/2+\varepsilon/2=\varepsilon.$$ Done!

Answer 5

To give yet another answer - for any $n$, we have $\sup_{k\geqslant n} x_n=1$, so that $\limsup_{n\to\infty}x_n = 1$, while $\inf_{k\geqslant n} x_n=-1$, so that $\liminf_{n\to\infty}x_n=-1$. Since $\limsup_{n\to\infty}x_n\neq\liminf_{n\to\infty} x_n$, the sequence does not converge.

Answer 6

I would shorten the proof.

Pick any $L \in \mathbb{R}$.

If $L \ge 0$, then we have $|x_n-L| \ge 1$ for all odd $n$. Hence $x_n$ does not converge to $L$.

If $L < 0$, then we have $|x_n-L| \ge 1$ for all even $n$. Hence $x_n$ does not converge to $L$.

Answer 7

We shall assume that the sequence $\{x_n\}$ converges and reach a contradiction. Let $L$ denote the number the sequence converges to, and set $\hat{\varepsilon}=\frac{1}{2}$. By definition there is $\hat{n} \in \mathbb{N}$ so that $|x_n-L|<\hat{\varepsilon}$ whenever $n \geq \hat{n}$. Hence $|x_{\hat{n}}-x_{\hat{n}+1}| \leq |x_{\hat{n}}-L|+|L-x_{\hat{n}+1}|<1$. This is a contradiction as clearly $|x_{\hat{n}}-x_{\hat{n}+1}|=2$. Therefore the sequence does not converge.

Answer 8

We have

$$ x_{2n} = (-1)^{2n} = 1 \to 1,$$

and

$$ x_{2n+1} = (-1)^{2n+1} = (-1)(-1)^{2n} = -1 \to -1.$$

Therefore, $x_n$ is divergent.