# Definition of “not converging” and proving $(-1)^n$ does not converge to $1$.

Remember that a sequence $x_n, n = 1,2,3\cdots$ is said to converge to $x$ as $n → ∞$ if for all $ε > 0$ there exists an $N ∈ \mathbb{N}$ such that $|x_n − x| < ε$ for all $n ≥ N$.

(a) Complete the following statement: “If the sequence $x_n, n = 1,2,3\cdots$ does not converge to $x$ as $n → ∞$, that means that there exists an $ε > 0$ such that...”

(b) Consider the sequence $x_n = (−1)^n, n = 1,2,3\cdots$ that is, the sequence is $(−1,1,−1,1,−1,...)$. Prove carefully, starting from your answer to part (a), that this sequence does not converge to 1.

I am confused with the first part and what epsilon represents!

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Convergence is stated as

"For every $\epsilon >0$ there exists a natural number $N$ such that $n\geq N$ implies $|x-x_n|<\epsilon$"

You might write it as $$\forall\epsilon >0\;\;\exists N\in\Bbb N \;\;\forall n\geq N \text{ we have } |x_n-x|<\epsilon$$

Now, we need to think, when can the above be false? We need just a "counterexample", that is, an $\epsilon >0$ for which no $N$ will every $|x-x_n|<\epsilon$, even though we make $n\geq N$. We might write this as

"There exists an $\epsilon >0$ such that for every natural number $N$, there exists an $n\geq N$ with $|x-x_n|\color{red}{\geq} \epsilon$."

Can you try and prove why $(-1)^n\not\to 1$? Hint: Take $\epsilon =1/2$ in the defintion.

ADD Alternatively, we can think about convergence as follows. Let's define the set

$$B(x,\epsilon)=\{y\in\Bbb R:|x-y|<\epsilon\}$$

This is usually called "the open ball with center $x$ and radius $\epsilon$. In $\Bbb R$ it is an open interval $(x-\epsilon,x+\epsilon)$, but in $\Bbb R^2$ it is a disk (with the Euclidean metric) and in $\Bbb R^3$ is a ball (a filled sphere). Now, we may state convergence as follows.

DEF Let $\langle x_n:n\in\Bbb N\rangle$ be a sequence in $\Bbb R$. Let $x\in \Bbb R$. We say that $x_n$ converges to $x$ if for each ball $B(x;\epsilon)$ we're given, there exists an $N$ such that the tail sequence

$$\langle x_n:n\geq N\rangle=\langle x_N,x_{N+1},\dots\rangle$$

is contained entirely in $B(x,\epsilon)$.

This definition helps in the sense that we can see convergence fails when we can find some $\epsilon>0$ such that no matter which "tail" ($N$ big) we take, some element of it will fail to be inside the ball $B(x;\epsilon)$. This directly generalizes to $\Bbb R^n$ with $$\|{\bf x}-{\bf y}\|:=\left(\sum_{i=1}^n (x_i-y_i)^2\right)^{1/2}$$

and $$B({\bf x};\epsilon):=\{{\bf y}\in\Bbb R^n:\|{\bf x}-{\bf y}\|<\epsilon\}$$

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Unfortunately no, but I can prove it with a graph showing that it does not converge but they require me to use part (a) in order to show this. Thank you. – Achchu Apr 5 '13 at 15:26
@Achchu You don't really have many choices. Take $N=23984$. Note that $$|1-x_n|$$ in $x_n=(-1)^n$ is either $0$ or $2$; yes? So if you take $\epsilon =1/2$, for example, you will have $|x-x_n|=2\geq 1/2$ for every odd greater that $N=23984$. So you will have just found an element that escapes your interval $(1-1/2,1+1/2)=(1/2,3/2)$ – Pedro Tamaroff Apr 5 '13 at 15:31
I'm not understanding why you took N=23984? – Achchu Apr 5 '13 at 15:35
@Achchu I was just giving you one example. The idea is that you can do it for any $N$. If $N$ is even, the $a_{N+1}=-1$ wont be in the interval. If $N$ is odd then $a_N=-1$ itself won't be in the interval. So you have just proven that there exists an $\epsilon>0$ which is $=1/2$, such that for no $N$, the tail sequence $x_N,x_{N+1},\dots$ is entirely contained in the interval. That is, you have shown the sequence doesn't converge to $1$. – Pedro Tamaroff Apr 5 '13 at 15:37

Logical Preliminaries

A biconditional $P\Leftrightarrow Q$ ($P$ iff $Q$) is equivalent to its contrapositive $\sim Q\Leftrightarrow\sim P$ (not $Q$ iff not $P$).

The negation of a universal $\sim\forall x,P$ (it is not the case that for all $x$, $P$) is $\exists x:\sim P$ (there is an $x$ such that not $P$).

The negation of an existential $\sim\exists x:P$ (it is not the case that there is an $x$ such that $P$) is $\forall x,\sim P$ (for all $x$, not $P$).

Application

The original statement is $$\left(\forall\epsilon\gt0,\exists N\in\mathbb{N}:\forall n\ge N,|x_n-x|\lt\epsilon\right)\iff\text{x_n converges to x as n\to\infty}$$ Its contrapositive is $$\text{x_n does not converge to x as n\to\infty}\iff\left(\exists\epsilon\gt0:\forall N\in\mathbb{N},\exists n\ge N:|x_n-x|\ge\epsilon\right)$$

Part b) is to apply the forgoing to show non-convergence.

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Here is (a).

(i.o means "infinitely often")

$$\sim \exists x \forall \epsilon > 0 \exists N {\rm s.t.} n\ge N \Rightarrow d(x_n, x) < \epsilon \iff \forall x \exists \epsilon > 0\; {\rm s.t.}\; d(x_n, x) \ge \epsilon\; {\rm i.o}$$

More digestibly, for any real $x$ there is an open interval $I$ containing $x$ so that $x_n\not\in I$ infinitely often.

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It appears he's doing it in $\mathbb{R}^1$ so it's possible he won't understand the "metric notation" – DanZimm Apr 5 '13 at 15:26
$d(x,y) = |x - y|$ – ncmathsadist Apr 5 '13 at 18:47