Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
add comment

3 Answers

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

share|improve this answer
    
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
add comment

This addresses part a):

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.

share|improve this answer
add comment

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.

share|improve this answer
1  
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
1  
$d(x,y) = |x - y|$ –  ncmathsadist Apr 5 '13 at 18:47
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.