Proof of Divergence: (-1)^n Sequence

[solakis1]

Prove that the sequence :\(\displaystyle (-1)^n\) diverges by using the ε-definition of the limit of a sequence

 

[HallsofIvy]

Ok, what is that definition? Can you give an explicit statement of the definition?

 

[solakis1]

Re: Sequences

Ok, what is that definition? Can you give an explicit statement of the definition?

Yes.
[sp] A sequence of real Nos, \(\displaystyle x_n\), converges in \(\displaystyle R \) iff

There exists \(\displaystyle m\in R\) such that ,\(\displaystyle \forall\epsilon>0\),there exists a natural No \(\displaystyle k\),such that :

\(\displaystyle \forall n\geq k\Longrightarrow |x_n-m|<\epsilon\)

.......OR in complete formalization.......

A sequence of real Nos, \(\displaystyle x_n\), converges in \(\displaystyle R \) iff

\(\displaystyle \exists m\forall\epsilon[\epsilon>0\Longrightarrow\exists k(k\in N\wedge\forall n(n\geq k\Longrightarrow |x_n-m|<\epsilon))]\)

Hence for diversion one ​may use the negation of the above definition[/sp]

 

[HallsofIvy]

Re: Sequences

Very good!

Spoiler

So, use a proof by contradiction. If this sequence converges to, say, m, then, given any $\epsilon> 0$, there exist N such that, if n> N, $|a_n- m|< \epsilon$. Suppose $\epsilon$ were equal to, say, 1/2. Since for any odd n, $a_n= -1$, and for any even n, $a_n= 1$, no matter what N is there will exist some even number, $n_e$, and some odd number, $n_o$, both larger than N such that $|a_{n_e}- m|= |1- m|< 1/2$ and $|a_{n_0}- m|= |-1- m|< 1/2$.

The first of those says $-1/2< 1- m< 1/2$ so that $-3/2< m< -1/2$ and the second that $-1/2< -1- m< 1/2$ so that $-3/2< -m< -1/2$ and then $1/2< m< 3/2$. $-3/2< m< -1/2$ and $1/2< m< 3/2$ cannot both be true.

Is there any reason why this pretty standard "Calculus I" homework problem was posted in "Challenge Questions and Puzzles"?

 

[Euge]

Re: Sequences

Here is my solution.

Spoiler

Given a real number $L$, set $\epsilon = \max\{\lvert 1 - \lvert L\rvert\rvert,2\}$. Then $\epsilon > 0$. If $N$ is a positive integer, let $n$ be $2N+1$ if $L \neq -1$, and $2N$ if $L=-1$. Then $n > N$ and $\lvert (-1)^n - L\rvert \ge \epsilon$. Hence, the sequence $\{(-1)^n\}$ is divergent.

 

[solakis1]

Re: Sequences

Very good!

Spoiler

So, use a proof by contradiction. If this sequence converges to, say, m, then, given any $\epsilon> 0$, there exist N such that, if n> N, $|a_n- m|< \epsilon$. Suppose $\epsilon$ were equal to, say, 1/2. Since for any odd n, $a_n= -1$, and for any even n, $a_n= 1$, no matter what N is there will exist some even number, $n_e$, and some odd number, $n_o$, both larger than N such that $|a_{n_e}- m|= |1- m|< 1/2$ and $|a_{n_0}- m|= |-1- m|< 1/2$.

The first of those says $-1/2< 1- m< 1/2$ so that $-3/2< m< -1/2$ and the second that $-1/2< -1- m< 1/2$ so that $-3/2< -m< -1/2$ and then $1/2< m< 3/2$. $-3/2< m< -1/2$ and $1/2< m< 3/2$ cannot both be true.

Is there any reason why this pretty standard "Calculus I" homework problem was posted in "Challenge Questions and Puzzles"?

[sp] if n>N,then is either even or odd and not both
For example if N =100 ,all the n>N are:
101 or 102 or 103 or 104 or 105 or 106 e.t.c,e.t.c

And not ,101 and,102 ,e.t.c,e.t.c , because n can take only one value

For n even we have :|1-m|<1/2 => 1/2<m<3/2 or

For n odd we have :|1+m|<1/2 => -3/2<m<-1/2

Hence if n>N => n is even or odd => (1/2<m<3/2)or (-3/2<m<-1/2) which is true

From propositional logic:

(pv q ,p=>r,q=>t) => r v t.............1

put:

p=n is even
q= n is odd
r= 1/2<m<3/2
t= -3/2<m<-1/2

Then by using the above law we end up with : (1/2<m<3/2) or (-3/2<m<-1/2) [/sp]

 

[solakis1]

Re: Sequences

Here is my solution.

Spoiler

Given a real number $L$, set $\epsilon = \max\{\lvert 1 - \lvert L\rvert\rvert,2\}$. Then $\epsilon > 0$. If $N$ is a positive integer, let $n$ be $2N+1$ if $L \neq -1$, and $2N$ if $L=-1$. Then $n > N$ and $\lvert (-1)^n - L\rvert \ge \epsilon$. Hence, the sequence $\{(-1)^n\}$ is divergent.

[sp] For n=2N+1, \(\displaystyle L\neq -1\) ,then n>N and \(\displaystyle |(-1)^n-L|=|-1-L|=|1+L|\)

So ε must be set :ε =min{|1+L|,2}

I think.[/sp]

 

[Euge]

Re: Sequences

[sp] For n=2N+1, \(\displaystyle L\neq -1\) ,then n>N and \(\displaystyle |(-1)^n-L|=|-1-L|=|1+L|\)

So ε must be set :ε =min{|1+L|,2}

I think.[/sp]

Spoiler

There was a typo, but $\min\{\lvert 1 + L\rvert, 2\}$ is not a valid choice for $\epsilon$ since then $\epsilon = 0$ when $L = -1$. In any case, I meant to have $\epsilon = \epsilon(L) := \lvert 1 + L\rvert$ for $L \neq -1$ and $2$ for $L = -1$. It is then clear that for each positive integer $n$, $\lvert (-1)^{2n} - (-1)\rvert = 2 = \epsilon(-1)$ and $\lvert (-1)^{2n+1} - L\rvert = \lvert 1 + L\rvert = \epsilon(L)$ (when $L \neq -1$)

 

[solakis1]

Re: Sequences

Spoiler

There was a typo, but $\min\{\lvert 1 + L\rvert, 2\}$ is not a valid choice for $\epsilon$ since then $\epsilon = 0$ when $L = -1$. In any case, I meant to have $\epsilon = \epsilon(L) := \lvert 1 + L\rvert$ for $L \neq -1$ and $2$ for $L = -1$. It is then clear that for each positive integer $n$, $\lvert (-1)^{2n} - (-1)\rvert = 2 = \epsilon(-1)$ and $\lvert (-1)^{2n+1} - L\rvert = \lvert 1 + L\rvert = \epsilon(L)$ (when $L \neq -1$)

[sp] Given L,choose ε=min{|1-L|,2},then for any natular No L,put n=2N for L=-1 and n=2N+2 for \(\displaystyle L\neq -1\).

Then we have for the 1st case n>N and \(\displaystyle |(-1)^n-L|=|1-(-1)|=2\geq\epsilon\) and thus \(\displaystyle |(-1)^n-L|\geq\epsilon\)
For the 2nd case n>N and \(\displaystyle |(-1)^n-L| =|1-L|\geq\epsilon\) thus \(\displaystyle |(-1)^n-L|\geq\epsilon\)

On the other hand if we choose ε=2 or ε=|1+L| ,since we also have L=-1 or \(\displaystyle L\neq -1\) we have to examine 4 cases as dictated by propositional logic.

(pvq)and (rvt)=> (p and r)v(p and t)v (q and r) v (q and t),where:

p= (ε=2)
q= (ε= |1+L|)

r=(L=-1)

t= (\(\displaystyle L\neq -1\)) [/sp]