1. ## Proving Divergence

The definition of a convergent squence is:
For every L in Real numbers, and all E>0 such that there exists N in Natural numbers n>/= N implies |xn-L|>E.
Negate this statement then use the negation to prove that the sequence xn=(-1)^n is not convergent.

I got the negation I think, but I'm not sure how to begin with the proof. We went over one example in class and it was for proving something converges. I got the negation as :
For all L in real numbers, there exists E>0 such that there exits N in natural numbers and n>/=N implies |xn-L|<E

2. ## Re: Proving divergence

Your definition of convergence is not correct. It should be

There exists L in Real numbers such that for all E>0 such that there exists N in Natural numbers such that n>/= N implies |xn-L|< E.

The negation of the definition of convergence should be: For every real number $\displaystyle L$ there exists a real number $\displaystyle \varepsilon>0$ such that for every natural number $\displaystyle N,$ there is a natural number $\displaystyle n\geqslant N$ such that $\displaystyle \left|x_n-L\right|\geqslant\varepsilon.$ Use this to prove the divergence of $\displaystyle \left(x_n\right)_{n=1}^\infty$ where $\displaystyle x_n=(-1)^n.$ Hint: Given $\displaystyle L,$ let $\displaystyle \varepsilon=\min\left\{|L+1|,|L-1|\right\}$ if $\displaystyle L\ne\pm1$ and $\displaystyle \varepsilon=2$ if $\displaystyle L=\pm1.$

3. ## Re: Proving Divergence

I don't understand how you are proving the negation. We just used the pieces of the definition to prove convergence in class. Is that what you are doing here?

4. ## Re: Proving Divergence

Originally Posted by renolovexoxo
I don't understand how you are proving the negation. We just used the pieces of the definition to prove convergence in class. Is that what you are doing here?

Is this true: $\displaystyle \left( {\forall n} \right)\left[ {\left| {{x_{n + 1}} - {x_n}} \right| = 2} \right]~?$
If that is true, how can the terms of the sequence $\displaystyle {\left( { - 1} \right)^n}$ get "close together"?

5. ## Re: Proving Divergence

I'm getting lost. I feel like there is something there that I'm not understanding.

6. ## Re: Proving Divergence

Originally Posted by renolovexoxo
I'm getting lost. I feel like there is something there that I'm not understanding.
I think that you need a live-sit-down with an instructor/lecturer.
We here just not equipped to aid with this kind profound confusion.