Negation of definition of convergence

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SUMMARY

The discussion centers on the negation of the definition of convergence in mathematical analysis. The original definition states that for every ε > 0, there exists an N such that |x_n - l| < ε for all n > N. The negation is correctly identified as there exists ε > 0 such that for all N, there exists n > N where |x_n - l| ≥ ε. The sequence x_n = (-1)^n n is proven to be divergent by demonstrating that it cannot converge to any limit L, as shown through specific ε choices and the behavior of the sequence.

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  • Understanding of limits and convergence in real analysis.
  • Familiarity with sequences and their properties.
  • Knowledge of mathematical notation and symbols used in analysis.
  • Ability to manipulate inequalities and understand ε-N definitions.
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  • Study the implications of the ε-N definition of convergence in real analysis.
  • Learn about divergent sequences and their characteristics.
  • Explore the concept of limits and continuity in mathematical functions.
  • Investigate proofs involving sequences and series in calculus.
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The definition of convergence is given by : ## \forall \epsilon > 0, \exists N \in \mathbb{R} ## such that ## |x_n - l | < \epsilon ## ## \forall n \in \mathbb{N} ## with ## n > N ##

negate this statement and prove that the sequence ## x_n = (-1)^nn ## is divergent using only the negation of the definition of convergence

ok here's my attempt of the negation of the statement:

## \exists \epsilon > 0 \forall N \in \mathbb{R} ## s.t. ## \exists n \in \mathbb{N} n > N ## and ## |x_n - l | \geq \epsilon ##

proof that ## x_n = (-1)^nn ## does not converge:

since the negation of the definition is "## \forall N ##" let N = 1, and find ## n_1 > 1 ## then let ## N = n_1 ## and find ## n_2 > N=n_1 ## etc leaving us with a sequence of indicies ## n_1 < n_2 < n_3.. ##

so let's prove that it does not converge to 1:

choose ## epsilon = 1/2 ## then for ## N>0 ## let n be the odd integers i.e. ## n_1 = 1 < n_2 = 3 < n_3 = 5 ... ## then n > N hence ## | (-1)^1(1) - 1 | >= 1/2 ## which is true hence ## x_n## does not converge to 1

take ## l > 1 ## and choose ##\epsilon ## s.t. ## \epsilon < l - 1 ## again, choosing ## N > 0 ## and n being all the odd integers greater than 0 we have n > N and ## |(-1)(1) - l | \geq \epsilon## which is true

take ## l < 1## and choose n to be all the odd integers above 0 then n > N and ## |-1 - l | >= \epsilon ## which is true therefore ## x_n ## does not converge

I think this works, but truth be told, I don't understand the negation of this statement. I mean, I understand HOW I got the negation, but I don't know what it means. The choosing N > 0 is from my lecture notes and I don't understand why that works, and why we can just choose N > 0 if the definition states for all N

any help please
 
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You are on the right track but I think it is a little simpler. Given the sequence ##a_n = n(-1)^n## you want to show it cannot have a limit. If it does have a limit L then for any ##\epsilon## there is an ##N## such that |##a_n -L##| whenever n > N.

Choosing ##\epsilon## = 1/2 is a great idea. Now what happens to ##a_n## as n gets large? How does each term relate to the previous one?

So you want to say something like: no matter what N I choose I can't get |##a_n -L##| < 1/2 for every n > N. Then explain why not.

The positive statement is I can find an N; the negation is I cannot find an N.
 

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