Proof verification: sequence a_n=(−1)^n does not converge

  • #1
CGandC
279
23
Theorem: Show that the sequence ## a_n = (-1)^n ## for all ## n \in \mathbb{N}, ## does not converge.
My Proof: Suppose that there exists a limit ##L## such that ## a_n \rightarrow L ##. Specifically, for ## \epsilon = 1 ## there exists ## n_0 ## s.t. for all ## n > n_0## then ##|(-1)^n-L|<1## , from this we can also infer that ##|(-1)^{n+1}-L|<1## for all ## n > n_0##.
Let ## n > n_0 ## be arbitrary, then ## 2 = | (-1)^n - (-1)^{n+1} | \leq | (-1)^n - L | + | L - (-1)^{n+1} | < 1 + 1 = 2 ##, so since ## n > n_0 ## was arbitrary , hence for all ## n>n_0 ## we have 2<2 which is false, hence we get a contradiction.

My question: Regarding to when I wrote " for all ## n>n_0 ##, we have 2<2 which is false, hence we get a contradiction. "
I'm trying to understand with respect to what there is a contradiction; we get that any statement of the form ## \forall n>n_0. P(n)## is false in my proof ( since we got ## \forall n>n_0 . 2<2 ) ## [ and ## P(n) ## is a statement that depends on ## n ## ], and since we already have in the proof the statement " for all ## n > n_0## then ##|(-1)^n-L|<1## " , then we get a contradiction. Is this correct?

I mean, in my proof I have ## \forall n>n_0 . 2<2 ## ( which is false ) and ## \forall n>n_0.|(−1)^n+1−L|<1 ## ( which is true by the assumption that there exists a limit ) , hence I get a contradiction and thus the assumption that the limit exists is false. So I wanted to know if my proof was correct.
 

Answers and Replies

  • #2
36,243
13,294
The proof is fine.
If the series would converge then there needs to be an n0 such that 2<2 is true for all larger n. That's obviously wrong. 2<2 can't be true for any n.
 

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