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Prove that the sequence ${(−1)^n}$ does not converge.

  • I know how to prove that the sequence diverges by contradiction, supposing the sequence converges to L. Then choosing $ε = 1$ such that $|(-1)^n - L| < 1$. Noting

$|(-1)^{2n} - L|$

$|1-L|<1$

$-1<1-L<1$

$-2<-L<0$

and

$|(-1)^{2n+1} - L|$

$|-1-L|<1$

$-1<-1-L<1$

$0<-L<2$

Hence, -L is bounded by both -2 and 0 and also by 0 and two which is a contradiction therefore the sequence diverges. But I do not know if I use this exact same idea to show that the sequence does not converge or if I have to change some things around. Any help would be much appreciated.

Sam
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    What is the difference between asserting that the sequence does not converge and asserting that it diverges? – José Carlos Santos Sep 26 '20 at 22:13
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    There are two conventions that I know of: the one where "divergent = not convergent" and the one where "divergent = divergent to infinity in absolute value". First convention is consistent with your work, but not with your question. The second convention is consistent with your question, but not with your work; and possibly it doesn't actually exist. –  Sep 26 '20 at 22:18
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    The sequence $(-1)^n$ admit two subsequences, precisely $(-1)^{2n};$ and $;(-1)^{2n+1}$, which admits distinct limits. – gpassante Sep 26 '20 at 22:24
  • Note that this is an immediate consequence of your previous question https://math.stackexchange.com/q/3841592/42969. – Martin R Sep 26 '20 at 22:31

2 Answers2

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A common early mistake most mathematicians make is they assume convergence means "tending to a finite value", and divergence means "tending to an infinite value".

The first definition is correct, the second is not. Divergence simply means "not tending to a finite value", the direct opposite of convergence.

With this is mind, you have proven that $(-1)^n$ does not converge to any single finite value, hence it is divergent, or non-convergent.

Rhys Hughes
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Alternative approach: if it converges it's Cauchy. But for any $N$ there are $n,m\ge N$ with $|a_n-a_m|=2$.