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Question taken from Calculus Early Transcendentals 7th edition textbook, section on infinite sequences:

Prove that if $\lim_\limits{n\to\infty}A_n=A\ne0$, then the sequence $(-1)^n \cdot A_n$ diverges.

So far I've attempted proof by contradiction, assuming that the above sequence does converge for $A \ne 0$. I'm now stuck in trying to find the contradiction using the precise limit definition. Of course, if anyone has any other methods, something more direct? I'd greatly appreciate it.

I also considered the idea that since I assumed it converges, I could break up $(-1)^n\cdot A_n$ into the product of sequences and state that $(-1)^n\dots$, but that just seemed too "handwave-y" and not rigorous.

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  • $\begingroup$ Special thanks to Aiden Chow for the edits. You're a big help! $\endgroup$
    – Malcolm
    Oct 4, 2020 at 3:26

2 Answers 2

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Hint : take two subsequences with even and odd terms respectively, then these two have different limits which are $A$ and $-A$.

Moreover, a convergent sequence can have exactly one subsequential limit.

Edit: for a real valued sequence,

Cauchy $\iff$ convergent.

So, take , $u_n=(-1)^{n} A_n$

Now, as, $n$ goes to $\infty$ ,$|u_{2n}-u_{2n-1}| \ge 2A $ , so, can't be a Cauchy sequence.

So, not converge.

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Consider two disjoint sub-sequences of the original sequence:

When $n$ is odd: this sub-sequence will converge to $-A$ because $(-1)^n = -1$ if $n$ odd

When $n$ is even: this sub-sequence will converge to A because $(-1)^n = 1$ if $n$ is even

The original sequence contains two sub-sequences with different limits, so it diverges.

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  • $\begingroup$ Thanks for the help. I don't have a lot of confidence when it comes to infinite sequences so wanted to try and squash it here. $\endgroup$
    – Malcolm
    Oct 4, 2020 at 3:36
  • $\begingroup$ Put the math between dollars. $\endgroup$ Oct 4, 2020 at 3:41

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