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

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.

• Special thanks to Aiden Chow for the edits. You're a big help! Oct 4 '20 at 3:26

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.

• 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. Oct 4 '20 at 3:36
• Put the math between dollars. Oct 4 '20 at 3:41

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.