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Prove or disprove that if $\sum \frac{n a_n}{2^{n-1}}$ diverges then $\sum (-1)^n a_n$ could be conditionally convergent.

Where to start? I don't see the connection between those two series.

btw, how do I write An like in recursive series? didn't see that on any guide.

Thank You!

PierreCarre
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Ori Yehuda
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    If $a_n$ converged to zero, $\sum \frac{n a_n}{2^{n-1}}$ would be convergent. So, we must have that $\lim a_n \ne 0$. Can you pick it up from here? – PierreCarre Jan 11 '23 at 10:27
  • No, I don't see where your'e going with this. Could you please expand? – Ori Yehuda Jan 11 '23 at 14:55

1 Answers1

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If $\sum (-1)^n a_n$ is convergent, we must have that $\lim a_n = 0$. However, if this is the case, we also have that $$ \left|\frac{n a_n}{2^{n-1}}\right|\leq \frac{M n}{2^{n-1}}, $$ which is convergent ($M$ is an upper bound for $|a_n|$). So, the conclusion is that claim is false.

PierreCarre
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