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I got this question:

Prove or disprove the following:

If the series $\sum_{k=1}^{\infty} a_k3^k$ diverges, Must the series $\sum_{k=1}^{\infty} a_k4^k$ diverge too?

I tried to find a couple of counterexamples but failed, I tried $a_k=1/k!$, $a_k=1/3^k$ and many more but wasn't able to find a counterexample. Then I tried to prove this statement but I wasn't able to proceed too.

Thanks for any hints.

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Yes. If the power series $\sum\limits_{k\ge1} a_k x^k$ has $R$ as a radius of convergence then for all $x$ such that $|x|>R$ the series is divergent and obviously if for $x=3$ the series is divergent and since $4>3\ge R$ then for $x=4$ the series is also divergent.

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  • $\begingroup$ Nice answer, Thanks. $\endgroup$ – MathNerd Jul 30 '14 at 7:52
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    $\begingroup$ You're welcome. $\endgroup$ – user63181 Jul 30 '14 at 7:53
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Yes. By the comparison test, $b_k = a_k3^k < a_k4^k = \tilde{b}_k$ for all $k \in \mathbb{N}$. Therefore, $\sum_{k=1}^\infty b_k < \sum_{k=1}^\infty \tilde{b}_k$ and since the left-hand side diverges, so does the right-hand side.

Edit: This assumes that $a_k \geq 0$ for all $k$, my mistake. See Sami Ben Romdhane's answer for a better, complete solution.

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    $\begingroup$ I think this is not enough since you assume $a_k$ to be positive. One could exclude absolutely converging series by your argument, but I think you need an additional idea for the general case. Alternatively, Sami Ben Romdhanes answer tackles the general case. $\endgroup$ – frog Jul 30 '14 at 7:42
  • $\begingroup$ Oh dear, you are absolutely right. The two examples he gave made me assume positivity. Sami's answer is indeed more complete. $\endgroup$ – Hubble Jul 30 '14 at 7:45

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