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

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.

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.
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.
• 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. – frog Jul 30 '14 at 7:42