Given a convergence series $\sum 3^na_n$. Prove/disprove: The series $\sum (-2)^na_n$ converges.

Question

Given a convergence series $\sum 3^na_n$.

Prove/disprove: The series $\sum (-2)^na_n$ converges.

I guess the arguement is true as this is a power series, but can it be proven formally>

Answer 1

Power series-less proof:

$\sum 3^na_n$ convergent $\implies$ $(3^na_n)_n\to 0$ $\implies$ $(3^na_n)_n$ bounded. Say $|3^na_n|\le K$. Then, $$|(-2)^na_n|\le(2/3)^nK$$ and as $\sum(2/3)^nK$ is convergent, $\sum(-2)^na_n$ is absolutely convergent by direct comparison test.

Answer 2

Consider the power series $\sum a_n x^n$. Since $\sum 3^na_n$ converges, the radius of convergence of the power series is $ \ge 3$. Hence, since $|-2| <3$, the series $\sum (-2)^na_n$ converges absolutely.