# If $\sum_{n=1}^{\infty}a_n3^n$ is convergent, then $\sum_{n=1}^{\infty}a_n(-2)^n$ is convergent

My class was given a set of practice problems and I can't figure this one out.

If $\sum_{n=1}^{\infty}a_n3^n$ is convergent, then $\sum_{n=1}^{\infty}a_n(-2)^n$ is convergent.

My intuition wants to say this is true, but I can't figure out why. I can't apply the ratio test between the two because $a_n$ could be alternating, and limit comparison test between the first sum and $\sum_{n=1}^{\infty}a_n2^n$ doesn't work because the limit comparison test doesn't prove absolute convergence. I don't see any way to apply the alternating series theorem.

Essentially, this problem seems trivial if I know $a_n$ is either always positive or always negative, but any sort of alternation and it breaks down.

• note that a series is convergent if it is absolutely convergent. – QED Nov 21 '17 at 5:30
• @AbishankaSaha The $a_n$'s are not promised to be non-negative. – Clement C. Nov 21 '17 at 5:32
• This is a special case of an important fact about power series: If $\sum_{n=0}^\infty a_n x^n$ converges for $x = x_0$ then it converges absolutely for all $x$ with $|x| < |x_0|$. – Martin R Nov 21 '17 at 6:17

If $\sum_{n=1}^{\infty}3^na_n$ converges then $\lim_{n\to\infty}3^na_n=0$, and so there is some $N$ such that $$|a_n|\leq 3^{-n}$$ for $n\geq N$. This then implies that $$|(-2)^na_n|\leq \Big(\frac{2}{3}\Big)^n$$ for $n\geq N$, and it follows that $\sum_n(-2)^na_n$ converges absolutely by comparison with a geometric series.
The power series $\sum_{n=1}^{\infty}a_n*x^n$ has radius of convergence $\ge 3$. Hence this power series converges for $x=-2$.