Power Series Convergence comparison

Question

Given $\sum_{n=0}^\infty c_n4^n$ is convergent, can this be used to find the convergence of $\sum_{n=0}^\infty c_n(-2)^n$?

Answer 1

Notice that $(|c_n|4^n)_{n\geqslant 1}$ is a bounded sequence, hence $|c_n|\leqslant M 4^{-n}$ for a universal constant $M$, and we deduce $|c_n|2^n\leqslant M2^{-n}$.

Answer 2

Outline: Since the first series converges, the terms $|c_n|4^n$ approach $0$. It follows by Comparison (geometric series) that the series $\sum (-1)^n c_n 2^n$ converges absolutely,and hence converges.

Answer 3

Because $|c_n 4^n|\rightarrow 0$ when $n\rightarrow \infty$, and $|c_n (-2)^n|=|c_n 2^n|=|c_n 4^n|\cdot \frac{1}{2^n}\leq \frac{1}{2^n} $ when $n$ is verylarge, so by M-test, $\sum c_n(-2)^n$ is convergent.