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

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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 $|c_n|2^n\leqslant M2^{-n}$.

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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.

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Just to confirm, are you using a comparison of two geometric series? What I mean to say is that are you comparing the $c_n(-2)^n$ series to another that is made up of the $4^n$ series to show convergence? – Guest Jul 20 '14 at 15:15
I think we call it Direct comparison test more correctly...... because $\sum \frac{1}{2^n} $ converges, so $\sum c_n(-2)^n$ absolutely converges. – Shine Jul 20 '14 at 15:19
Omit finite terms, we compare $|c_n (-2)^n|$ and $\frac{1}{2^n}$, we don't use $c_n4^n$ any more because it is smaller than 1. – Shine Jul 20 '14 at 15:24

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.

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