Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question

3 Answers 3

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

share|improve this answer

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.

share|improve this answer
    
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 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 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 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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.