Given Cn4^n is convergent as n goes from 0 to infinity:

(i) how is Cn(-2)^n as n goes from 0 to infinity convergent?

(ii) how is Cn(-4)^n as n goes from 0 to infinity not convergent?

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- Oct 5th 2010, 03:49 PMTaurus3comparing series
Given Cn4^n is convergent as n goes from 0 to infinity:

(i) how is Cn(-2)^n as n goes from 0 to infinity convergent?

(ii) how is Cn(-4)^n as n goes from 0 to infinity not convergent? - Oct 5th 2010, 08:07 PMtonio
- Oct 5th 2010, 09:08 PMTaurus3
cn is just a constant, or a coefficient of the series.

- Oct 6th 2010, 08:13 AMKrizalid
even being a constant $\displaystyle 4^n$ doesn't converge.

- Oct 6th 2010, 04:38 PMtonio
- Oct 7th 2010, 05:27 AMHallsofIvy
The power series $\displaystyle \sum_{n=0}^\infty C_n x^n$ always has a "radius of convergence", R, and converges inside that radius- on the interval [-R, R] while diverging outside that interval. Since $\displaystyle \sum_{n=0}^\infty C_n4^n$ converges, 4 is either inside that interval or is an endpoint (R= 4).

In either case, since -4< -2< 4, -2 is inside the interval of convergence and $\displaystyle \sum_{n=0}^\infty C_n(-2)^n$ converges.

I had first thought that "if $\displaystyle \sum_{n=0}^\infty C_n4^n= \sum_{n=0}^\infty |C_n(-4)^n|$ converges, then the series $\displaystyle \sum_{n=0}^\infty C_n(-4)^n$ must converge but then realized that I was assuming that $\displaystyle C_n$ is**positive**- and that is not given. If the radius of convergence is 4, knowing that the power series converges at one endpoint of the interval of convergence does not tell us whether the series converges at the other endpoint.

If we know only that $\displaystyle \sum_{n=0}^\infty C_n4^n$ converges, then we know that $\displaystyle \sum_{n=0}^\infty C_n(-2)^n$ converges but do not know whether $\displaystyle \sum_{n=0}^\infty C_n(-4)^n$ converges or not.

If we**also**know that $\displaystyle C_n\ge 0$ for all n,**then**we know that $\displaystyle \sum_{n=0}^\infty C_n(-4)^n$ converges (absolutely).