1. ## comparing 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?

2. Originally Posted by Taurus3
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?

Whawt do you mean by Cn4^n and etc.???

Tonio

3. cn is just a constant, or a coefficient of the series.

4. even being a constant $4^n$ doesn't converge.

5. Originally Posted by Krizalid
even being a constant $4^n$ doesn't converge.

I'm afraid he must be meaning something else, and if $c_n=0\,\,\forall n\in \mathbb{N}$ , then $c_n4^n=0$ does converge...but he must be on something else.

Tonio

6. The power series $\sum_{n=0}^\infty C_n x^n$ always has a "radius of convergence", R, and converges inside that radius- on the interval [-R, R] while diverging outside that interval. Since $\sum_{n=0}^\infty C_n4^n$ 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 $\sum_{n=0}^\infty C_n(-2)^n$ converges.

I had first thought that "if $\sum_{n=0}^\infty C_n4^n= \sum_{n=0}^\infty |C_n(-4)^n|$ converges, then the series $\sum_{n=0}^\infty C_n(-4)^n$ must converge but then realized that I was assuming that $C_n$ 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 $\sum_{n=0}^\infty C_n4^n$ converges, then we know that $\sum_{n=0}^\infty C_n(-2)^n$ converges but do not know whether $\sum_{n=0}^\infty C_n(-4)^n$ converges or not.

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

,

,

### if cn4^n is convergent

Click on a term to search for related topics.