# power series justification

• Apr 11th 2011, 10:35 AM
Kuma
power series justification
Hi.

Quick question.
Given a power series of the form:

sigma cn(x+3)^n. It is given that this series converges when x=0 and diverges when
x = -8

Now does it converge when the series is

sigma cn2^n

The answer is yes but not sure how to justify.
• Apr 11th 2011, 10:46 AM
FernandoRevilla
Hint:

As $\displaystyle \sum_{n=0}^{+\infty}c_n3^n$ converges, the radius of convergence of the series $\displaystyle \sum_{n=0}^{+\infty}c_nx^n$ is $\displaystyle \geq 3$
• Apr 11th 2011, 10:52 AM
Kuma
Quote:

Originally Posted by FernandoRevilla
Hint:

As $\displaystyle \sum_{n=0}^{+\infty}c_n3^n$ converges, the radius of convergence of the series $\displaystyle \sum_{n=0}^{+\infty}c_nx^n$ is $\displaystyle \geq 3$

I was trying to figure out the radius of convergence. This is what i have.

we know that |z-a| < R means that the series converges absolutely for some number R.
Here i have a as (-3) and z=x=0 when it converges.
so |-3| < R => |3| < R

now |3| can be + or - 3...so I guess is R > -3 or 3?
I dunno if im doing this right.