Determining the radius of convergence

[porroadventum]

1. Determine the raius of convergence and interval of convergence of the power series $\sum$ from n=1 to $\infty$ (3+(-1)ⁿ)ⁿxⁿ.



2. Usually when finding the radius of convergence of a power series I start off by using the ratio test: lim_{n$\rightarrow$∞}|((3+(-1)ⁿ⁺¹)ⁿ⁺¹xⁿ⁺¹/ (3+(-1)ⁿ)ⁿxⁿ|

But this limit does not exist since this equation is just oscillating between 0 and +infinity.

Is there a radius of convergence?

 

[HallsofIvy]

1. Determine the raius of convergence and interval of convergence of the power series $\sum$ from n=1 to $\infty$ (3+(-1)ⁿ)ⁿxⁿ.



2. Usually when finding the radius of convergence of a power series I start off by using the ratio test: lim_{n$\rightarrow$∞}|((3+(-1)ⁿ⁺¹)ⁿ⁺¹xⁿ⁺¹/ (3+(-1)ⁿ)ⁿxⁿ|

But this limit does not exist since this equation is just oscillating between 0 and +infinity.

Is there a radius of convergence?

Every power series has a radius of convergence- but it is possible that the radius of convergence is 0 or infinity.

Here, you have
$$\frac{(3+(-1)^{n+1})^{n+1}}{(3+(-1)^n)^n}|x|$$

The simplest way to handle this is to look at n even and odd separately:
1) if n is even, then n+1 is odd so $(-1)^n= 1$ and $(-1)^{n+1}= -1$ so that we have
$$\frac{(2^{n+1}}{4^n}|x|= (2)(1/2)^n|x|$$
and, as n goes to infinity that goes to 0.

2) if n is odd, then n+1 is even so $(-1)^n= -1$ and $(-1)^a{n+1}= 1$ so that we have
$$\frac{4^{n+1}}{2^n}|x|= 4(2^n)|x|$$
which does not converge. The series converges only for x= 0 and the radius of convergence is 0.

 

[Dick]

Every power series has a radius of convergence- but it is possible that the radius of convergence is 0 or infinity.

Here, you have
$$\frac{(3+(-1)^{n+1})^{n+1}}{(3+(-1)^n)^n}|x|$$

The simplest way to handle this is to look at n even and odd separately:
1) if n is even, then n+1 is odd so $(-1)^n= 1$ and $(-1)^{n+1}= -1$ so that we have
$$\frac{(2^{n+1}}{4^n}|x|= (2)(1/2)^n|x|$$
and, as n goes to infinity that goes to 0.

2) if n is odd, then n+1 is even so $(-1)^n= -1$ and $(-1)^a{n+1}= 1$ so that we have
$$\frac{4^{n+1}}{2^n}|x|= 4(2^n)|x|$$
which does not converge. The series converges only for x= 0 and the radius of convergence is 0.

You've only shown that the limit in the ratio test doesn't exist. That means the ratio test is inconclusive. You can't say anything about the radius of convergence from that. You want to split the series into two other series consisting of even and odd terms of the original series. Those both have a well defined radius of convergence.

 

[porroadventum]

I split the original series into even and odd terms and got the radius of convergence =1/4 for even terms and =1/2 for odd terms. Is this correct?

 

[Dick]

I split the original series into even and odd terms and got the radius of convergence =1/4 for even terms and =1/2 for odd terms. Is this correct?

Yes, so what do you say for the radius of convergence of the whole series?

 

[porroadventum]

...3/4?

 

[Dick]

...3/4?

Think about it again. One of the series diverges for |x|>1/4.

 

[porroadventum]

Oh so it will the orginal series will have a radius of convergence of 1/4?!

 

[Dick]

Oh so it will the orginal series will have a radius of convergence of 1/4?!

Yes, it will be the smaller of the two radii.

 

[porroadventum]

Does that mean the interval of convergence is (-1/4,1/4)? Because when I let x=1/4,=-1/4 I get 1 and +/-1 respectively which are divergent series. Therefore it is absolutely convergent in (-1/4,1/4) and is divergent everywhere else.

THank you for your help on this.

 

[Dick]

Does that mean the interval of convergence is (-1/4,1/4)? Because when I let x=1/4,=-1/4 I get 1 and +/-1 respectively which are divergent series. Therefore it is absolutely convergent in (-1/4,1/4) and is divergent everywhere else.

THank you for your help on this.

That's it. You're welcome.