Determining the radius of convergence

In summary: So, in summary, the power series \sum from n=1 to \infty (3+(-1)n)nxn has a radius of convergence of 1/4 and an interval of convergence of (-1/4, 1/4). This is found by splitting the series into even and odd terms and finding the radius of convergence for each, then taking the smaller radius. The series is absolutely convergent in the interval (-1/4, 1/4) and divergent everywhere else.
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: limn$\rightarrow$∞|((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?

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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: limn$\rightarrow$∞|((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.

HallsofIvy said:
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.

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?

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?

...3/4?

...3/4?

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

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

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.

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.

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.

1. What is the radius of convergence?

The radius of convergence is a mathematical concept used to determine the interval within which a power series converges. It is represented by the variable "R" and is the distance from the center of the power series to the closest point where the series converges.

2. How is the radius of convergence calculated?

The radius of convergence can be calculated using the ratio test or the root test. The ratio test involves taking the limit of the ratio of consecutive terms in the series, while the root test involves taking the limit of the nth root of the absolute value of the terms in the series.

3. What does it mean if the radius of convergence is infinite?

If the radius of convergence is infinite, it means that the power series converges for all values of the variable within the interval of convergence. This is also known as a "convergent power series".

4. Can the radius of convergence be negative?

No, the radius of convergence cannot be negative. It represents a distance and therefore must be a positive value. However, it is possible for the interval of convergence to include negative values.

5. How is the radius of convergence used in real-world applications?

The radius of convergence is used in various fields of science and engineering to approximate functions and solve differential equations. It is also used in numerical analysis to determine the accuracy and convergence of numerical methods.

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