Radius/Open Interval of Convergence for the Series

[jegues]

Homework Statement

Find the radius of convergence and the open interval of convergence for the series

\sum^{\infty}_{n=0} \frac{(-1)^{n}\sqrt{(2n)!}}{2^{n}n!}(x-3)^{4n}

Homework Equations

The Attempt at a Solution

See figure attached.

I'm just curious if I've done the radius of convergence correct or not. If I haven't done this part right then I certainly won't get the interval of converge right as well.

I found that,

R_{x} = \infty

So my open interval of convergence would be,

-\infty < (x-3)^4 < \infty \rightarrow -\infty < x < \infty

Did I make a mistake in calculating R_{y}?

Edit: I found my mistake. I forgot a factorial on the bottom term. If I put it in,

2lim_{n \rightarrow \infty} \frac{(n+1)}{\sqrt{(2n+2)}\sqrt{(2n+1)}} = 1

Therefore,

R_{x} = (x-3)^{4} = 1 \rightarrow R_{x} = 4

Is this correct? Or should it be something like,

|x-3| < 1

Depending on which of the above is correct, then the open interval of convergence is,

-4 < x < 4

or

2 < x < 4

Which one is correct?

2nd Edit: After reviewing I think the latter case is correct so,

\framebox{R_{x} = 1}

\text{Open Interval of Convergence: }2 < x < 4

 

[jegues]

Updated.

 

[Mark44]

You're doing some things wrong, but because your limit was 1, they didn't matter this time.

In the ratio test you should be calculating
\frac{a_{n+1}}{a_n}
but you seem to be calculating this
\frac{a_n}{a_{n+1}}
As I said, since the limit is 1, it didn't matter.

The radius of convergence is indeed 1, and the interval of convergence is 2 < x < 4. It's possible that the series is convergent at one or both endpoints of this interval, which you probably should check. That's pretty easy to do: just substitute x = 2 and x = 4 into the series and see if it converges at each.

 

[jegues]

You're doing some things wrong, but because your limit was 1, they didn't matter this time.

In the ratio test you should be calculating
\frac{a_{n+1}}{a_n}
but you seem to be calculating this
\frac{a_n}{a_{n+1}}
As I said, since the limit is 1, it didn't matter.

The radius of convergence is indeed 1, and the interval of convergence is 2 < x < 4. It's possible that the series is convergent at one or both endpoints of this interval, which you probably should check. That's pretty easy to do: just substitute x = 2 and x = 4 into the series and see if it converges at each.

I disagree with your version of the ratio test.

Straight from my textbook,

"The radius of convergence of a power series, \sum_{n=0}^{\infty} a_{n}x^{n} is given by,

R = lim_{n \rightarrow \infty} \left|\frac{a_{n}}{a_{n+1}}\right|

or

R = lim_{n \rightarrow \infty}\frac{1}{\sqrt[n]{\left|a_{n}\right|}}

provided that either limit exists or is equal to infinity.


and the question only asked for the open interval of convergence so we don't need to look at the end points of interval.

 

[Mark44]

All of my calculus books (below) show the ratio in the Ratio Test as I wrote it. What you wrote was a formula for the radius of convergence, which is different.

• Calculus and Analytic Geometry, Schwartz, 2nd Ed.
• Calculus and Analytic Geometry, Thomas/Finney, 8th Ed.
• Calculus Concepts and Contexts, Stewart
• Advanced Engineering Mathematics, Kreyszig, 3rd Ed.

 

[jegues]

All of my calculus books (below) show the ratio in the Ratio Test as I wrote it. What you wrote was a formula for the radius of convergence, which is different.

• Calculus and Analytic Geometry, Schwartz, 2nd Ed.
• Calculus and Analytic Geometry, Thomas/Finney, 8th Ed.
• Calculus Concepts and Contexts, Stewart
• Advanced Engineering Mathematics, Kreyszig, 3rd Ed.

I don't think we've ever learned the ratio test. Why do I need the ratio test in this problem anyways?

I'm assuming we don't if, after all, we haven't been taught it in my course. (Unless I'm mixing something up?)

For my purposes, isn't the fomrula for the radius of convergence enough?

Sorry for the confusion if any.

 

[Mark44]

It doesn't make much sense to talk about the radius of convergence of a series if you haven't established that the series actually converges. The Ratio Test and others (Integral Test, n-th Term Test for Divergence, and so on) are tools that enable you to distinguish between series that converge and those that diverge.

 

[jegues]

It doesn't make much sense to talk about the radius of convergence of a series if you haven't established that the series actually converges. The Ratio Test and others (Integral Test, n-th Term Test for Divergence, and so on) are tools that enable you to distinguish between series that converge and those that diverge.

So where does this leave me? What you're saying makes perfect sense, however on these type of problems we've always just jumped right into finding the radius of convergence, followed by the open interval of convergence without checking whether or not the series actually converged.

Could there be a reason my professor doing the problems in this manner? Will we obtain a radius of convergence if the series doesn't converge? If not, that could be another manner of testing for convergence.

What are your thoughts?

 

[Mark44]

If you're working with power series (series in powers of (x - a)), the series will always converge for at least one value of x, x = a. This would make the radius of convergence equal to 0, and the interval of convergence being {a}.

As to why your professor is doing things this way, I have no idea. All of the calculus texts I've taught out of work with series of numbers first, then power series, and then talk about the radius of convergence after that.

 

[jegues]

If you're working with power series (series in powers of (x - a)), the series will always converge for at least one value of x, x = a. This would make the radius of convergence equal to 0, and the interval of convergence being {a}.

As to why your professor is doing things this way, I have no idea. All of the calculus texts I've taught out of work with series of numbers first, then power series, and then talk about the radius of convergence after that.

We did not cover series of numbers in our course.

We looked at sequences of numbers and functions, then series of functions. I'm not entirely sure why, but maybe there is some reasoning behind it.