Radius and Interval of Convergence of Power Series

#10.13 Radius and Interval of Convergence of Power Series

Oooookay, that title definitely had a lot of buzzwords… namely, radius of convergence, interval of convergence, and power series. You haven’t seen them in any of the previous study guides, either, so they’re definitely new to you. Let’s define them one by one! 😉

#👊 What’s a Power Series?

A power series is a series of the form $\sum^{\infty}_{n=0}a_n(x-r)^n$, where n is a non-negative integer, $a_n$ is a sequence of real numbers, and r is a real number.

In this case, $a_n$ can be any sequence, most of which you’ve already seen in previous study guides! r refers to where we center our power series function (e.g., a “power series centered at x = 3” will give us (x - 3) at that part of the series).

#🔵 Radius and Interval of Convergence

One of the questions we have about power series approximations of functions is where the approximation is valid, or in other words, where the power series converges. For a given x, we can find the radius, and then the interval of convergence for a power series.

For a Taylor series centered at x = r, the only place where we are entirely sure that it converges to is at x = r, but we can expand this to a greater range using our knowledge of the ratio test. Let’s make an example to demonstrate this!

#Recap: The Ratio Test

For a series $\sum a_n$, let $L=|\lim\limits_{n→ \infty}\frac{a_{n+1}}{a_n}|$.

• If L < 1, then the series converges.
• If L > 1, then the series diverges.
• If L = 1, then the test is indeterminate. To review the ratio test in depth, check out 10.8 Ratio Test for Convergence.

#❓ Power Series Practice Question

Find the interval, radius of convergence, and the center of the interval of convergence for the series:

$$\sum^{\infty}_{n=0}\frac{2^n}{n}(4x-8)^n$$

First, we use the Ratio Test and identify our $a_n$ and $a_{n+1}$:

$$a_{n}=\frac{2^{n}}{n}(4x-8)^{n}$$

$$a_{n+1}=\frac{2^{n+1}}{n+1}(4x-8)^{n+1}=\frac{2^n*2}{n+1}(4x-8)^n(4x-8)$$

Then, we find L:

$$L=\lim\limits_{n→ \infty}|\frac{a_{n+1}}{a_n}|=\lim\limits_{n→ \infty}|\frac{\frac{2^n*2}{n+1}(4x-8)^n(4x-8)}{\frac{2^{n}}{n}(4x-8)^{n}}|$$

Seems complicated at first but we actually cancel out a couple terms: $2^n,(4x-8)^n$. This leaves us with:

$$L=\lim\limits_{n→ \infty}|\frac{2n}{n+1}(4x-8)|$$

Since (4x - 8) is not in terms of n, we can factor it out of the limit term. We, then, evaluate the limit using L’hopital’s Rule, which gives us 2:

$$L=(4x-8)*\lim\limits_{n→ \infty}|\frac{2n}{n+1}|=(4x-8)*2$$

Remember, the series converges when L < 1, which we can simplify:

$$|L|<1$$

$$|2(4x-8)|<1$$

$$|2*4(x-2)|<1$$

$$|x-2|<\frac{1}{8}$$

This gives us a radius of converge of 1/8, a center of the interval of convergence at x = 2

To find the interval of convergence, we use our knowledge of absolute values and the expression $|x-2|<\frac{1}{8}$ to get:

$$x-2<\frac{1}{8}=x<\frac{1}{8}+\frac{16}{8}=x<\frac{17}{8}$$

$$x-2>-\frac{1}{8}=x>-\frac{1}{8}+\frac{16}{8}=x>\frac{15}{8}$$

One last thing: we need to test these endpoints—namely, the two extreme points of a line segment or interval—by plugging the values into the original series to see if they are included in our solution or not:

At x = 15/8:

$$\sum^{\infty}_{n=0}\frac{2^n}{n}(4(\frac{15}{8})-8)^n$$

$$\sum^{\infty}_{n=0}\frac{2^n}{n}(\frac{15}{2}-8)^n$$

$$\sum^{\infty}_{n=0}\frac{2^n}{n}(\frac{-1}{2})^n$$

$$\sum^{\infty}_{n=0}\frac{2^n}{n}(\frac{(-1)^n}{2^n})$$

$$\sum^{\infty}_{n=0}\frac{(-1)^n}{n}$$

From our previous encounter with the alternate harmonic series above, we can say that the series converges at x = 15/8. In other words, 15/8 is included in our interval of convergence. What about x = 17/8?

At x = 17/8:

$$\sum^{\infty}_{n=0}\frac{2^n}{n}(4(\frac{17}{8})-8)^n$$

$$\sum^{\infty}_{n=0}\frac{2^n}{n}(\frac{17}{2}-8)^n$$

$$\sum^{\infty}_{n=0}\frac{2^n}{n}(\frac{1}{2})^n$$

$$\sum^{\infty}_{n=0}\frac{2^n}{n}(\frac{1}{2^n})$$

$$\sum^{\infty}_{n=0}\frac{1}{n}$$

Another familiar face: the harmonic series! We can, thus, say that the series diverges at x = 17/8. In other words, 17/8 is not included in our interval of convergence.

Altogether, our interval of convergence is $\frac{15}{8},\frac{17}{8})$ or $\frac{15}{8}\leq x <\frac{17}{8}.$

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#⭐ Summing Up Power Series

To summarize what we introduced above:

To make things easier for you, here’s a quick guide on what you should do when you encounter a power series problem that asks you to find the radius & interval of convergence:

1. Apply the ratio test!
2. Use the ratio test to find your radius of convergence and endpoints.
3. Plug endpoints back into your original series to see if they are included in the solution or not… this’ll help you finalize your interval of convergence!

• If the endpoint series converges, that endpoint is included.
• If the endpoint series diverges, that endpoint is not included.
• Use your “series and tests” (study guides [10.3 to 10.9) toolkit! That’s it! While the journey to the answer seems long and arduous, you’ll notice that the building blocks from earlier study guides and math courses like the p-series test, the ratio test, and even absolute values all come together in the concept of power series. As always, mastery comes with practice, and becoming an expert at this topic will help you brush up on the prior concepts in this unit as well.

Good luck! 🎈