# Power Series in X & the Interval of Convergence

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• 0:04 Power Series in X
• 1:42 Interval of Convergence
• 5:25 Lesson Summary
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Instructor
Gerald Lemay

Gerald has taught engineering, math and science and has a doctorate in electrical engineering.

Expert Contributor
Robert Ferdinand

Robert Ferdinand has taught university-level mathematics, statistics and computer science from freshmen to senior level. Robert has a PhD in Applied Mathematics.

A power series is often used in physics and engineering to approximate functions. In this lesson, we'll explore the power series in x and show how to find the interval of convergence.

## Power Series in X

My favorite part of the hardware store is the power tool section, especially the battery-powered tools. Speaking of power and tools, there's this mathematical tool called the power series in x.

A power series in x is an infinite sum of terms where each term is a factor multiplying an x plus a constant raised to a power.

### Building a Power Series

Infinity is a lot of terms to add! What if we start with the first 4 terms? The sum of those four terms could look like:

We let x take on a value. Then, we add the four terms and get a result. What if we let x have values from -5 to -1 ? Then we could substitute these values for x, sum the four terms, and save the results. Plotting these results as a function of x:

Can you predict what the next term in this series looks like? If you said (x + 3)^4 divided by 5, then you are correct! What if we sum the first 10 terms of the series? Let's see, the tenth term will be (x + 3)^9 divided by 10. Again, substitute x values from -5 to -1, sum the ten terms, and save the results. If these results for 10 terms are plotted with the results for 4 terms we see:

There seems to be some values for x where the results have changed a lot when we went from 4 terms to 10 terms. The values to the left of -4 and the values to right of -2 are quite different. The sum is staying pretty much the same between x = -4 and x = -2. This is getting interesting.

While our battery continues to charge, let's build on these ideas.

## Interval of Convergence

What's the general term we have been summing? Is it (x + 3)^n divided by n + 1? Let's check this out. If n = 2, we have (x + 3)^2 divided by 3. Looking back when we summed four terms, this was our third term.

What about the first term, which is a 1? No problem. Letting n = 0 gives us (x + 3)^0 divided by 0 + 1. The (x + 3)^0 is 1 and 0 + 1 is 1. So we have 1 divided by 1, which is 1.

To express the infinite sum starting from n = 0, we write:

The capital sigma means to take the sum from n = 0 to n = infinity of the general term. We're going to use the symbol An for the general term.

The ratio test says a series converges if

We know the An general term:

To get the An+1 expression, we replace the n with n + 1:

Let's work out the absolute value part first:

Since (n + 1)/(n + 2) is always greater than 0, we can take it outside of the absolute value:

Now, we deal with the limit part of the ratio test:

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## Power Series and Convergence: Practice Problems

#### Key Terms

• Power Series
• Interval of Convergence

• Paper
• Pencil

### Practice Problems -- Show All Your Work

(a) Find the interval of convergence of the power series whose n-th term is An = (2x)n.

(b) Find the interval of convergence of the power series whose n-th term is An = (3x -2)n/n.

(a) Find the limit, as n approaches infinity, of the absolute value of:

A(n+1)/A(n) = (2x)(n+1)/(2x)n = 2x.

Convergence happens when absolute value of 2x is less than 1 which leads to:

-1 < 2x < 1 or -1/2 < x < 1/2.

At the end point x = -1/2, we get the series whose n-th term is An = (-1)n, which does not converge. At the end point of x = 1/2, we get the series whose n-th term is An = (1)n, which does not converge either.

Hence, the interval of convergence is -1/2 < x < 1/2.

(b) Find the limit, as n approaches infinity, of the absolute value of:

A(n+1)/A(n) = (3x-2)(n+1)/(n+1) (n)/ (3x-2)(n+1) = (3x - 2) (n)/(n+1)

Convergence happens when absolute value of (3x - 2) is less than 1, which gives us:

-1 < 3x - 2 < 1.

Solving the inequality above gives us -1 + 2 < 3x < 1 + 2 or 1 < 3x < 3 or 1/3 < x < 1.

Endpoint x = 1/3 yields a series whose n-th term is An = (-1)n/n. From the alternating series test, the absolute values of the terms of this series are positive, decreasing, and approach zero as n approaches infinity. Hence, we have convergence at x = 1/3.

Endpoint x = 1 gives a series whose n-th term is An = (1)n/n. This is a divergent harmonic series. Hence, we DO NOT have convergence at x = 1.

The interval of convergence is therefore,

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