Power Series and Radius of Convergence Worksheet

Question 1

Find the radius of convergence and interval of convergence for the power series ∑ from n = 0 to ∞ of (x - 3)^n / 5^n.

Accepted Answer

This is a geometric series with ratio (x - 3)/5. It converges when |x - 3| < 5, so the radius of convergence is 5. The endpoints x = -2 and x = 8 both make the terms fail to approach 0, so the interval of convergence is (-2, 8).

Question 2

Find the radius of convergence for the power series ∑ from n = 0 to ∞ of n!(x + 1)^n.

Accepted Answer

Using the ratio test, the ratio of consecutive terms approaches (n + 1)|x + 1|, which grows without bound unless x = -1. Therefore, the radius of convergence is 0, and the series converges only at its center x = -1.

Question 3

Find the radius and interval of convergence for ∑ from n = 1 to ∞ of (x - 2)^n / (n^2 4^n).

Accepted Answer

The ratio test gives |x - 2|/4 < 1, so the radius of convergence is 4. At x = 6, the series becomes ∑ 1/n^2, which converges. At x = -2, the series becomes ∑ (-1)^n/n^2, which also converges absolutely. The interval of convergence is [-2, 6].

Question 4

Use the ratio test to find the radius and interval of convergence for ∑ from n = 1 to ∞ of n x^n / 3^n.

Accepted Answer

The ratio test gives a limiting ratio of |x|/3, so the series converges when |x| < 3. The radius of convergence is 3. At x = 3, the terms are n, and at x = -3, the terms are n(-1)^n, so the terms do not approach 0 at either endpoint. The interval of convergence is (-3, 3).

Question 5

Write a power series representation for 1/(1 - x/2), and state its radius of convergence.

Accepted Answer

Since 1/(1 - r) = ∑ r^n, we can use r = x/2. The power series is ∑ from n = 0 to ∞ of (x/2)^n. It converges when |x/2| < 1, so the radius of convergence is 2.

Question 6

Write 1/(3 - x) as a power series centered at 0, and state the radius of convergence.

Accepted Answer

Rewrite the function as 1/(3 - x) = (1/3)/(1 - x/3). The power series is ∑ from n = 0 to ∞ of x^n/3^(n + 1). It converges when |x/3| < 1, so the radius of convergence is 3.

Question 7

For the power series ∑ from n = 1 to ∞ of (2^n/n)(x - 4)^n, find the radius and interval of convergence.

Accepted Answer

The coefficients grow like 2^n, so the series converges when |2(x - 4)| < 1. The radius of convergence is 1/2. At x = 9/2, the series becomes ∑ 1/n, which diverges. At x = 7/2, the series becomes ∑ (-1)^n/n, which converges conditionally. The interval of convergence is [7/2, 9/2).

Question 8

Find the center, radius, and interval of convergence for ∑ from n = 1 to ∞ of (3x - 6)^n/n.

Accepted Answer

Rewrite 3x - 6 as 3(x - 2), so the center is x = 2. The series converges when |3(x - 2)| < 1, so the radius is 1/3. At x = 7/3, the series is ∑ 1/n, which diverges. At x = 5/3, the series is ∑ (-1)^n/n, which converges conditionally. The interval of convergence is [5/3, 7/3).

Question 9

A power series has interval of convergence (-4, 10). What are its center and radius of convergence?

Accepted Answer

The center is the midpoint of the interval, which is (-4 + 10)/2 = 3. The radius is the distance from the center to either endpoint, which is 7. Therefore, the center is 3 and the radius of convergence is 7.

Question 10

Find the radius of convergence for the Maclaurin series ∑ from n = 0 to ∞ of x^n/n!.

Accepted Answer

Using the ratio test, the ratio of consecutive terms is |x|/(n + 1), which approaches 0 for every real value of x. Therefore, the series converges for all real x, and the radius of convergence is infinite.

Question 11

The series for ln(1 + x) is ∑ from n = 1 to ∞ of (-1)^(n + 1)x^n/n. Find its radius and interval of convergence.

Accepted Answer

The main convergence condition is |x| < 1, so the radius of convergence is 1. At x = 1, the series becomes the alternating harmonic series, which converges. At x = -1, the series becomes -∑ 1/n, which diverges. The interval of convergence is (-1, 1].

Question 12

A power series is centered at x = -2 and has radius of convergence 6. Write the open interval where the series is guaranteed to converge before checking endpoints.

Accepted Answer

The series is guaranteed to converge when |x - (-2)| < 6, or |x + 2| < 6. Solving gives -8 < x < 4. The open interval is (-8, 4).

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