Power Series and Radius of Convergence
Using ratio tests, centers, and endpoints
Power Series and Radius of Convergence
Using ratio tests, centers, and endpoints
Math - Grade 9-12
- 1
Find the radius of convergence and interval of convergence for the power series ∑ from n = 0 to ∞ of (x - 3)^n / 5^n.
Compare the series to the geometric series ∑ r^n, which converges when |r| < 1.
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). - 2
Find the radius of convergence for the power series ∑ from n = 0 to ∞ of n!(x + 1)^n.
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. - 3
Find the radius and interval of convergence for ∑ from n = 1 to ∞ of (x - 2)^n / (n^2 4^n).
After finding the radius, test both endpoints separately.
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]. - 4
Use the ratio test to find the radius and interval of convergence for ∑ from n = 1 to ∞ of n x^n / 3^n.
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). - 5
Write a power series representation for 1/(1 - x/2), and state its radius of convergence.
Use the geometric series formula 1/(1 - r) = 1 + r + r^2 + r^3 + ...
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. - 6
Write 1/(3 - x) as a power series centered at 0, and state the radius of convergence.
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. - 7
For the power series ∑ from n = 1 to ∞ of (2^n/n)(x - 4)^n, find the radius and interval of convergence.
At the endpoints, check whether the series becomes harmonic or alternating harmonic.
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). - 8
Find the center, radius, and interval of convergence for ∑ from n = 1 to ∞ of (3x - 6)^n/n.
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). - 9
A power series has interval of convergence (-4, 10). What are its center and radius of convergence?
The radius is half the length of the interval.
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. - 10
Find the radius of convergence for the Maclaurin series ∑ from n = 0 to ∞ of x^n/n!.
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. - 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.
The endpoints are not decided by the radius alone. Substitute x = 1 and x = -1.
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]. - 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.
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).