Question

A power series is given. (a) Find the radius of convergence. (b) Find the interval of convergence. $$\sum_{n=0}^{\infty} \frac{x^{n}}{2^{n}}$$

Short answer

(a) \( R = 2 \); (b) Interval: \((-2, 2)\).

Step-by-step solution

Identify the Power Series

We are given the power series \( \sum_{n=0}^{\infty} \frac{x^{n}}{2^{n}} \). This series is already in the standard form of a power series \( \sum_{n=0}^{\infty} a_n x^n \), where \( a_n = \frac{1}{2^n} \).

Apply the Ratio Test to Find Radius of Convergence

To find the radius of convergence, we apply the Ratio Test:\[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{1/2^{n+1}}{1/2^n} \right| = \lim_{n \to \infty} \frac{1}{2} = \frac{1}{2}. \]Using the Ratio Test, the radius of convergence \( R \) is given by \( \frac{1}{R} = \frac{1}{2} \). Therefore, \( R = 2 \).

Find the Interval of Convergence

Since the radius of convergence is 2, the interval of convergence centered at 0 is given by:\[ -2 < x < 2. \]We need to check convergence at the endpoints \( x = -2 \) and \( x = 2 \).

Test Endpoint x = -2

Substitute \( x = -2 \) into the series:\[ \sum_{n=0}^{\infty} \left(\frac{-2}{2}\right)^{n} = \sum_{n=0}^{\infty} (-1)^n, \]which is an alternating series that does not converge since the terms do not approach zero. So, \( x = -2 \) is not included in the interval of convergence.

Test Endpoint x = 2

Substitute \( x = 2 \) into the series:\[ \sum_{n=0}^{\infty} \left(\frac{2}{2}\right)^{n} = \sum_{n=0}^{\infty} 1^n = \sum_{n=0}^{\infty} 1, \]which is a divergent series since the terms are all 1. Therefore, \( x = 2 \) is not included in the interval of convergence.

Conclude the Interval of Convergence

The radius of convergence is 2, and neither endpoint converges. Thus, the interval of convergence is:\[ (-2, 2) \].

Key concepts

Radius of Convergence

Understanding the radius of convergence is essential when dealing with power series. In simple terms, it tells us how far from the center value (often at zero) a power series still converges. When working with the power series given in the exercise \( \sum_{n=0}^{\infty} \frac{x^{n}}{2^{n}} \), we apply the Ratio Test to compute this radius.

As seen in the solution step, the Ratio Test requires us to evaluate the limit of \( \frac{a_{n+1}}{a_n} \) as \( n \to \infty \). For this series, we have \( a_n = \frac{1}{2^n} \), leading us to find:

\[ \lim_{n \to \infty} \left| \frac{1/2^{n+1}}{1/2^n} \right| = \lim_{n \to \infty} \frac{1}{2} = \frac{1}{2}. \]

This provides the reciprocal of the radius of convergence; hence the radius of convergence \( R \) is \( 2 \). This means the series will converge for all \( x \) values between \( -2 \) and \( 2 \).

In summary:

• The radius of convergence is determined using the Ratio Test.
• The value calculated represents how far the power series converges from the center point.

Interval of Convergence

Identifying the interval of convergence is crucial, as it clarifies the exact range of \( x \) values for which the power series converges. After we determine the radius of convergence, the next step is to assess whether the series converges at the endpoints of that interval:

For the series \( \sum_{n=0}^{\infty} \frac{x^{n}}{2^{n}} \), we start with an interval based on the radius of convergence: \( -2 < x < 2 \).

However, to declare the complete interval, we need to test convergence at the endpoints \( x = -2 \) and \( x = 2 \).

• At \( x = -2 \), substituting into the series provides \( \sum_{n=0}^{\infty} (-1)^n \), known not to converge.
• At \( x = 2 \), substituting gives \( \sum_{n=0}^{\infty} 1 \), again a divergent series.

Since neither series converges at these endpoints, they are not included in the interval. The final interval of convergence is \( (-2, 2) \).

Remember:

• The interval of convergence includes all \( x \) where the series converges.
• Endpoint testing is critical to decide if they should be part of the interval.

Ratio Test

The Ratio Test is a powerful tool for determining the convergence of a series. It's especially useful with power series, like the one given in the exercise. The test involves calculating:

\[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|. \]

If this limit is less than 1, the series converges absolutely. If it's greater than 1, the series diverges. If it equals 1, the test is inconclusive. For our power series \( \sum_{n=0}^{\infty} \frac{x^{n}}{2^{n}} \), the Ratio Test has been applied as follows:

The sequence \( a_n = \frac{1}{2^n} \) gives the limit:

\[ \lim_{n \to \infty} \left| \frac{1/2^{n+1}}{1/2^n} \right| = \frac{1}{2}, \] which is less than 1, indicating convergence if \( \left| x \right| < 2 \).

Utilizing the Ratio Test:

• Helps pinpoint the radius of convergence.
• Is simple to implement by directly substituting terms.
• Quickly shows if the series converges based on the calculated limit.

Remember, the Ratio Test is straightforward for series with terms that are easily factorable, making it ideal for textbook problems like this one.