Question

Find the interval of convergence of each of the following power series; be sure to investigate the endpoints of the interval in each case. $$\sum_{n=1}^{\infty} \frac{(x-1)^{n}}{2^{n}}$$

Short answer

The interval of convergence is \((-1, 3)\).

Step-by-step solution

Identify the Power Series Format

The given power series is \(\sum_{n=1}^{\infty} \frac{(x-1)^{n}}{2^{n}}\). It is of the form \(\sum_{n=1}^{\infty} a_n (x - c)^n\). Here, \(a_n = \frac{1}{2^n}\) and \(c = 1\).

Determine the Radius of Convergence Using the Ratio Test

Use the Ratio Test to find the radius of convergence. Compute \[L = \lim_{{n \to \infty}} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{{n \to \infty}} \left| \frac{\frac{1}{2^{n+1}}}{\frac{1}{2^n}} \right| = \lim_{{n \to \infty}} \left| \frac{1}{2} \right| = \frac{1}{2}.\]

Find the Radius of Convergence

The radius of convergence \(R\) is the reciprocal of \(L\): \[R = \frac{1}{L} = \frac{1}{\frac{1}{2}} = 2.\]

Write the Interval of Convergence

The interval of convergence is centered around \(c = 1\) and extends \(R = 2\) units to either side. Thus, \(x\) must satisfy: \[1 - 2 < x < 1 + 2\] or \[-1 < x < 3.\]

Test the Endpoints

Check the series at the endpoints \(x = -1\) and \(x = 3\). For \(x = -1\): \[\sum_{n=1}^{\infty} \frac{(-1 - 1)^n}{2^n} = \sum_{n=1}^{\infty} \frac{(-2)^n}{2^n} = \sum_{n=1}^{\infty} (-1)^n\]This series does not converge since the terms do not tend to zero.For \(x = 3\): \[\sum_{n=1}^{\infty} \frac{(3 - 1)^n}{2^n} = \sum_{n=1}^{\infty} \frac{2^n}{2^n} = \sum_{n=1}^{\infty} 1\]This series also does not converge.

State the Final Interval of Convergence

Since the series does not converge at either endpoint, the interval of convergence is \((-1, 3)\).

Key concepts

power series

A power series is a series of the form \( \sum_{n=0}^{\infty} a_n (x - c)^n \). It is an infinite sum involving a variable, typically denoted as x, raised to varying powers. The coefficients \( a_n \) are constants. The term (x - c) represents a shift, where c is a fixed value called the center of the power series. Power series are similar to polynomials but with an infinite number of terms. They converge over a certain interval around this center.

ratio test

The ratio test is a method to determine the convergence of a series. For a series \( \sum a_n \), the test involves finding the limit of the absolute value of the ratio of consecutive terms: \[ L = \lim_{{n \to \infty}} \left| \frac{a_{n+1}}{a_n} \right| \]. If L < 1, the series converges; if L > 1, it diverges. If L = 1, the test is inconclusive. Applying this to our power series, we compute the limit and find L = \( \frac{1}{2} \), which means the series converges.

radius of convergence

The radius of convergence R for a power series \( \sum a_n (x - c)^n \) is the distance from the center c to the boundary within which the series converges. It is found using the ratio test: \( R = \frac{1}{L} \). For our example, we calculated L = \( \frac{1}{2} \), so \( R = 2 \). This indicates the power series converges for all x within 2 units of the center (1), i.e., x in \[-1, 3\].

endpoints

Endpoints are crucial when determining the interval of convergence for a power series. Although a series may converge inside an interval, it might behave differently at the endpoints. For the series \(\sum_{n=1}^{\infty} \frac{(x-1)^{n}}{2^{n}}\), the calculations showed that neither endpoint, x = -1 or x = 3, results in a converging series. Thus, they are excluded from the interval of convergence, giving us \( (-1, 3) \). Always verify endpoint behavior to ensure accurate intervals.