Chapter 1: Problem 16

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

Expert verified

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 $(-1, 3)$.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.