Chapter 1: Problem 13

Determine the radius and interval of convergence of the following infinite series: a. \(\sum_{n=1}^{\infty}(-1)^{n} \frac{(x-1)^{n}}{n}\) b. \(\sum_{n=1}^{\infty} \frac{x^{n}}{2^{n} n !}\). c. \(\sum_{n=1}^{\infty} \frac{1}{n}\left(\frac{x}{5}\right)^{n}\). d. \(\sum_{n=1}^{\infty}(-1)^{n} \frac{x^{n}}{\sqrt{n}}\).

Short Answer

Expert verified

Radii and intervals of convergence for the series are: a) \(R = 1, (0, 2)\); b) \(R = \infty, (-\infty, \infty)\); c) \(R = 5, (-5, 5)\); d) \(R = \infty, (-\infty, \infty)\).

Step by step solution

Infinite series a

Start with series a, \(\sum_{n=1}^{\infty}(-1)^{n} \frac{(x-1)^{n}}{n}\). Here, \(a_{n} = (-1)^{n} \frac{(x-1)^{n}}{n}\), and we note the series is centered at \(x=1\). We use the ratio test for determining the convergence: \(\lim_{n\to\infty} \left | \frac{a_{n+1}}{a_{n}} \right | < 1\). Substituting \(a_{n}\) and \(a_{n+1}\), we find that the radius of convergence (\(R\)) is 1. The interval of convergence would be determined by substituting \(x = 0\) and \(x = 2\) (the endpoints of the interval \((1-R, 1+R)\)) into the series and testing for convergence.

Infinite series b

For the series \(\sum_{n=1}^{\infty} \frac{x^{n}}{2^{n} n !}\), we note that the series is centered at zero. The radius and interval of convergence is \( R = \infty\) and \((- \infty, \infty)\) respectively, because for this series, factorials in the denominator cause the terms to decrease rapidly, leading to convergence for all real values of \(x\).

Infinite series c

For the series \(\sum_{n=1}^{\infty} \frac{1}{n}\left(\frac{x}{5}\right)^{n}\), note that the series is centered at zero. The terms form a geometric progression with common ratio \(\frac{x}{5}\). Here, radius of convergence is \(R = 5\), for which \(-5 < x < 5\), because for the geometric series to converge, the common ratio must satisfy \(-1 < \frac{x}{5} < 1\). Thus interval of convergence is \((-5,5)\).

Infinite series d

For the series \(\sum_{n=1}^{\infty}(-1)^{n} \frac{x^{n}}{\sqrt{n}}\), it is centered at zero. The radius of convergence (\(R\)) is \(\infty\) and the interval of convergence is \((- \infty, \infty)\), because the root test for convergence \(\lim_{n\to\infty} \left | a_{n}^{1/n} \right | < 1\) which results in \(x \in (-\infty , \infty)\).