Chapter 1: Problem 4

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

Short Answer

Expert verified

The interval of convergence is \[ [-\sqrt{2}, \sqrt{2}] \].

Step by step solution

Identify the general term of the series

The general term of the power series is given by $a_n = \frac{x^{2n}}{2^n n^2}$.

Apply the Ratio Test

To find the radius of convergence, apply the Ratio Test: $\frac{a_{n+1}}{a_n} = \frac{\frac{x^{2(n+1)}}{2^{n+1} (n+1)^2}}{\frac{x^{2n}}{2^n n^2}} = \frac{x^{2(n+1)} \cdot 2^n n^2}{x^{2n} \cdot 2^{n+1} (n+1)^2} $. Simplifying this expression gives us $L = \lim_{n \to \infty} \left| \frac{x^2}{2} \cdot \frac{n^2}{(n+1)^2} \right| = \left| \frac{x^2}{2} \right| \cdot \lim_{n \to \infty} \frac{n^2}{(n+1)^2} $. Since $\lim_{n \to \infty} \frac{n^2}{(n+1)^2} = 1$, we obtain $L = \left| \frac{x^2}{2} \right|.$

Define the interval of convergence from Ratio Test

For the series to converge, the Ratio Test requires $L < 1$: \ \left| \frac{x^2}{2} \right| < 1.\ This simplifies to $ |x^2| < 2 $ which simplifies further to $ |x| < \sqrt{2} $. Therefore, the radius of convergence is $ \sqrt{2} $ and the interval of convergence is $ -\sqrt{2} < x < \sqrt{2} $.

Test the endpoints of the interval

Test the endpoints $x = \pm \sqrt{2}$: $x = \sqrt{2}$ gives $ \sum_{n=1}^{\infty} \frac{(\sqrt{2})^{2n}}{2^n n^2} = \sum_{n=1}^{\infty} \frac{2^n}{2^n n^2} = \sum_{n=1}^{\infty} \frac{1}{n^2} $, which converges by the p-series test. Similarly, $x = -\sqrt{2}$ gives $ \sum_{n=1}^{\infty} \frac{(-\sqrt{2})^{2n}}{2^n n^2} = \sum_{n=1}^{\infty} \frac{2^n}{2^n n^2} = \sum_{n=1}^{\infty} \frac{1}{n^2} $, which also converges.

State the complete interval of convergence

Since the series converges at both endpoints, the interval of convergence is \[ [-\sqrt{2}, \sqrt{2}] \].

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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 an infinite series of the form $ \sum_{n=0}^{ \infty} a_n (x - c)^n $, where $a_n$ are coefficients, and $c$ is a constant called the center of the series. Power series can be used to represent a variety of functions and have intervals over which they converge to a particular value. Understanding power series is critical as they provide a way to precisely approximate complex functions using polynomials.Key features of power series:

They include terms involving $x$ raised to increasing powers.
Each term is multiplied by a coefficient $a_n$.
The center $c$ defines the point about which the series is expanded.

The convergence of a power series depends significantly on the behavior of the coefficients and often involves testing endpoints to determine the full interval where the series holds true.

Ratio Test

The Ratio Test is a powerful tool used to determine the convergence or divergence of an infinite series. It's especially useful when dealing with factorials or exponential functions.How to apply the Ratio Test:

Given a series $\sum a_n$, consider the ratio of successive terms $\frac{a_{n+1}}{a_n}$.
Compute the limit $L$ as $n$ approaches infinity.
If $L < 1$, the series converges.
If $L > 1$, the series diverges.
If $L = 1$, the test is inconclusive.

Example:In the given problem, the series is $\sum \frac{x^{2n}}{2^n n^2}$. By applying the Ratio Test, we simplify the ratio of $\frac{a_{n+1}}{a_n}$ and find the limit $L = \left| \frac{x^2}{2} \right|$. Since $L < 1$ leads to convergence, we use the Ratio Test to confirm the series' behavior relative to its radius of convergence.

p-series test

The p-series test is a method used to determine the convergence of series in the form of $\sum \frac{1}{n^p}$, where $p$ is a positive constant.Main points:

A p-series converges if the exponent $p > 1$.
It diverges if $p \leq 1$.

In the given example, once the ratio test identifies potential convergence intervals, we use the p-series test to evaluate specific endpoints. For instance, testing the endpoint $x = \sqrt{2}$, we transform the series to $\sum \frac{1}{n^2}$, which is a p-series with $p = 2$ and therefore converges.
This comprehensive approach ensures we understand not just the radius, but also the full interval of convergence including its endpoints.

Radius of Convergence

The radius of convergence for a power series is the distance from the series' center to the boundary within which the series converges.Steps to Find the Radius:

Apply the Ratio Test to the general term of the series.
Identify the critical radius $R$ where the limit $L$ equals 1.

In formula form, when applying the Ratio Test, we often find a relationship like $L < 1$ leading to $ |x - c| < R $. For the given series, the series converges within $ |x| < \sqrt{2} $, meaning the radius of convergence is $R = \sqrt{2}$. After finding $R$, always test the endpoints separately to determine if the series converges or diverges at those points.
This process identifies the interval of convergence as $[-\sqrt{2}, \sqrt{2}]$.