Problem 11
Question
In Exercises \(7-28,\) find the interval of convergence of the power series. (Be sure to include a check for convergence at the endpoints of the interval.) $$ \sum_{n=0}^{\infty} \frac{x^{n}}{n !} $$
Step-by-Step Solution
Verified Answer
The interval of convergence for the given power series is all real numbers.
1Step 1: Apply The Ratio Test
The Ratio Test for a given series \(\sum_{n=0}^{\infty} a_n\) involves calculating the limit as \(n\) goes to infinity of the absolute value of \(\frac{a_{n+1}}{a_n}\). If this limit is less than 1, the series converges absolutely, and if it is greater than 1, the series diverges. In this exercise, we are dealing with the power series \(\sum_{n=0}^{\infty} \frac{x^{n}}{n !}\). Thus, \(a_n = \frac{x^{n}}{n !}\) and \(a_{n+1} = \frac{x^{n+1}}{(n+1) !}\). The absolute value of the ratio \(\frac{a_{n+1}}{a_n}\) is then \(\frac{x}{n+1}\). Taking the limit of this as \(n\) goes to infinity, we get 0.
2Step 2: Interpreting The Result
Since 0 is less than 1, the power series converges absolutely for all real numbers \(x\). The radius of convergence is infinity.
3Step 3: Checking The Endpoints
In this instance, since the radius of convergence is infinite, the interval of convergence is all real numbers regardless of the behavior at the endpoints. As a result, there is no need to check the endpoints for convergence in this case.
Key Concepts
Power SeriesRatio TestRadius of Convergence
Power Series
When we talk about a power series, we are looking at an infinite series in the form of \( \sum_{n=0}^\infty a_n x^n \), where \( a_n \) are coefficients, and \( x \) can be any variable. It's essential to understand that a power series is like a polynomial with infinitely many terms and can represent functions within a certain interval, known as the interval of convergence.
Power series can be centered around different points, commonly \( x = 0 \), but the central point, often referred to as \( c \), can be any real number, which would give the series the form \( \sum_{n=0}^\infty a_n (x - c)^n \). The power series expansion can provide accurate approximations for functions and can converge or diverge depending on the value of \( x \). Therefore, determining where it converges—that is, the set of values for \( x \) for which the series sums to a finite number—is a critical aspect of working with power series.
Power series can be centered around different points, commonly \( x = 0 \), but the central point, often referred to as \( c \), can be any real number, which would give the series the form \( \sum_{n=0}^\infty a_n (x - c)^n \). The power series expansion can provide accurate approximations for functions and can converge or diverge depending on the value of \( x \). Therefore, determining where it converges—that is, the set of values for \( x \) for which the series sums to a finite number—is a critical aspect of working with power series.
Ratio Test
The Ratio Test is a vital tool used in calculus to determine the convergence or divergence of an infinite series. To perform the Ratio Test, you calculate the limit of the absolute value of the ratio of successive terms, \( \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_{n}}\right| \).
If this limit is less than 1, the series converges absolutely; if it is greater than 1, the series diverges; and if the limit is exactly 1, the test is inconclusive. In the case of the power series \( \sum_{n=0}^\infty \frac{x^{n}}{n!} \), applying the Ratio Test simplifies to the limit of \( \frac{x}{n+1} \) as \( n \) approaches infinity, which is 0 for any finite value of \( x \). Hence, the series converges for all real numbers.
If this limit is less than 1, the series converges absolutely; if it is greater than 1, the series diverges; and if the limit is exactly 1, the test is inconclusive. In the case of the power series \( \sum_{n=0}^\infty \frac{x^{n}}{n!} \), applying the Ratio Test simplifies to the limit of \( \frac{x}{n+1} \) as \( n \) approaches infinity, which is 0 for any finite value of \( x \). Hence, the series converges for all real numbers.
Radius of Convergence
The radius of convergence is the distance from the center of the power series expansion, within which the series converges. It is a numerical value that dictates how 'wide' the interval of convergence is on the number line. To find this, you can use various tests, including the Ratio Test.
In the exercise provided, the Ratio Test showed that the power series \( \sum_{n=0}^\infty \frac{x^{n}}{n!} \) converges for all real numbers \( x \), which implies that the radius of convergence is infinite. This is a particular characteristic of some power series, especially those, like the exponential series, that converge for every real-number input. In most other cases, the radius will be a finite value, and you would then need to check for convergence at the endpoints to articulate the complete interval of convergence.
In the exercise provided, the Ratio Test showed that the power series \( \sum_{n=0}^\infty \frac{x^{n}}{n!} \) converges for all real numbers \( x \), which implies that the radius of convergence is infinite. This is a particular characteristic of some power series, especially those, like the exponential series, that converge for every real-number input. In most other cases, the radius will be a finite value, and you would then need to check for convergence at the endpoints to articulate the complete interval of convergence.
Other exercises in this chapter
Problem 11
Verify that the infinite series diverges. $$ \sum_{n=1}^{\infty} \frac{n^{2}}{n^{2}+1} $$
View solution Problem 11
Use the Integral Test to determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty} \frac{\arctan n}{n^{2}+1} $$
View solution Problem 11
In Exercises \(7-18\), find the Maclaurin polynomial of degree \(n\) for the function. $$ f(x)=\sin x, \quad n=5 $$
View solution Problem 11
In Exercises 11-14, prove that the Maclaurin series for the function converges to the function for all \(x\). $$ f(x)=\cos x $$
View solution