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Chapter 11: Problem 7

Find the radius of convergence and interval of convergence of the series. $$\sum_{n=0}^{\infty} \frac{x^{n}}{n !}$$

Short Answer

Expert verified
The radius is infinite, and the interval is \((-\infty, \infty)\).

Step by step solution

01

Recognize the Series

The given series is \( \sum_{n=0}^{\infty} \frac{x^n}{n!} \). This is a power series in terms of \( x \), centered at \( x = 0 \). It resembles the Maclaurin series for the exponential function \( e^x \), which converges for all real \( x \).
02

Apply the Ratio Test for Convergence

To determine the radius of convergence, use the Ratio Test, which involves the limit \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \) where \( a_n = \frac{x^n}{n!} \). Calculate \[ \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{x^{n+1}}{(n+1)!} \cdot \frac{n!}{x^n} \right| = \left| \frac{x}{n+1} \right|. \]
03

Evaluate the Limit of the Ratio Test

Evaluate the limit \( \lim_{n \to \infty} \left| \frac{x}{n+1} \right| \). As \( n \to \infty \), \( \frac{x}{n+1} \to 0 \) for all \( x \). Since the limit is 0, which is less than 1, the series converges for all \( x \).
04

Determine the Radius of Convergence

The Ratio Test result implies the series converges for all real \( x \). Hence, the radius of convergence \( R \) is infinite, \( R = \infty \).
05

Define the Interval of Convergence

Since the radius of convergence is infinite, the interval of convergence is \( (-\infty, \infty) \). The series converges for all real \( x \).

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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 essentially an infinite series that functions much like a polynomial, but with infinitely many terms. It takes the form: \[ \sum_{n=0}^{\infty} a_n (x - c)^n \] where each term of the series is a power of \( x \), multiplied by a coefficient \( a_n \). The constant \( c \) is the center of the series, and in many cases, \( c = 0 \), which specifically forms a Maclaurin series. Power series can converge, which means that as \( n \) goes to infinity, the sum of the series approaches a specific value. However, the behavior of convergence depends on \( x \) and the series itself.
  • If \( x \) approaches certain values, the series might diverge (not settle to a specific sum).
  • The values of \( x \) where the series converges form the interval of convergence.
Ratio Test
The Ratio Test is a powerful tool to determine the convergence of an infinite series, especially a power series. It involves calculating the limit: \[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \] For a series \( \sum a_n \), where \( a_n \) is the general term of the series, if this limit is:
  • Less than 1, the series converges absolutely.
  • Greater than 1, or infinity, the series diverges.
  • Equal to 1, the test is inconclusive.
In the case of the exercise series \( \sum \frac{x^n}{n!} \), each term reduces to \( \left| \frac{x}{n+1} \right| \). As \( n \to \infty \), this expression approaches zero for all \( x \), confirming the series converges for all real numbers.
Maclaurin Series
The Maclaurin series is a special case of the power series that is centered at 0. It looks like this: \[ \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n \] where each term's coefficient involves derivatives of a function \( f \) evaluated at 0.
  • The Maclaurin series for \( e^x \) is particularly simple: \( \sum_{n=0}^{\infty} \frac{x^n}{n!} \).
  • This series converges for every possible value of \( x \), demonstrating its powerful approximation ability for the exponential function \( e^x \).
The exercise series \( \sum \frac{x^n}{n!} \) is a perfect example of a Maclaurin series, which is why it converges for all \( x \).
Interval of Convergence
The interval of convergence of a power series is the set of \( x \)-values for which the series converges to a finite limit. Here's how it's determined:
  • By finding the radius of convergence \( R \), which dictates how far from the center \( c \) the convergence extends.
  • For a radius \( R \), the interval is \( (c-R, c+R) \), but endpoints may need individual checking.
However, for the series \( \sum \frac{x^n}{n!} \), the limit derived from the Ratio Test was 0 for all \( x \) since \( \lim_{n \to \infty} \frac{x}{n+1} = 0 \). Thus, the radius of convergence is infinite, \( R = \infty \), making the interval of convergence \( (-\infty, \infty) \). This means it converges everywhere without exception.

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Most popular questions from this chapter

Show that the sequence defined by $$ a_{1}=2 \quad a_{n+1}=\frac{1}{3-a_{n}} $$ satisfies \(0< a_{n} \leqslant 2\) and is decreasing. Deduce that the sequence is convergent and find its limit.

(a) Approximate \(f\) by a Taylor polynomial with degree \(n\) at the number a. (b) Use Taylor's Inequality to estimate the accuracy of the approximation \(f(x) \approx T_{n}(x)\) when \(x\) lies in the given interval. (c) Check your result in part (b) by graphing \(\left|R_{n}(x)\right|\) $$f(x)=x \sin x, \quad a=0, \quad n=4, \quad-1 \leqslant x \leqslant 1$$

Let $$f(x)=\sum_{n=1}^{\infty} \frac{x^{n}}{n^{2}}$$ Find the intervals of convergence for \(f, f^{\prime},\) and \(f^{\prime \prime}\)

(a) Approximate \(f\) by a Taylor polynomial with degree \(n\) at the number a. (b) Use Taylor's Inequality to estimate the accuracy of the approximation \(f(x) \approx T_{n}(x)\) when \(x\) lies in the given interval. (c) Check your result in part (b) by graphing \(\left|R_{n}(x)\right|\) $$f(x)=x^{2 / 3}, \quad a=1, \quad n=3, \quad 0.8 \leqslant x \leqslant 1.2$$

(a) Approximate \(f\) by a Taylor polynomial with degree \(n\) at the number a. (b) Use Taylor's Inequality to estimate the accuracy of the approximation \(f(x) \approx T_{n}(x)\) when \(x\) lies in the given interval. (c) Check your result in part (b) by graphing \(\left|R_{n}(x)\right|\) $$f(x)=e^{x^{2}}, \quad a=0, \quad n=3, \quad 0 \leqslant x \leqslant 0.1$$
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