/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 32 Find the radius of convergence a... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the radius of convergence and the interval of convergence. $$\sum_{k=0}^{\infty} \frac{k !}{2^{k}} x^{k}$$

Short Answer

Expert verified
Radius of convergence \( R = 0 \); interval of convergence is \( x = 0 \).

Step by step solution

01

Recognize the Power Series

The given series is \( \sum_{k=0}^{\infty} \frac{k !}{2^{k}} x^{k} \). This is a power series of the form \( \sum_{k=0}^{\infty} a_k x^k \) where \( a_k = \frac{k!}{2^k} \).
02

Apply the Ratio Test

To find the radius of convergence, apply the ratio test. The ratio test states the series \( \sum a_k x^k \) converges if \( \lim_{k \to \infty} \left | \frac{a_{k+1}}{a_k} x \right | < 1 \). Compute \( \lim_{k \to \infty} \left | \frac{(k+1)!}{2^{k+1}} \cdot \frac{2^k}{k!} \right | \cdot |x| \)
03

Simplify the Ratio

The expression simplifies to \( \lim_{k \to \infty} \left | \frac{(k+1)}{2} \right | \cdot |x| \). Further simplify to \( \frac{|x|}{2} \cdot (k+1) \).
04

Evaluate the Limit

Evaluate the limit as \( k \) approaches infinity: \( \lim_{k \to \infty} \frac{|x|}{2} \cdot (k+1) = \infty \) unless \( x = 0 \).
05

Determine the Radius of Convergence

The result is \( \infty \), indicating divergence except possibly at \( x = 0 \), where the series converges. Thus, the radius of convergence \( R = 0 \).
06

Find the Interval of Convergence

Since the radius of convergence \( R = 0 \), the interval of convergence is \( x = 0 \).

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Key Concepts

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

Understanding the Ratio Test for Power Series
When examining a power series like \( \sum_{k=0}^{\infty} \frac{k !}{2^{k}} x^{k} \), it is crucial to determine where this series will converge. One useful tool for this is the ratio test.
The ratio test helps us figure out whether a series converges by comparing the limit of consecutive terms.
For a power series of the form \( \sum a_k x^k \), the ratio test states that the series will converge if:
  • \( \lim_{k \to \infty} \left | \frac{a_{k+1}}{a_k} x \right | < 1 \)
Applying this to our series means calculating \( \lim_{k \to \infty} \left | \frac{(k+1)!}{2^{k+1}} \cdot \frac{2^k}{k!} \right | \cdot |x| \).
After substitution and simplification, the ratio becomes \( \left(\frac{|x|}{2} \right) (k+1)\).
As \( k \) approaches infinity, the expression goes to infinity unless \( x = 0 \). This means the series will converge only at that point.
Exploring the Interval of Convergence
The interval of convergence tells us the specific values of \( x \) for which a power series converges.
Once we know the radius of convergence, finding this interval becomes easier.
In our example, after using the ratio test, the series only converges when \( x = 0 \), meaning the radius of convergence is \( R = 0 \).
The interval of convergence is then simply the single point \( x = 0 \).
This means the power series does not converge for any other values of \( x \), because the ratio test indicated divergence elsewhere.
Thus, understanding the radius of convergence directly tells us the limited interval where the series is convergent.
Introduction to Power Series
A power series is a series of the form \( \sum_{k=0}^{\infty} a_k x^k \), where each term is a power of \( x \) multiplied by a coefficient \( a_k \).
Power series are important in mathematics because they can represent functions over certain intervals.
  • They offer ways to approximate more complex functions.
  • Many functions can be expressed as a power series.
In our specific problem, the series \( \sum_{k=0}^{\infty} \frac{k !}{2^{k}} x^{k} \) is a power series centered around \( x = 0 \).
We describe a power series in terms of its coefficients \( a_k \) and radius of convergence.
The radius is crucial as it tells us the extent to which values of \( x \) in the real number line this series converges.
Power series can also help in solving differential equations and modeling physical phenomena.

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Most popular questions from this chapter

Exercise will show how a partial sum can be used to obtain upper and lower bounds on the sum of a series when the hypotheses of the integral test are satisfied. This result will be needed in Exercises. It was stated in Exercise 35 that $$\sum_{k=1}^{\infty} \frac{1}{k^{4}}=\frac{\pi^{4}}{90}$$ (a) Let \(s_{n}\) be the \(n\) th partial sum of the series above. Show that $$ s_{n}+\frac{1}{3(n+1)^{3}} < \frac{\pi^{4}}{90} < s_{n}+\frac{1}{3 n^{3}} $$ (b) We can use a partial sum of the series to approximate \(\pi^{4} / 90\) to three decimal-place accuracy by capturing the sum of the series in an interval of length 0.001 (or less). Find the smallest value of \(n\) such that the interval containing \(\pi^{4} / 90\) in part (a) has a length of 0.001 or less. (c) Approximate \(\pi^{4} / 90\) to three decimal places using the midpoint of an interval of width at most 0.001 that contains the sum of the series. Use a calculating utility to confirm that your answer is within 0.0005 of \(\pi^{4} / 90\).

Apply the divergence test and state what it tells you about the series. Apply the divergence test and state what it tells you about the series. $${(a)} \sum_{k=1}^{\infty} \frac{k^{2}+k+3}{2 k^{2}+1}$$ $${(b)}\sum_{k=1}^{\infty}\left(1+\frac{1}{k}\right)^{k}$$ $${(c)}\sum_{k=1}^{\infty} \cos k \pi$$ $${(d)}\sum_{k=1}^{\infty} \frac{1}{k !}$$

(a) We will see later that the polynomial \(1-x^{2} / 2\) is the "local quadratic" approximation for \(\cos x\) at \(x=0.\) Make a conjecture about the convergence of the series $$\sum_{k=1}^{\infty}\left[1-\cos \left(\frac{1}{k}\right)\right]$$ by considering this approximation. (b) Try to confirm your conjecture using the limit comparison test.

Find the Taylor polynomials of orders \(n=0,1,2,3\) and 4 about \(x=x_{0},\) and then find the \(n\) th Taylor polynomial for the function in sigma notation. $$\frac{1}{x} ; x_{0}=-1$$

Find the first four distinct Taylor polynomials about \(x=x_{0},\) and use a graphing utility to graph the given function and the Taylor polynomials on the same screen. $$f(x)=\cos x ; x_{0}=\pi$$

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