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Find the radius of convergence and the interval of convergence. $$ \sum_{k=0}^{\infty} \frac{3^{k}}{k !} x^{k} $$

Short Answer

Expert verified
Radius of convergence is \(\infty\), interval is \((-\infty, \infty)\).

Step by step solution

01

Identify the Series Type

The given series is \( \sum_{k=0}^{\infty} \frac{3^k}{k!} x^k \). This is a power series of the form \( \sum_{k=0}^{\infty} c_k x^k \) where \( c_k = \frac{3^k}{k!} \). Notice that it resembles a Maclaurin series and has an exponential component.
02

Use the Ratio Test for Convergence

Apply the ratio test to determine the radius of convergence \( R \). The ratio test requires evaluating:\[L = \lim_{k \to \infty} \left| \frac{c_{k+1}}{c_k} \right| \times x = \lim_{k \to \infty} \left| \frac{3^{k+1}}{(k+1)!} \times \frac{k!}{3^k} \right| \times |x|\]
03

Simplify the Ratio Test

Simplify the expression from the ratio test:\[L = \lim_{k \to \infty} \left| \frac{3^{k+1}}{3^k} \times \frac{k!}{(k+1)!} \right| \times |x| = \lim_{k \to \infty} \left| \frac{3}{k+1} \right| \times |x| \]As \( k \to \infty \), \( \frac{3}{k+1} \to 0 \), so \( L = 0 \).
04

Determine the Radius of Convergence

Since the ratio \( L = 0 \) for any value of \( x \), the series converges for all \( x \). Thus, the radius of convergence \( R \) is \( \infty \).
05

Find the Interval of Convergence

Since the radius of convergence \( R = \infty \), the interval of convergence is \((-\infty, \infty)\), meaning that the series converges for all real numbers \( x \).

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Key Concepts

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

Interval of Convergence
When working with power series, finding the interval of convergence is a key task. The interval of convergence is the range of values for which a power series converges. Every power series has an associated radius of convergence, denoted by \( R \).
If \( R \) is infinite, the series converges for all real numbers, resulting in an interval of convergence from \( -\infty \) to \( \infty \). If \( R \) is finite, then the interval of convergence is \((c - R, c + R)\), where \( c \) is the center of the series.
  • Calculate \( R \) using the ratio or root test.
  • Consider endpoints separately as they might converge or diverge.
  • Express the interval with inequalities or interval notation.
In the given exercise, since \( R = \infty \), the series converges for all values of \( x \), leading to an interval of convergence of \((-\infty, \infty)\). This means that no matter what value you choose for \( x \), the series will sum to a finite number.
Power Series
A power series is a type of series that takes the form \( \sum_{k=0}^{\infty} c_k x^k \). This series essentially represents a polynomial with infinitely many terms centered around a particular point, often at \( x = 0 \), making it a Maclaurin series.
The coefficients \( c_k \) play a crucial role in determining the behavior and convergence of the series. These values dictate how rapidly the terms decrease, impacting convergence.
  • A power series resembles a finite polynomial but contains terms of all powers of \( x \).
  • It can represent various functions, especially when converging over a particular interval.
  • The nature and form of \( c_k \) influence the series' convergence.
In this exercise, \( c_k = \frac{3^k}{k!} \), exhibiting exponential behavior due to the factorial in the denominator. This particular form is related closely to the exponential function.
Ratio Test
The ratio test is a technique used to determine the convergence of a series, especially efficient for power series. It involves calculating the limit of the absolute ratio of consecutive terms. Formally, it states that a series \( \sum a_k \) converges absolutely if\[ L = \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| < 1 \]
If \( L > 1 \), the series diverges. If \( L = 1 \), the test is inconclusive. The ratio test helps identify the radius of convergence \( R \) by finding \( R = \frac{1}{L} \) when \( L \) is between 0 and 1.
  • Calculate \( L \) using the precise formula and simplify as needed.
  • Check if \( L \) is less than 1 for convergence.
  • Recognize that \( \frac{1}{L} \) determines \( R \), unless \( L = 0 \) indicating infinite convergence.
In the original problem, the ratio \( L \) is 0, indicating convergence for all \( x \), hence \( R = \infty \). This application of the ratio test simplifies understanding convergence behavior effectively.

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