Chapter 9: Problem 14
Find the radius of convergence. $$\sum_{n=0}^{\infty}\left(2^{n}+n^{2}\right) x^{n}$$
Short Answer
Expert verified
The radius of convergence is \( \frac{1}{2} \).
Step by step solution
01
Identify form of power series
The series given is \( \sum_{n=0}^{\infty} (2^n + n^2) x^n \). This series can be rewritten in the form \( \sum_{n=0}^{\infty} a_n x^n \) where \( a_n = 2^n + n^2 \).
02
Apply the Ratio Test
The Radius of Convergence \( R \) of a power series is found using the ratio test. The ratio test is applied using the formula: \[ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|. \]
03
Compute the ratio of consecutive terms
Substitute \( a_n \) with \( 2^n + n^2 \): \( a_{n+1} = 2^{n+1} + (n+1)^2 \).Compute the ratio: \[ \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{2^{n+1} + (n+1)^2}{2^n + n^2} \right| = \left| \frac{2 \cdot 2^n + n^2 + 2n + 1}{2^n + n^2} \right|. \]
04
Simplify the limit of the ratio
Factor out \( 2^n \) from both numerator and denominator: \[ \left| \frac{2(1 + \frac{n^2 + 2n + 1}{2^{n+1}})}{1 + \frac{n^2}{2^n}} \right|. \]As \( n \to \infty \), \( \frac{n^2}{2^n} \to 0 \) and \( \frac{n^2 + 2n + 1}{2^{n+1}} \to 0 \).Thus, \( L = \lim_{n \to \infty} 2 = 2 \).
05
Determine the Radius of Convergence
The Radius of Convergence \( R \) is given by \( \frac{1}{L} \). Since \( L = 2 \), \( R = \frac{1}{2} \).
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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 a type of infinite series that takes the form:
The given series is:
Power series allow the representation of a wide variety of functions once the range (radius) of the variable \( x \) for which the series converges is known.
- \( \sum_{n=0}^{\infty} a_n x^n \)
The given series is:
- \( \sum_{n=0}^{\infty} (2^n + n^2) x^n \)
Power series allow the representation of a wide variety of functions once the range (radius) of the variable \( x \) for which the series converges is known.
Ratio Test
The ratio test is a popular tool for determining the convergence of infinite series. It works by examining the ratio of successive terms in a series, which can give insight into the behavior of the series as it progresses. For any series \( \sum a_n \), the ratio test involves evaluating:
- \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \)
- Compute the ratio \( \frac{a_{n+1}}{a_n} \).
- Simplify the expression and take the limit as \( n \to \infty \).
- If \( L < 1 \), the series converges absolutely.
- If \( L > 1 \), the series diverges.
- If \( L = 1 \), the test is inconclusive.
- The ratio \( \frac{a_{n+1}}{a_n} \) was simplified to \( 2 \).
Convergence
Convergence refers to whether or not the sum of an infinite series approaches a finite number. Understanding convergence is crucial in determining if a series like our power series effectively represents a function over a certain interval. By using the ratio test, we calculated the limit \( L = 2 \), which allowed us to determine that:
- If \( x \) is within the radius of convergence, the series converges.
- \( R = \frac{1}{2} \)