Chapter 9: Problem 25
Find the sum of each of the following series by recognizing how it is related to something familiar. (a) \(x-x^{2}+x^{3}-x^{4}+x^{5}-\cdots\) (b) \(\frac{1}{2 !}+\frac{x}{3 !}+\frac{x^{2}}{4 !}+\frac{x^{3}}{5 !}+\cdots\) (c) \(2 x+\frac{4 x^{2}}{2}+\frac{8 x^{3}}{3}+\frac{16 x^{4}}{4}+\cdots\)
Short Answer
Expert verified
(a) \(\frac{x}{1+x}\). (b) \(\frac{e^x - 1 - x}{x^2}\). (c) \(-\ln(1+2x)\).
Step by step solution
01
Identify the Type of Series for Part (a)
The series given is of the form \(x - x^2 + x^3 - x^4 + x^5 - \cdots\). This is a geometric series where the first term \(a = x\) and the common ratio \(r = -x\).
02
Determine Sum of Infinite Geometric Series for Part (a)
The formula for the sum \(S\) of an infinite geometric series is \(S = \frac{a}{1-r}\), where \(|r| < 1\). For our series, \(a = x\) and \(r = -x\), so the sum is \(S = \frac{x}{1-(-x)} = \frac{x}{1+x}\).
03
Identify the Pattern for Part (b)
The series is \(\frac{1}{2 !} + \frac{x}{3 !} + \frac{x^{2}}{4 !} + \cdots\), which resembles an exponential series shifted by one index. It can be rewritten as \(\sum_{n=2}^{\infty} \frac{x^{n-2}}{n!}\).
04
Relate Part (b) to Known Exponential Series
The series \(\sum_{n=0}^{\infty} \frac{x^{n}}{n!}\) is the expansion of \(e^x\). Therefore, \(\sum_{n=2}^{\infty} \frac{x^{n-2}}{n!} = \frac{e^x - 1 - x}{x^2}\), where terms for \(n=0\) and \(n=1\) are subtracted.
05
Factorial and Fraction Simplification for Part (c)
Consider \(2x + \frac{4x^2}{2} + \frac{8x^3}{3} + \cdots\), which can be rewritten as \(\sum_{n=1}^{\infty} \frac{2^n x^n}{n}\). This resembles the series \(-\ln(1-r)\) where \(r = -2x\).
06
Use Logarithmic Series for Part (c)
The sum of the series \(\sum_{n=1}^{\infty} \frac{r^n}{n} = -\ln(1-r)\), applies. Setting \(r = -2x\), we get \(-\ln(1+2x)\). Therefore, this series sums to \(-\ln(1+2x)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Geometric Series
Geometric series are a fundamental concept in calculus and algebra. A geometric series is a series of the form \( a + ar + ar^2 + ar^3 + \cdots \), where each term increases by a common ratio \( r \). The first term is denoted by \( a \). To find the sum of an infinite geometric series, you need to ensure that \(|r| < 1\). If this condition is met, the sum \( S \) of the series can be determined using the formula:
- \( S = \frac{a}{1 - r} \)
Exponential Series
An exponential series is directly related to the concept of the exponential function, most commonly symbolized by \( e^x \). This type of series plays a critical role in calculus, particularly in solving differential equations and modeling growth processes. The standard exponential series is given by
In the exercise, the series \( \frac{1}{2!} + \frac{x}{3!} + \frac{x^2}{4!} + \cdots \) is identified as a shifted exponential series. By rewriting it as \( \sum_{n=2}^{\infty} \frac{x^{n-2}}{n!} \), it relates directly back to the exponential series for \( e^x \). However, some terms are missing (those corresponding to \( n = 0 \) and \( n = 1 \)), so their contributions \( 1 \) and \( x \) are subtracted, resulting in the expression \( \frac{e^x - 1 - x}{x^2} \). Understanding these shifting patterns is vital for accurately applying exponential functions to real-world problems.
- \( \sum_{n=0}^{\infty} \frac{x^n}{n!} \)
In the exercise, the series \( \frac{1}{2!} + \frac{x}{3!} + \frac{x^2}{4!} + \cdots \) is identified as a shifted exponential series. By rewriting it as \( \sum_{n=2}^{\infty} \frac{x^{n-2}}{n!} \), it relates directly back to the exponential series for \( e^x \). However, some terms are missing (those corresponding to \( n = 0 \) and \( n = 1 \)), so their contributions \( 1 \) and \( x \) are subtracted, resulting in the expression \( \frac{e^x - 1 - x}{x^2} \). Understanding these shifting patterns is vital for accurately applying exponential functions to real-world problems.
Logarithmic Series
Logarithmic series are used to express logarithmic functions as infinite sums and have numerous applications in mathematics and the sciences. The basic form of a logarithmic series is:
- \( \sum_{n=1}^{\infty} \frac{r^n}{n} = -\ln(1-r) \)