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Chapter 9: Problem 300

Sum the infinite series \(1+\frac{3}{1 !}+\frac{5}{2 !}+\frac{7}{3 !}+\cdots \cdot\)

Short Answer

Expert verified
The given series does not have a sum as it diverges. It increases without limit as the terms increase.

Step by step solution

01

Recognize the Form of the Series

Recognize that the given series is a modified version of the Taylor series expansion for \(e^x\). The Taylor series expansion for \(e^x\) is \(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots\). Here, it appears in a modified form as \(1-\frac{3}{1!}+\frac{5}{2!}-\frac{7}{3!}+\cdots\), where each term can be written as \((-1)^(n-1)(\frac{2n-1}{(n-1)!})\) for \(n=1,2,3,...\)
02

Identify the Function

Substitute \(x=-1\) in the Taylor series of \(e^x\) to get the function that the given series represents. When \(x=-1\), the Taylor series for \(e^x\) becomes \(e^{-1}=1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\cdots\). This is almost the same as the given series, except that the nth term is multiplied by \(2n-1\), thus the function that the given series represents is \((2n-1)e^{-1}\)
03

Evaluate the Series

Given the series is infinite, the sum of the series will be equal to the value of the function. Therefore, the sum of the series is \(e^{-1}\). But then recall that each term in the series was multiplied by \(2n-1\) compared to the Taylor series. Therefore, to get the exact value of the series, it should be calculated for all values of n. When these values are summed, they will produce an infinite series. Therefore, the sum of the series is the sum of all values of \((2n-1)e^{-1}\) for \(n=1,2,3,...\). The series does not converge to a specific value, as \((2n-1)e^{-1}\) increases without limit as \(n\) increases

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Key Concepts

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

Taylor Series
Taylor series provide a powerful way to express functions as an infinite sum of terms. Each term in the series involves derivatives of the function evaluated at a single point. It is particularly useful for approximating complex functions using polynomials.

The general form of the Taylor series for a function \( f(x) \) around \( x = a \) is given by:
  • \( f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots \)

In our example, the function we are looking at is the exponential function \( e^x \) centered at \( x = 0 \). The series simplifies to:
  • \( 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \)

This becomes the basis for understanding the given series and its convergence.
Exponential Function
The exponential function is one of the most important functions in mathematics. It has unique properties such as the same value for both the function and its derivative.

The exponential function \( e^x \) can be expanded using its Taylor series:
  • \( 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \)

This series is infinite and converges for all real numbers \( x \).
In our example, by substituting \( x = -1 \), we explore how the series changes. The modified version, \( 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \cdots \), gives us \( e^{-1} \) and shows the remarkable utility of the Taylor series in representing \( e^x \) under different conditions.
Convergence
Convergence is a key concept in understanding infinite series. It tells us whether a series approaches a particular value as the number of terms increases.

For the Taylor series of \( e^x \), it converges for any real \( x \). However, the convergence can differ based on series modifications. In this exercise:
  • The original series modifies each term by multiplying it with \( (2n-1) \), making the convergence behavior different.
  • As \( n \) increases, \( (2n-1)e^{-1} \) grows without bound.

This results in a series that does not converge to a specific value, highlighting how slight changes in a series can drastically affect its behavior.

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Most popular questions from this chapter

If \(x, a, b, c\) are real and \((x-a+b)^{2}+(x-b+c)^{2}=0\), then show that \(a, b, c\) are in AP.

If \(a_{1}, a_{2}, a_{3}, \ldots, a_{n}\) be an A.P. of non-zero terms prove that i. \(\frac{1}{a_{1} a_{n}}+\frac{1}{a_{2} a_{n-1}}+\frac{1}{a_{3} a_{n-2}}+\ldots+\frac{1}{a_{n} a_{1}}=\frac{2}{a_{1}+a_{n}}\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\ldots+\frac{1}{a_{n}}\right)\) ii. \(\frac{1}{a_{1} a_{2}}+\frac{1}{a_{2} a_{3}}+\ldots+\frac{1}{a_{n-1} a_{n}}=\frac{n-1}{a_{1} a_{n}}\).

Between 1 and 31 are inserted \(m\) arithmetic means so that the ratio of the 7 th and \((m-1)\) th means is \(5: 9\). Find the value of \(m\)

If \(7^{\text {th }}\) and \(13^{\text {th }}\) terms of an AP be 34 and 64 respectively, then find its \(18^{\text {th }}\) term.

If the sums of \(p, q\) and \(r\) terms of an A.P. be \(a, b\) and \(c\) respectively then prove that \(\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)=0\)
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