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Chapter 9: Problem 296

Find \(\sum_{n=1}^{\infty} \frac{1}{(2 n-1) !}\)

Short Answer

Expert verified
The sum of the series \(\sum_{n=1}^{\infty} \frac{1}{(2 n-1) !}\) is \(\frac{1}{2}(e^{-1} + e^{1}) - 1\)

Step by step solution

01

Recognize the Taylor series for exponential function

The Taylor series for the function \(e^x\) at \(x = -1\) and \(x = 1\) can be written as: \[e^{-1} = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!}\] \[e^{1} = \sum_{n=0}^{\infty} \frac{1^n}{n!}\] These series include terms for both odd and even values of \(n\).
02

Divide the series into even and odd terms

For \(e{-1}\) and \(e{1}\) split the series into two, one for even \(n\) and another for odd \(n\), using the fact that \(n = 2k\) for even \(n\) and \(n = 2k-1\) for odd \(n\), where \(k\) is a natural number. Now, note that the sum of \(e^{-1}\) and \(e^{1}\) will cover all the odd factorial terms needed.
03

Determine even and odd terms series

\[e^{-1} + e^{1} = \sum_{k=0}^{\infty} \frac{(-1)^{2k}}{(2k)!} + \sum_{k=0}^{\infty} \frac{1^{2k}}{(2k)!} + \sum_{k=0}^{\infty} \frac{(-1)^{2k+1}}{(2k+1)!} + \sum_{k=0}^{\infty} \frac{1^{2k+1}}{(2k+1)!}\] The series for odd factorials starting from \(k=1\) to infinity is the series we need to find. So, the sum of \(\frac{1}{(2 n-1) !}\) from \(n=1\) to infinity will be: \[\sum_{n=1}^{\infty} \frac{1}{(2 n-1) !} = \frac{1}{2}(e^{-1} + e^{1}) - 1\]

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Key Concepts

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

Taylor Series
A Taylor series is a way to represent a function as an infinite sum of terms. These terms are derived from the values of a function's derivatives at a single point. The general formula for a Taylor series centered around zero (also known as a Maclaurin series) is:
  • \[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \dots \]
  • Each term in the series involves the function's derivatives, factorials, and powers of \( x \).
In the context of the problem, the Taylor series for the exponential function \(e^x\) is used to find the sum of the series of odd factorials. Understanding the Taylor series allows us to break down complex functions into sums that are easier to analyze and compute.
Exponential Function
The exponential function, denoted as \(e^x\), is a mathematical function of the form:
  • \[ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \]
  • This series can be expanded for both positive and negative values of \(x\).
For example, the exponential function at \(x=1\) and \(x=-1\) are:
  • \[ e^{1} = \sum_{n=0}^{\infty} \frac{1^n}{n!} \]
  • \[ e^{-1} = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} \]
These expansions allow us to compute the exponential function value using a series of simple calculations.In this exercise, these series are split to distinguish between odd and even terms, ultimately reaching the desired sum of odd factorial terms.
Factorials
Factorials, denoted by an exclamation point (!), represent the product of all positive integers up to a given number. For example:
  • \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \)
  • \(0!\) is a special case which equals 1 by definition.
Factorials play a crucial role in the Taylor series and the exponential function as they appear in the denominators of the series terms.By splitting the series of the exponential function into even and odd indexed terms, we work with only odd factorials.This allows the calculation of sums involving these odd factorial terms, as seen in the original exercise.
Odd Terms
Odd terms in a series are those associated with odd indices. For instance, from the Taylor series expansion:
  • The term \( x^1/1! \) is an odd term.
  • The term \( x^3/3! \) is another example of an odd term.
In the exercise, the focus is on extracting all odd terms from the exponential function's Taylor series.This extraction is critical, as we are interested in the sum of odd factorial terms:
  • Using the properties of even and odd functions, the solution breaks them into simpler, distinct parts for calculation.
  • Through this method, we establish the relationship of odd terms to the respective exponential functions at specific points.
Understanding this separation helps clarify how seemingly complex infinite series sums are evaluated effectively.

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Most popular questions from this chapter

If \(a_{1}, a_{2}, a_{3}, \ldots, a_{n}\) are in A.P. where \(a_{i}>0\) for all \(i\), show that \(\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\cdots \cdots+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_{n}}}=\frac{n-1}{\sqrt{a_{1}}+\sqrt{a_{n}}}\)

Find three numbers in G.P. whose sum is 65 and whose product is 3375 .

If the sum of \(n\) terms of an AP be \(3 n^{2}-n\), then find its first term and common difference.

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. \(\frac{1}{a_{1} a_{2}}+\frac{1}{a_{2} a_{3}}+\ldots+\frac{1}{a_{n-1} a_{n}}=\frac{n-1}{a_{1} a_{n}}\).

The product of three numbers in G.P. is 125 and sum of their products taken in pairs is \(87 \frac{1}{2}\). Find them.
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