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Chapter 9: Problem 303

Sum the infinite series \(\frac{2}{3 !}+\frac{4}{5 !}+\frac{6}{7 !}+\frac{8}{9 !}+\cdots\)

Short Answer

Expert verified
The sum of the infinite series is \(蟺 - 2\).

Step by step solution

01

Identify the Pattern

Observe the series to determine a pattern. It is seen that the numerator is 2k and the denominator is (2k+1)! for k = 1,2,3,... . This pattern will help in the next steps.
02

Simplicity the fraction

The term can be simplified by writing it as 2k / (2k+1)!. Factorial (2k+1)! is the same as (2k+1) * (2k)!.Thus, the term becomes 2k / ((2k+1) * (2k)!), which simplifies to 2 / (2k+1).
03

Sum the Simplified Series

Now that the series is simplified, it becomes 鈭� [2 / (2k+1)] for k = 1 to infinity. This series is already known and its sum is simply 蟺 - 2. Therefore, the sum of the original series 鈭� [2k / (2k+1)!] is also 蟺 - 2.

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Key Concepts

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

Understanding Factorials
A factorial, denoted by an exclamation mark (!), is a product of an integer and all the integers below it down to 1. For example, 5 factorial, or \(5!\), is calculated as \(5 \times 4 \times 3 \times 2 \times 1 = 120\). Factorials grow very quickly and are crucial in many areas of mathematics, particularly in permutations and combinations.
In our series, the denominator includes a factorial, like \((2k+1)!\). This means you multiply all integers from \(1\) up to \((2k+1)\). Understanding how these grow helps us deal with the infinite series by recognizing how large these terms become.
Pattern Recognition in Series
Recognizing patterns in a series is like solving a puzzle. It allows you to simplify a given series and often aids in finding its sum. In our exercise, the terms were expressed differently between numerator and denominator. By examining them closely, a pattern can be deduced:
  • The numerator follows a sequence of even numbers: \(2k\).
  • The denominator follows a sequence of odd factorial numbers: \((2k+1)!\).
This recognition leads to a simplified form: \(\frac{2k}{(2k+1)!}\) can be rewritten using the pattern \(\frac{2}{2k+1}\), which is much simpler to work with as it reveals a familiar form.
Simplifying Fractions in Series
Simplifying fractions in a series is key to calculating sums efficiently. This involves breaking down the terms to their simplest forms. In the given problem, we simplify \(\frac{2k}{(2k+1)!}\) by recognizing that \((2k+1)!\) equals \((2k+1) \cdot (2k)!\). This lets us cancel terms, reducing the series to \(\frac{2}{2k+1}\).
This simpler form provides insight into summing the series. Often, a known result (like \(\pi - 2\) in this case) relates to these simplified terms, allowing for straightforward calculation. Simplifying fractions converts complex series into manageable forms, revealing deeper mathematical truths.

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Most popular questions from this chapter

The sum of \(n\) terms of the two series \(3+10+17+\ldots \ldots\) and \(63+65+67+\ldots \ldots\) are equal, then find the value of \(n\).

If \(S_{1}, S_{2}, S_{3}, \ldots, S_{m}\) are the sums of \(n\) term of \(m\) A.P.'s whose first terms are \(1,2,3, \ldots, m\) and common differences are \(1,3,5, \ldots ., 2 m-1\) respectively. Show that \(S_{1}+S_{2}+S_{3}+\ldots+S_{m}=\frac{1}{2} m n(m n+1)\).

For what value of \(n, \frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}}\) is the arithmetic mean of \(a\) and \(b ?\)

If the ratio of the sum of \(m\) term and \(n\) terms of an A.P. be \(m^{2}: n^{2}\), prove that the ratio of its \(m\) th and \(n\) th terms will be \(2 m-1: 2 n-1\).

If \(a, b, c, x\) are all real numbers, and \(\left(a^{2}+b^{2}\right) x^{2}-2 b(a+c) x+\left(b^{2}+c^{2}\right)=0\) then show that \(a, b, c\) are in G.P., and \(x\) is their common ratio.
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