Problem 303 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

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

Step-by-Step Solution

Answer

The sum of the infinite series is \(π - 2\).

Step 1: 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.

Step 2: 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).

Step 3: Sum the Simplified Series

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

Key Concepts

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.