Problem 313 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Sum the infinite series \(1+\frac{1}{3 \cdot 2^{2}}+\frac{1}{5 \cdot 2^{4}}+\frac{1}{7 \cdot 2^{6}}+\cdots\)

Step-by-Step Solution

Answer

The sum of the given infinite series is \( \frac{4}{3}\)

Step 1: Identify the series

First notice that the series \(1+\frac{1}{3 \cdot 2^{2}}+\frac{1}{5 \cdot 2^{4}}+\frac{1}{7 \cdot 2^{6}}+\cdots\) is a geometric series. The terms are presented in a specific pattern. Each term can be rewritten as \(\frac{1}{(2n-1) \cdot 2^{2n}}\) where 'n' starts from 0 and increases by 1 each time.

Step 2: Identify the common ratio and first term

Upon rewriting the terms, it's noticed that in the nth term \(\frac{1}{(2n-1) \cdot 2^{2n}}\), the coefficient of n in the denominator keeps increasing by 2 for each new term. Hence the common ratio 'r' is \(\frac{1}{2^{2}}\) and first term 'a' is 1.

Step 3: Apply the formula for sum of an infinite geometric series

Now that values of 'a' and 'r' are known, the formula for the sum of an infinite geometric series which is \(\frac{a}{1-r}\) can be applied. Plug in the values to get the sum, \(\frac{1}{1-\frac{1}{4}}\).

Key Concepts

Geometric Series

A geometric series is a series where each term after the first is found by multiplying the previous term by a constant called the common ratio. This means if you know the first term and the common ratio, you can determine any term in the series.
For example, in the series

• First term (\(a\)) = 1
• Second term = \(\frac{1}{3 \cdot 2^{2}}\)
• Third term = \(\frac{1}{5 \cdot 2^{4}}\)

The formula to find the nth term is given by \(\frac{1}{(2n-1) \cdot 2^{2n}}\).
Notice how each term involves a multiplication by a constant factor from the previous term.

Sum of Series

The sum of a series is the total value when you add up all the terms in the series. For a finite series, you simply add each term one by one. However, infinite series require a different approach.
For finite geometric series, the sum can be calculated using a specific formula depending on the common ratio being positive or negative. But for an infinite geometric series, you would use:

• \( S = \frac{a}{1-r} \)

where \(a\) is the first term and \(r\) is the common ratio. This formula can be used when \(|r| < 1\) to find the sum of an infinite series.

Infinite Geometric Series

An infinite geometric series extends indefinitely but retains the fundamental characteristic of multiplication by a common ratio. The common ratio must be such that its absolute value is less than one. This ensures that the series converges, meaning it adds up to a finite sum.
For instance, consider the series provided:

• First term (\(a\)) = 1
• Common ratio (\(r\)) = \(\frac{1}{4}\)

Restating this information into the infinite series sum formula, we can calculate the sum as \(\frac{1}{1 - \frac{1}{4}}\).

Series Pattern

Recognizing the pattern in a series is crucial to understanding and solving it. A series may follow an arithmetic, geometric, or another complex formation pattern. For the above problem, the series follows a geometric pattern, specifically in a structured form where each term is given by \(\frac{1}{(2n-1) \cdot 2^{2n}}\), with \(n\) starting at 0.

This pattern allows us to systematically determine each subsequent term. By comprehending the series' pattern, it becomes easier to analyze and apply the appropriate formulas. This understanding directly helps in evaluating the sum, especially for infinite series, where direct summation isn't feasible.