Problem 55 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

\(\lim _{n \rightarrow \infty}\left(\tan \theta+\frac{1}{2} \tan \frac{\theta}{2}+\frac{1}{2^{2}} \tan \frac{\theta}{2^{2}}+\ldots+\frac{1}{2^{n}} \tan \frac{\theta}{2^{n}}\right)=\) (A) \(\frac{1}{\theta}\) (B) \(\frac{1}{\theta}-2 \cot 2 \theta\) (C) \(2 \cot 2 \theta\) (D) None of these

Step-by-Step Solution

Answer

The limit is \(2 \cot 2 \theta\), so the answer is (C).

Step 1: Identify the Series Components

The series given in the problem can be written as:\[\tan \theta + \frac{1}{2} \tan \frac{\theta}{2} + \frac{1}{2^2} \tan \frac{\theta}{2^2} + \ldots + \frac{1}{2^n} \tan \frac{\theta}{2^n}\]Recognize that this is an infinite series with terms of the form \(\frac{1}{2^k} \tan \frac{\theta}{2^k}\) for \(k = 0, 1, 2, \ldots, n\). Our task is to find the limit of the sum as \(n\) approaches infinity.

Step 2: Express the Series Using Sum Notation

The series can also be expressed using summation notation:\[S_n = \sum_{k=0}^{n} \frac{1}{2^k} \tan \frac{\theta}{2^k}\]We are interested in finding \(\lim_{n \to \infty} S_n\).

Step 3: Sum of Infinite Series

To find the sum of the infinite series, let's examine what happens as \(n\) approaches infinity. Notice that for large \(k\), the term \(\tan \frac{\theta}{2^k}\) converges to \(\tan 0 = 0\). Thus, the value of each successive term becomes smaller and approaches zero.

Step 4: Express Using Telescoping Series

Consider the identity for tan difference to simplify our series:\[\tan A - \tan B = \frac{\sin(A-B)}{\cos A \cos B}\]With trigonometric transformations, \(\tan \theta = 2 \cot (2\theta)\) can be derived when full expansion includes transformations using cotangent properties.

Step 5: Conclude the Simplified Series

Using trigonometric identities and telescopic method, the sum can be transformed and simplified to:\[S = 2 \cot 2\theta\]This expression arises as most intermediate terms cancel each other out when infinitely expanded.

Key Concepts

Infinite Series

An infinite series is essentially a sum of an endless list of numbers, where each number is called a term. In the given exercise, we had an infinite series represented by \[ \tan \theta + \frac{1}{2} \tan \frac{\theta}{2} + \frac{1}{2^2} \tan \frac{\theta}{2^2} + \ldots + \frac{1}{2^n} \tan \frac{\theta}{2^n} \]This series has terms that decrease in magnitude as the power of 2 in the denominator increases. This allows us to determine the pattern of the series:

• Each term is of the form \( \frac{1}{2^k} \tan \frac{\theta}{2^k} \).
• As \( n \to \infty \), these terms shrink towards zero, suggesting the series converges.

To solve the original problem, we find the limit of the series as \( n \) reaches infinity. This requires recognizing the form of the sequence and realizing that its progressive terms will lead to a finite sum, even though there are infinite terms.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that hold true for all values of the variables within their domains. They simplify complex trigonometric expressions, making it easier to solve problems involving trigonometric limits or sums. In the context of the exercise:

• The identity \( \tan A - \tan B = \frac{\sin(A-B)}{\cos A \cos B} \) provides valuable simplification for the series.
• Using transformations, \( \tan \theta \) can be expressed as \( 2 \cot(2\theta) \), helping to simplify the expression further when looking for the limit.

These identities help to reduce the series into a form where telescoping and limit evaluation techniques become applicable. Mastering these identities is crucial for solving such infinite series problems effectively.

Telescoping Series

A telescoping series is one where many terms cancel out with subsequent terms, simplifying the sum to involve only a few remaining terms. In our exercise, we used this method to find the limit of the given series:

• The key is using the trigonometric identity to express the terms in a way such that when summed, intermediate terms cancel out.
• Upon rearranging using \( \tan \theta = 2 \cot(2\theta) \), the series telescopes, leaving a sum of the outermost terms after others cancel.

This results in a cleaner, simpler expression: \( S = 2 \cot 2\theta \), where most individual terms diminish, allowing the infinite nature of the series to resolve into a finite, comprehensible answer. Understanding the telescoping process is vital in series problem-solving, providing insight into the often hidden simplicity of infinite sums.