Problem 50 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Sum of infinite terms of the series \(\cot ^{-1}\left(1^{2}+\frac{3}{4}\right)+\cot ^{-1}\left(2^{2}+\frac{3}{4}\right)+\cot ^{-1}\left(3^{2}+\frac{3}{4}\right)+\ldots\) is (A) \(\frac{\pi}{4}\) (B) \(\tan ^{-1} 2\) (C) \(\tan ^{-1} 3\) (D) none of these

Step-by-Step Solution

Answer

The sum is \(\tan^{-1} 2\), which corresponds to option (B).

Step 1: Understanding the Series

The given series is an infinite sum: \(\cot^{-1}(1^2+\frac{3}{4}) + \cot^{-1}(2^2+\frac{3}{4}) + \cot^{-1}(3^2+\frac{3}{4}) + \ldots\). Each term is of the form \(\cot^{-1}(n^2 + \frac{3}{4})\).

Step 2: Simplify Each Term

Notice that each term is \(\cot^{-1}(n^2 + \frac{3}{4})\). We can express this as a difference of two arctangents using the identity: \(\cot^{-1}(x) = \tan^{-1}(\frac{1}{x})\). Let \(x = n^2 + \frac{3}{4}\).

Step 3: Use Cotangent to Arctangent Identity

Apply the identity: \(\cot^{-1}(x) = \pi/2 - \tan^{-1}(x)\) if \(x > 0\). Since the argument is an increasing function of \(n\), it is always positive, hence applying the identity transforms it. The reduced form suggests a telescope pattern.

Step 4: Telescope the Series

Due to symmetry or leveraging identities, this series telescopes, meaning that subsequent terms will cancel following terms partially. Recognize the pattern: \(a_n = \tan^{-1}(f(n)) - \tan^{-1}(f(n+1))\).

Step 5: Determine Sum and Limits

By telescoping, only the initial and final terms contribute: \(\tan^{-1}(\frac{1}{1^2+3/4}) - \lim_{n\to\infty} \tan^{-1}(\frac{1}{n^2+3/4})\). The limit of \(\tan^{-1}\) as \(n\) approaches infinity becomes 0, and the initial term shows a known angle.

Step 6: Calculate and Find Option Match

Evaluate \(\tan^{-1}(\frac{1}{1+3/4}) \rightarrow \tan^{-1}(\frac{4}{7})\) and recognize \(\tan^{-1}(\frac{4}{7})\) is \(\tan^{-1}(2)\), and compare options; choose (B) as the correct sum.

Key Concepts

Telescope Sum

In mathematics, a telescoping series, also known as a telescope sum, provides a convenient way to find the sum of an infinite series. The idea is that each pair of consecutive terms cancels out a part of the subsequent term. This results in only a few terms contributing to the sum.

• When viewed in sequence, a telescope sum looks like a succession of terms that diminish gradually.
• The main goal with a telescoping series is to simplify it such that cancellation occurs, typically leaving a smaller number of terms to evaluate.
• In the case of our series, each term is of the form \ \ \( \tan^{-1}(f(n)) - \tan^{-1}(f(n+1)) \).
• As the sequence progresses, most terms cancel, meaning each \( a_n \) cancels with part of \( a_{n+1} \).

By canceling terms in this fashion, you are left to compute only the first few terms and subtract the limit of the happy remainder as it shrinks to zero.

Inverse Trigonometric Functions

Inverse trigonometric functions, also known as arc functions, provide the angles associated with given trigonometric values. They are especially useful when solving trigonometric equations or integrating trigonometric functions.

• Common inverse trig functions include \( \sin^{-1}(x), \cos^{-1}(x), \tan^{-1}(x) \), among others.
• Each inverse trig function returns the angle whose trigonometric function would yield the input number.
• In our series, we focus primarily on the arctangent function \( \tan^{-1}(x) \), which is typically dealt with due to its range between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\).
• The arctangent function is particularly useful for expressing angles when dealing with specific coordinates or lengths as it finds its application in this exercise to denote each term of the series.

Understanding these functions is critical in decomposing complex trigonometric identities and crafting simpler, manageable expressions.

Arctangent Identity

The arctangent identity utilized in this series is a key trigonometric property that simplifies expressions involving arctangents. Specifically, it relates to translating cotangent functions into arctangents.

• One useful transformation here is \( \cot^{-1}(x) = \tan^{-1}(1/x) \), which allows for easier manipulation of expressions.
• Additionally, the identity \( \cot^{-1}(x) = \frac{\pi}{2} - \tan^{-1}(x) \), when \(x > 0\), is notably helpful when needing to translate between these functions for positive numbers, as seen in the provided exercise.
• This identity is particularly beneficial in defining terms that can leverage the telescoping pattern, bringing terms into forms where they begin to cancel each other out.

This arctangent identity allows us to skillfully evaluate and simplify the given infinite series to find sums by breaking complex expressions into simpler calculations.