Problem 26 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

Find the sum of \(\sum_{n=1}^{\infty} \tan ^{-1}\left(\frac{8 n}{n^{4}-2 n^{2}+5}\right)\)

Step-by-Step Solution

Answer

The sum of the series is \(\tan ^{-1}(8)-\tan ^{-1}(4)+ \frac{\pi}{2}\)

Step 1: Simplify arctangent expression

The first step is to simplify the expression inside the arctangent function. Fortunately, \(\tan ^{-1}(x)-\tan ^{-1}(y)=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)\) can be used for this purpose. Hence, \(\tan ^{-1}\left(\frac{8 n}{n^{4}-2 n^{2}+5}\right)\) can be rewritten as \(\tan ^{-1}(4 (n+1))-\tan ^{-1}(4 n)\).

Step 2: Use telescoping series

It can be observed that the series simplifies tremendously due to the telescoping nature (a series where most terms cancel out with subsequent terms), where \(-\tan ^{-1}(4 n)\) cancels \(\tan ^{-1}(4 (n+1))\) for all \(n \geq 1\). Thus, only first and last terms will survive.

Step 3: Find and compute limits

The limits \(n=1\) and \(n \to \infty\) are plugged in. Doing so, the series simplifies to \(\tan ^{-1}(8)-\tan ^{-1}(4)+\lim _{n \to \infty}[-\tan ^{-1}(4 n)]\). Note that \(-\tan ^{-1}(4 n)\) as \(n \to \infty\), goes to \(-\pi / 2\).

Step 4: Compute the sum

Finally, compute the sum by adding and subtracting the relevant arctangent values, which simplifies to \(\tan ^{-1}(8)-\tan ^{-1}(4)+ \frac{\pi}{2}\). This is the sum of the given infinite series.

Key Concepts

Arctangent Function

The arctangent function, denoted as \( \tan^{-1}(x) \), is an inverse trigonometric function used to find the angle whose tangent is \( x \). It is especially useful in calculus for simplifying and solving series and integrals.

Key properties of the arctangent function include:

• The range of \( \tan^{-1}(x) \) is from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\).
• \( \tan^{-1}(x) \) is an odd function, meaning \( \tan^{-1}(-x) = -\tan^{-1}(x) \).
• It helps convert complex expressions into simpler forms. As seen in the given exercise, the identity \( \tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1}\left(\frac{x-y}{1+xy}\right) \) is used to break down arctangent expressions.

Understanding these properties allows the manipulation of expressions for solving complex series. Recognizing how arctangent values subtract or add up is crucial in simplifying series through techniques like telescoping.

Telescoping Series

A telescoping series is a mathematical series where most terms cancel each other out, leaving only a few terms for computation. This simplification effectively reduces complex series calculations to simpler forms.

In the provided solution:

• The initial expression is reformulated using the arctangent identity to produce terms that nearly cancel each other when expanded, such as \(-\tan^{-1}(4n)\) and \(\tan^{-1}(4(n+1))\).
• The process results in cancellation, where all intermediate terms vanish, leaving only the first and the last terms of the series to evaluate.

Such series are powerful in solving challenging problems as they constrain endless evaluations into a simple subtraction of boundary terms. Understanding telescoping can significantly simplify the approach to infinite series.

Limits in Calculus

Limits are a foundational concept in calculus, necessary for handling infinite series and understanding the behavior of functions as variables approach specific values. They help in predicting the behavior of functions or terms at bounds.

In the context of this exercise:

• As \( n \to \infty \), the term \(-\tan^{-1}(4n)\) approaches \(-\frac{\pi}{2}\), indicating how \( \tan^{-1} \) behaves with large inputs, hinting at horizontal asymptotes which aid in determining series bounds.
• Limits transform a series evaluation problem into simpler terms, reflecting the nature of infinity and providing exact solutions for sums of infinitely extending series.

With this understanding of limits, we analyze how large inputs affect series outcomes, enabling precise solution formulations for otherwise complex infinite series.