Problem 11 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

Let \(S=\sum_{r=1}^{n} \cot ^{-1}\left(2^{r+1}+\frac{1}{2^{r}}\right)\), then find \(\lim _{n \rightarrow \infty}(S)\).

Step-by-Step Solution

Answer

Therefore, \(\lim_{n \rightarrow \infty} S = \pi * (n-1) - \arccot(2.5)\).

Step 1: Simplify the cotangent inverse

The cotangent inverse (arccotangent) function can be expressed in terms of the natural logarithm (logarithm with base e). Therefore, rewrite the expression inside the cotangent inverse to prepare it for simplification. So, \(2^{r+1}+\frac{1}{2^{r}} = 2^r*2+\frac{1}{2^{r}} = 2*2^r+2^{-r}\).

Step 2: Verify the given series

The range of arccot (x) is \( (0, \pi) \). For all x > 0, arccot (2x) + arccot (x/2) = \(\pi\). So, \(\arccot\left(2^2 *2^r+2^{-r}\right) + \arccot\left(2 * 2^{r-1} + 2^{-(r-1)}\right) = \pi\). Therefore, the given series is a telescoping series, which is a series whose partial sums eventually only have a fixed number of terms after cancellation. This will simplify the calculation of the limit.

Step 3: Calculate the limit

Knowing that the series is telescoping, we can express the sum as \(\sum_{r=1}^{n} \arccot\left(2^{r+1}+\frac{1}{2^{r}}\right) = [\arccot(4+1) + \arccot(0.5+2) - \arccot(2+0.5)] + [\arccot(8+0.5) + \arccot(0.25+4) - \arccot(4+0.5)] + ... + [\arccot\left(2^{n+2}+\frac{1}{2^{n}}\right) - \arccot\left(2^{n+1}+\frac{1}{2^{n-1}}\right)]\). After simplification, the sum S turns into \(\pi * (n-1) + \arccot\left(2^{n+2}+\frac{1}{2^{n}}\right) - \arccot(2.5)\). As n approaches to infinity, \(\lim _{n \rightarrow \infty} \arccot\left(2^{n+2}+\frac{1}{2^{n}}\right) = 0\). This is because the arccot function approaches 0 as its argument approaches to infinity.

Key Concepts

Telescoping Series

A telescoping series is a fascinating concept that simplifies the process of finding a series' sum. When we talk about a telescoping series, think of it like a series of dominoes. If you push the first domino, it knocks over the next, until all but a few are left standing.
In mathematical terms, a telescoping series is a series of terms that, after performing some algebraic operations, collapse into only a few remaining terms. This drastic reduction is what makes solving them much simpler.

• The key is identifying a pattern where terms cancel out.
• Most of the terms in the sum get eliminated by the next term in the sequence.

For our problem, each part of the series has terms that cancel when the series is expanded, leaving only a small number of terms to sum. This canceling effect makes it much more manageable to evaluate the sum or its limit, as seen in our solution.

Limits

Understanding limits is crucial when dealing with sequences or series that go on indefinitely. A limit helps us understand the behavior of a function as it approaches a certain point or extends towards infinity.
In our context, after simplifying the telescoping series, we look at the term that approaches a limit as the number of terms, denoted by \( n \), goes to infinity. The limit will often tell us what the sum converges to, even if the series itself isn’t fully calculable.

• Limits demonstrate where a series or sequence gets 'closer and closer' without necessarily reaching a specific value.
• They encapsulate the behavior at the 'ends' of a function or sequence.

In the presented solution, the arccotangent term approaches 0 as its argument becomes infinitely large, explaining why \( S \) converges to a specific value as \( n \) approaches infinity.

Infinite Series

An infinite series is essentially a summation of an endless sequence of numbers or expressions. Understanding how infinite series work helps us solve complex mathematical problems.
The most intriguing part about infinite series is how they can sum to a finite number even though they continue endlessly. In mathematics:

• Each term in the series represents a step in an unending sequence.
• The concept of convergence is vital, as it shows whether the infinite series approaches a finite limit or not.

In our example, even though we are summing an infinite number of terms, the series can still converge. Working with telescoping series within an infinite sum shows this beautifully. By identifying that only a finite number of terms affect the result, we don't have to add up to infinity to understand the series' behavior.