Problem 94 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

\(\lim _{n \rightarrow \infty} \sum_{r=1}^{n} \cot ^{-1}\left(\frac{r^{3}-r+\frac{1}{r}}{2}\right)\) is equal to (A) 0 (B) \(\pi\) (C) \(\frac{\pi}{2}\) (D) None of these

Step-by-Step Solution

Answer

(A) 0

Step 1: Understanding the Function

We need to evaluate \( \lim_{n \rightarrow \infty} \sum_{r=1}^{n} \cot^{-1}\left(\frac{r^{3}-r+\frac{1}{r}}{2}\right) \). The function involves an inverse cotangent and a summation.

Step 2: Simplification of the Expression

Consider the expression inside \( \cot^{-1} \), \( \frac{r^{3} - r + \frac{1}{r}}{2} \). For large \( r \), the \( \frac{1}{r} \) becomes negligible as it approaches zero.

Step 3: Analyze Large r Approximation

When \( r \) is large, \( r^{3} \) dominates \( r \) and \( \frac{1}{r} \), making \( \frac{r^{3} - r + \frac{1}{r}}{2} \approx \frac{r^{3}}{2} \). The cotangent inverse for large expressions tends to zero: \( \cot^{-1}(x) \to 0 \) as \( x \to \infty \).

Step 4: Converging Series

As \( n \rightarrow \infty \), the series is a sum of terms that become increasingly smaller. Hence the entire series tends to a constant multiplied by large but decreasing fractions, converging to zero.

Step 5: Final Limit Evaluation

As all terms \( \cot^{-1}\left(\frac{r^{3}-r+\frac{1}{r}}{2}\right) \to 0 \) as \( r \to \infty \), the sum \( \lim_{n \to \infty} \sum_{r=1}^{n} \cot^{-1}\left(\frac{r^{3}-r+\frac{1}{r}}{2}\right) \to 0 \).

Key Concepts

Inverse Trigonometric Functions

Understanding inverse trigonometric functions is crucial when dealing with limits in calculus. These functions, also known as arcus functions, provide the angles of right-angled triangles when the trigonometric ratios are known. One widely used function within this category is the inverse cotangent, denoted as \( \cot^{-1}(x) \). It gives the angle whose cotangent is \( x \). The range of \( \cot^{-1}(x) \) is typically from \( 0 \) to \( \pi \), giving it a unique position in solving many types of limit problems.

When evaluating limits such as \( \lim_{n \to \infty} \sum_{r=1}^{n} \cot^{-1}\left(\frac{r^{3}-r+\frac{1}{r}}{2}\right) \), it becomes crucial to understand the behavior of \( \cot^{-1}(x) \) as \( x \) approaches infinity.

• For large values of \( x \): \( \cot^{-1}(x) \to 0 \)
• For small values of \( x \): The function approaches \( \frac{\pi}{2} \) as \( x \to 0 \).

This phenomenon makes \( \cot^{-1}(x) \) ideal for analyzing limits where terms simplify significantly for large \( r \) values.

Series Convergence

Series convergence is a vital concept in calculus, particularly when evaluating limits of sequences and summations. It helps determine whether a series sums up to a finite value, which is critical in analyzing infinite processes. For a series \( \sum_{r=1}^{n} a_r \), convergence implies that the sum becomes stable as \( n \) grows.

To recognize convergence, consider the following aspects:

• Diminishing Terms: Each term must eventually lessen in value, dwindling toward zero as \( r \) increases.
• Bound Testing: If terms diminish fast enough relative to the series comparisons, then convergence is viable.

In our problem, each term \( \cot^{-1}\left(\frac{r^{3}-r+\frac{1}{r}}{2}\right) \) tends towards zero for large \( r \). Therefore, despite the sum having multiple terms, the decreasing value ensures that the series converges towards a stable result.

This converging characteristic allows us to simplify complex calculations by predicting how the sum behaves as \( n \to \infty \), ultimately contributing to the solution that the series converges to zero.

Large r Approximation

The technique of making a large \( r \) approximation simplifies complex expressions by focusing on the dominant terms for large values. This method is immensely useful when dealing with limits involving sums or integrals. By ignoring insignificant terms, we can better grasp the behavior of a function as it scales.

In the expression \( \frac{r^{3}-r+\frac{1}{r}}{2} \), the approximation is straightforward:

• For large \( r \), \( \frac{1}{r} \) becomes negligible.
• The \( r^{3} \) term overshadows \( -r \) due to its power.
• Thus the expression simplifies to \( \frac{r^{3}}{2} \).

This simplification expedites calculations significantly because for large \( r \), any smaller terms become minimal in impact, aiding in understanding the function's scaling behavior.

In this problem, the inverse cotangent of such a large value trends towards zero, reinforcing that each term in the summation is vanishingly small as \( n \to \infty \). Therefore, applying large \( r \) approximations is key in deducing that the entire series will converge to zero.