Problem 72 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

\(\lim _{n \rightarrow \infty} \frac{1}{n}\left[1+\frac{n^{2}}{n^{2}+1^{2}}+\frac{n^{2}}{n^{2}+2^{2}}+\ldots+\frac{n^{2}}{n^{2}+(n-1)^{2}}\right]\) is equal to (A) \(\frac{\pi}{2}\) (B) \(\frac{\pi}{3}\) (C) \(\frac{\pi}{4}\) (D) \(\frac{\pi}{6}\)

Step-by-Step Solution

Answer

The limit is \( \frac{\pi}{4} \), matching option (C).

Step 1: Understand the Problem

We need to find the limit of a sequence as \( n \) approaches infinity. The sequence is a sum of terms, where each term has the form \( \frac{n^2}{n^2 + k^2} \) for \( k = 0, 1, 2, \ldots, n-1 \), and this sum is multiplied by \( \frac{1}{n} \).

Step 2: Rewriting the Sum

Rewrite the expression as \( \frac{1}{n} \sum_{k=0}^{n-1} \frac{n^2}{n^2 + k^2} \). Notice this resembles a Riemann sum, which can approximate an integral as \( n \rightarrow \infty \).

Step 3: Simplifying the Expression

Consider \( \frac{n^2}{n^2 + k^2} = 1 - \frac{k^2}{n^2 + k^2} \). So we rewrite the sum as \( \sum_{k=0}^{n-1} \left( 1 - \frac{k^2}{n^2 + k^2} \right)\). This simplifies to \( n - \sum_{k=0}^{n-1} \frac{k^2}{n^2+k^2} \).

Step 4: Approximate the Sum as an Integral

Now consider the term \( \frac{k^2}{n^2+k^2} \), which can be approximated by \( \frac{x^2}{1+x^2} \) when we let \( x = \frac{k}{n} \), giving us the integral \( \int_0^1 \left( 1 - \frac{x^2}{1+x^2} \right) \, dx \).

Step 5: Calculate the Integral

The integral \( \int_0^1 \left( 1 - \frac{x^2}{1+x^2} \right) \ dx \) simplifies to \( \int_0^1 \frac{1}{1+x^2} \ dx \), which is \( \left[ \arctan x \right]_0^1 \). Evaluating this gives \( \arctan(1) - \arctan(0) = \frac{\pi}{4} \).

Step 6: Conclusion

Therefore, after evaluating the integral, the original limit expression evaluates to \( \frac{1}{n}( n - n \cdot \text{integral value} ) = \frac{\pi}{4} \) for large \( n \).

Key Concepts

Riemann Sum

A Riemann sum is a method used in calculus to approximate the value of an integral. It is particularly helpful when trying to understand the concept of integration in situations where the exact integral is difficult to compute.
The basic idea behind a Riemann sum is to divide a region into small slices, and then sum up the area (or volume) of those slices to approximate the total area (or volume) under a curve.

• We divide the interval over which we want to integrate into smaller sub-intervals.
• Each sub-interval has a representative point, often taken as the left endpoint, right endpoint, or the midpoint.
• We calculate the value of the function at this representative point and multiply by the width of the sub-interval.
• The sum of these products gives us the Riemann sum.

As the number of sub-intervals increases (or equivalently, the width of each sub-interval decreases), the Riemann sum becomes more accurate. When the number of sub-intervals becomes infinite, under normal circumstances, the Riemann sum converges to the exact value of the integral.

Integral Approximation

Integral approximation is the process of estimating the value of a definite integral from a function given over an interval. When the function is continuous over the interval, a common method for approximating the integral is through the use of Riemann sums, as described previously. In our exercise, using a Riemann sum transforms our sum into an integral approximation because:

• The terms in the sum describe areas under the curve of a function, specifically the function related to the arctan function.
• By expressing the sum using a limit process as described (with \( n \) tending to infinity), the Riemann sum is a more accurate reflection of the integral of the function.
• The limit of such sums represents the precise area under the curve, especially as the intervals become infinitesimally small.

The accuracy of integral approximation increases as you use more partitions (increasing \( n \)). In this exercise, the conclusion that \( \frac{\pi}{4} \) is the value shows just how effective this method can be when \( n \) approaches infinity.

Arctan Function

The arctan function, also known as the inverse tangent function, is a crucial component in both trigonometry and calculus. It is the inverse of the tangent function, represented as \( \arctan x \) or \( \tan^{-1} x \). This function is particularly interesting in calculus because:

• It maps real numbers from its domain to the range \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \).
• The derivative of \( \arctan x \) is \( \frac{1}{1+x^2} \), which is the integrand in our solution process.
• This integral yields the arctan function when evaluated between our limits \( 0 \) and \( 1 \).
• Interestingly, this application helps in solving integrals that result from limits of Riemann sums and confirms the nature of \( \arctan \) as a foundational calculus tool.

Moreover, in this exercise, the outcome of integrating \( \frac{1}{1+x^2} \) from 0 to 1 is \( \arctan(1) - \arctan(0) = \frac{\pi}{4} \), demonstrating how it plays a vital role in connecting integral calculus with trigonometric functions.