Problem 44 - Calculus Early Transcendentals: Pearson New International Edition

Question

Show that $$ \lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left[\frac{1}{1+(k / n)^{2}}\right] \frac{1}{n}=\frac{\pi}{4} $$

Step-by-Step Solution

Answer

The limit is \( \frac{\pi}{4} \) as it represents the integral \( \int_{0}^{1} \frac{1}{1+x^2} \, dx \).

Step 1: Recognize the Integral Representation

Notice that the expression resembles a Riemann sum for the integral of a function. The terms in the sum \[ \sum_{k=1}^{n} \left[ \frac{1}{1+(k/n)^{2}} \right] \frac{1}{n} \] indicate the integral of the function \( f(x) = \frac{1}{1+x^2} \) on the interval \([0,1]\). If \( x = \frac{k}{n} \), then \( \frac{1}{n} \) acts as \( dx \). This is important as we use this fact to translate the sum into an integral form.

Step 2: Identify the Corresponding Integral

Translate the Riemann sum into the definite integral: \[ \int_{0}^{1} \frac{1}{1+x^2} \, dx \]This integral representation shows that finding the limit of the Riemann sum will be equivalent to evaluating this integral.

Step 3: Evaluate the Integral

The integral \( \int_{0}^{1} \frac{1}{1+x^2} \, dx \) is known to be an arc tangent, since the derivative of \( \tan^{-1}(x) \) is \( \frac{1}{1+x^2} \). Thus, the evaluation proceeds as follows:\[ \left[ \tan^{-1}(x) \right]_{0}^{1} = \tan^{-1}(1) - \tan^{-1}(0) \]From trigonometry, \( \tan^{-1}(1) = \frac{\pi}{4} \) and \( \tan^{-1}(0) = 0 \), so \[ \tan^{-1}(1) - \tan^{-1}(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4} \].

Step 4: Conclude the Limit

Since the sum \( \sum_{k=1}^{n} \left[ \frac{1}{1+(k/n)^{2}} \right] \frac{1}{n} \) is a Riemann sum that represents the integral \( \int_{0}^{1} \frac{1}{1+x^2} \, dx \) and the integral evaluates to \( \frac{\pi}{4} \), we conclude that the limit is \( \frac{\pi}{4} \).

Key Concepts

Limit of a Sequence

When dealing with Riemann sums, the limits of sequences transform a sum into an integral as the number of subintervals (or sequence terms) approaches infinity. In our exercise, the sequence we focus on is:

• \[ \sum_{k=1}^{n} \left[ \frac{1}{1+(k/n)^{2}} \right] \frac{1}{n} \]

As \( n \rightarrow \infty \), the idea is that the width of each rectangle (represented by \( \frac{1}{n} \)) becomes infinitesimally small. Consequently, the Riemann sum seamlessly transitions into a definite integral.
When we calculate the limit, we are essentially checking that the infinite sequence of sums approached by the Riemann sum resolves to the exact area under the curve of the function \( \frac{1}{1+x^2} \) between 0 and 1. Thus, the limit of this sequence equals \( \frac{\pi}{4} \).

Definite Integral

A definite integral represents the exact area under a curve over a specified interval. In the case of our current exercise, the integral to be calculated is:

• \[ \int_{0}^{1} \frac{1}{1+x^2} \, dx \]

This integration serves as the continuous analog of the discrete Riemann sum we've considered. It transforms the approximation of area (via finite sums) into an exact number.
The evaluation of this integral involves recognizing that the integrand, \( \frac{1}{1+x^2} \), is the derivative of the inverse tangent function, \( \tan^{-1}(x) \). This direct connection allows us to use the fundamental theorem of calculus straightforwardly. Calculating it over our interval [0, 1], we find the integral evaluates to \( \frac{\pi}{4} \), which coincidentally also equals the sequence's limit.

Trigonometric Functions

Trigonometric functions often arise in calculus due to their smooth periodic nature and fundamental properties. In this exercise, the tangent function plays a crucial role.
The integral \( \int_{0}^{1} \frac{1}{1+x^2} \, dx \) hinges on trigonometric identities. Recognizing that \( \frac{1}{1+x^2} \) has a derivative counterpart in \( \tan^{-1}(x) \) allows us to turn this problem into an easier evaluation of:

• \[ \left[ \tan^{-1}(x) \right]_{0}^{1} \]

Using well-known trigonometric values, such as:

• \( \tan^{-1}(1) = \frac{\pi}{4} \)
• \( \tan^{-1}(0) = 0 \)

We determine the definite integral as \( \frac{\pi}{4} \), succinctly connecting this problem to trigonometric functions' useful properties in calculus.