Problem 45 - Calculus Early Transcendentals: Pearson New International Edition

Question

For \(k \geq 0\), find $$ \lim _{n \rightarrow \infty} \frac{1^{k}+2^{k}+\cdots+n^{k}}{n^{k+1}} $$ Hint: Though this has the \(\infty / \infty\) form, l'Hôpital's Rule is not helpful. Think of a Riemann sum.

Step-by-Step Solution

Answer

The limit is \( \frac{1}{k+1} \).

Step 1: Understanding the Problem Statement

We are given a limit expression \( \lim _{n \rightarrow \infty} \frac{1^{k}+2^{k}+\cdots+n^{k}}{n^{k+1}} \). This is indeed an indeterminate form \( \frac{\infty}{\infty} \). Our task is to find the limit using the concept of a Riemann sum instead of l'Hôpital's Rule.

Step 2: Rewriting the Summation as a Riemann Sum

The numerator \( 1^k + 2^k + \cdots + n^k \) can be represented as \( \sum_{i=1}^{n} i^k \). This can approximate the integral \( \int_{0}^{1} x^k \, dx \) by considering it as a Riemann sum when divided by \( n^{k+1} \).

Step 3: Expressing the Riemann Sum

We express \( \sum_{i=1}^{n} i^k \) as a function of the integral:\[ \frac{1}{n^{k+1}} \sum_{i=1}^{n} i^k \approx \int_{0}^{1} x^k \, dx \text{ as } n \to \infty. \] The Riemann sum \( \frac{i^k}{n^{k+1}} \) represents small increments \((\Delta x)\) over \([0,1]\).

Step 4: Evaluating the Integral

The integral \( \int_{0}^{1} x^k \, dx \) becomes:\[ \frac{1}{k+1} \text{ since the integral of } x^k \text{ is } \frac{x^{k+1}}{k+1}. \] Therefore, the Riemann sum approaches \( \frac{1}{k+1} \) as \( n \to \infty \).

Step 5: Concluding the Limit

Thus, the limit \( \lim _{n \rightarrow \infty} \frac{1^{k}+2^{k}+\cdots+n^{k}}{n^{k+1}} \) is equal to the value of the integral, hence \( \frac{1}{k+1} \).

Key Concepts

Riemann Sum

A Riemann Sum is a method used to approximate the total value of a function, often to understand the concept of integration. It divides the function into small segments that approximate areas under the curve. The total sum of these segments gives an approximation of the integral.
To understand Riemann Sums, it's helpful to think about them visually: the graph of the function is divided into many thin rectangles. Each rectangle's area is calculated and summed up.
Subdivide the interval \(a, b\) into small segments of width \(\Delta x\). Each sub-interval \(i\) contributes \(f(x_i) \Delta x\) to the sum, where \(f(x_i)\) is the height at a selected point \(x_i\) in the segment. The Riemann Sum becomes more accurate as the number of segments \(n\) increases.

• It is essentially the sum of the areas of the rectangles under the curve.
• Best applies to functions that are continuous over the interval \(a, b\).
• The more rectangles used, the closer the approximation to the exact integral.

Riemann Sums convert sums into integrals as segment partitioning improves, giving a framework to approximate definite integrals.

Limits

The concept of limits is fundamental in calculus. A limit helps us understand the behavior of a function as it approaches a specific point or as a variable approaches infinity or negative infinity.
Consider \(\lim_{x \to a} f(x)\), which indicates where the function \(f(x)\) tends to as \(x\) approaches \(a\). Limits paradoxically allow us to discuss outcomes at points not necessarily present in the function.
In this problem, we consider the expression: \(\lim_{n \rightarrow \infty} rac{1^{k}+2^{k}+\cdots+n^{k}}{n^{k+1}}\). Both the numerator and the denominator tend to infinity individually, creating an indeterminate form \(\frac{\infty}{\infty}\).

• Understanding limits helps in determining waypoints in evaluating complex expressions.
• Limits are crucial in analyzing function behavior at boundary points or continuity.
• They also define the basis for derivatives and integrals through infinitesimal approaches.

By employing limits, we realize the behavior of infinite series and other continuous behaviors in functions as \(n \) gets infinitely large.

Integration

Integration, an essential calculus concept, involves finding the total size or value, such as area under the curve for a given function. It is the inverse operation of differentiation and can be seen as accumulating quantities.
The definite integral, denoted as \( \int_{a}^{b} f(x) \, dx \), evaluates the total area under the curve \(f(x)\) from \(a\) to \(b\). The fundamental theorem of calculus links derivatives and integrals, showing how accumulation and rates of change are connected.
In our problem, integration helps simplify the infinite series \(1^k + 2^k + \cdots + n^k\), through the expression of \( \int_{0}^{1} x^k \, dx \).

• Integration builds on Riemann Sums, refining area approximations to exact measurements.
• It supports solving physical problems like finding distances, areas, or volumes.
• Integration methods include substitution, parts, and partial fractions, among others.

By understanding integration, we fathom the underlying essence of accumulation in mathematical modeling and real-world scenarios.

Indeterminate Forms

Indeterminate forms arise in calculus when the limits of functions result in undefined expressions such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). These forms require special techniques, since traditional algebraic manipulation cannot resolve them.
In our exercise, we encounter \(\frac{\infty}{\infty}\), which demands clever techniques like substituting a sum with a Riemann Sum rather than defaulting to l'Hôpital's Rule.
These tricky expressions remind us that the world of calculus isn't always straightforward; sometimes we need alternate pathways.

• Understanding these forms is vital since they appear frequently in calculus problems.
• l'Hôpital's Rule is a standard but not always sufficient tool for \(\frac{0}{0}\) and \(\frac{\infty}{\infty}\).-type indeterminacies.
• Riemann Sums and substitution often provide alternative methods to tackle these problems.

Handling indeterminate forms allows deeper insight into challenging calculus problems, compelling us to explore beyond conventional methods.