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: Understand the sum

The expression \( 1^k + 2^k + \cdots + n^k \) can be thought of as the sum of the \( k \)-th powers of the first \( n \) natural numbers.

Step 2: Express as a Riemann Sum

Rewrite the sum \( 1^k + 2^k + \cdots + n^k \) as \( \sum_{i=1}^{n} i^k \). To connect this to a Riemann sum, note that \( i^k \) can be treated as a function evaluated at \( x_i = \frac{i}{n} \cdot n \) where the width of each subinterval is \( \frac{1}{n} \).

Step 3: Approximate as an integral

The sum \( \sum_{i=1}^{n} i^k \) is approximately equal to \( n^{k+1} \int_{0}^{1} x^k \, dx \) when \( n \rightarrow \infty \). This arises from thinking of \( x_i^k \) as \( \left(\frac{i}{n}\right)^k \) and noting the limit forms a Riemann sum.

Step 4: Evaluate the integral

Calculate the integral \( \int_{0}^{1} x^k \, dx \). This is \( \frac{1}{k+1} \), found by applying the power rule for integrals.

Step 5: Apply the integral to the limit

Considering the agreement with the definition of the integral, the expression simplifies as \( \lim _{n \rightarrow \infty} \frac{\sum_{i=1}^{n} i^k}{n^{k+1}} = \lim _{n \rightarrow \infty} \frac{n^{k+1} \int_{0}^{1} x^k \, dx}{n^{k+1}} \), which equals the integral \( \int_{0}^{1} x^k \, dx = \frac{1}{k+1} \).

Step 6: Conclude the limit

Thus, the limit \( \lim_{n \rightarrow \infty} \frac{1^k + 2^k + \cdots + n^k}{n^{k+1}} \) equals \( \frac{1}{k+1} \), realized since the division by \( n^{k+1} \) yields a form of the integral.

Key Concepts

Riemann sum

Understanding the concept of a Riemann sum is essential when relating discrete sums to continuous integrals. In mathematics, a Riemann sum is used to estimate the total area underneath a curve on a graph, which is often referred to as the integral of a function.
The Riemann sum breaks down this problem into small, manageable pieces—each piece can be seen as a rectangle with one side lying on the function.

• It involves partitioning the interval of interest into smaller sub-intervals.
• Each sub-interval is evaluated by sampling the function at a specific point, typically at the endpoint, midpoint, or any point within the interval.
• The sum of the areas of all these rectangles gives an approximation of the integral.

When dealing with our problem, the sum \( 1^k + 2^k + \cdots + n^k \) can be thought of as a Riemann sum.
By setting each \( i^k \) as the function evaluated at a certain point, we see that this setup leads us to the interval \( [0, 1] \), which becomes clearer as \( n \rightarrow \infty \).
This approximation helps transition from a sum to an integral, allowing for easier computation of limits.

Integral calculus

Integral calculus focuses on the accumulation of quantities and how they combine to solve for areas under curves, among other applications. The problem \( \lim _{n \rightarrow \infty} \frac{1^k + 2^k + \cdots + n^k}{n^{k+1}} \) shows how an integrated approach can simplify complex expressions.
The key here is to understand how to move from a discrete sum to a continuous integral which is a fundamental operation in calculus.

• The expression \( 1^k + 2^k + \cdots + n^k \) transforms to an approximate continuous form \( \int_{0}^{1} x^k \, dx \) as \( n \rightarrow \infty \).
• Using the power rule for integration, we know \( \int_{0}^{1} x^k \, dx = \frac{1}{k+1} \).

Integral calculus helps us to understand and solve limits of sums by considering these sums as Riemann sums representing integrals.
This is why l'Hôpital's Rule was not advised here; instead, thinking in terms of integral calculus allowed for a straightforward evaluation of the limit.

Limits

Limits are a central concept in calculus, describing what a function approaches as the input approaches a certain value. In this problem, as \( n \rightarrow \infty \), we sought the limit of \( \frac{1^k + 2^k + \cdots + n^k}{n^{k+1}} \).

• While it initially presents as an indeterminate form \( \frac{\infty}{\infty} \), transforming the sum into an integral simplifies the evaluation.
• By recognizing the sum as a Riemann sum, we express it as the integral \( \int_{0}^{1} x^k \, dx \).
• Thus, \( \lim_{n \rightarrow \infty} \frac{1^k + 2^k + \cdots + n^k}{n^{k+1}} = \frac{1}{k+1} \), through the application of integral calculus.

Understanding limits in this context underscores the interconnected nature of different calculus concepts like sums, integrals, and limits.
It allows us to transition from discrete calculations to finding solutions in the continuous domain efficiently.