Problem 47 - Essential Calculus

Question

Determine the convergence or divergence of the sequence with the given \(n\) th term. If the sequence converges, find its limit. \(a_{n}=\frac{n^{p}}{e^{n}}, p>0\)

Step-by-Step Solution

Answer

The sequence converges and its limit is 0.

Step 1: Apply the Limit Test

To determine whether a sequence converges, check if the limit of the sequence as \(n\) approaches infinity exists. In other words, calculate \(\lim_{n\to\infty}\frac{n^{p}}{e^{n}}\).

Step 2: Application of L'Hopital's rule

Given that it's the limit of type \(0/0\) or \(\infty/\infty\), apply L'Hopital's rule. The rule states that \(\lim_{x\to c}\frac{f(x)}{g(x)}\) is equivalent to \(\lim_{x\to c}\frac{f'(x)}{g'(x)}\) when either limit involves an indeterminate form.

Step 3: Finding the derivative

The derivative of \(n^{p}\) is \(p \cdot n^{p-1}\) and the derivative of \(e^{n}\) is \(e^{n}\). Therefore, the new limit becomes \(\lim_{n\to\infty}\frac{p \cdot n^{p-1}}{e^{n}}\). Since the denominator increases exponentially faster than the numerator, this limit equals zero.

Step 4: Determine convergence or divergence

A sequence converges if the limit exists and is finite. In this case, the limit is 0, hence the sequence converges.

Key Concepts

Limit of a Sequence

The concept of the limit of a sequence is essential to determine convergence or divergence. In simple terms, a sequence is a set of numbers in a specific order. Mathematically, a sequence \( \{a_n\} \) converges to a limit \( L \) as \( n \) approaches infinity if for every positive number \( \varepsilon \), there is an integer \( N \) such that for all \( n \geq N \), the terms \( a_n \) are within \( \varepsilon \) of \( L \).

This means that as you go further along the sequence, the numbers get closer and closer to \( L \). If a sequence doesn't get closer to any single number, it diverges.

For example, consider a sequence where the \( n \)th term is \( \frac{n^p}{e^n} \). To determine if this sequence converges, we calculate the limit \( \lim_{n \to \infty} \frac{n^p}{e^n} \). If this limit results in a finite number, the sequence converges, otherwise it diverges.

L'Hopital's Rule

L'Hopital's Rule is a powerful tool for finding limits that involve indeterminate forms. When you encounter limits of the form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), you can apply this rule.

L'Hopital's Rule states that if you have a limit \( \lim_{x \to c} \frac{f(x)}{g(x)} \), and both \( f(c) \) and \( g(c) \) give zero or are infinite, then this limit is equivalent to \( \lim_{x \to c} \frac{f'(x)}{g'(x)} \), provided the latter limit exists.

In our sequence \( \frac{n^p}{e^n} \), as \( n \to \infty \), both numerator and denominator grow very large, forming the \( \frac{\infty}{\infty} \) type. Thus, by applying L'Hopital's Rule, we find \( \lim_{n \to \infty} \frac{p \cdot n^{p-1}}{e^n} \). This subsequent limit helps us evaluate the convergence of the sequence.

Exponential Functions

Exponential functions are a crucial part of calculus, characterized by rapid growth rates. An exponential function takes the form \( a^x \), where \( a \) is a constant and \( x \) is the variable. In our sequence problem, the function \( e^n \) is exponential.

The base \( e \approx 2.718 \) is special, known as Euler's number, and it appears frequently due to its properties in growth processes and calculus.

When comparing exponential functions to polynomial functions, exponentials grow much faster. This growth rate is why in the sequence \( \frac{n^p}{e^n} \), as \( n \) gets very large, \( e^n \) quickly surpasses \( n^p \), and the fraction approaches zero. Understanding this relationship between exponential and polynomial growth helps us determine that the sequence converges, with its limit being zero.