Problem 36 - Single Variable Calculus

Question

Determine whether the sequence \(\left\\{a_{n}\right\\}\) converges or diverges. If it converges, find its limit. \(a_{n}=\frac{n^{p}}{e^{n}}, \quad p>0\)

Step-by-Step Solution

Answer

The sequence \(a_{n} = \frac{n^{p}}{e^{n}}\), where \(p > 0\), converges and its limit is 0.

Step 1: Use limit test for convergence

We will find the limit of the sequence \(a_{n}\) as \(n\) approaches infinity. If the limit exists and is finite, the sequence converges. If the limit does not exist or is infinity, the sequence diverges.

Step 2: Finding the limit of the sequence

To find the limit of the sequence, we need to find the limit of the function \(a_{n} = \frac{n^{p}}{e^{n}}\) as \(n\) approaches infinity. \[\lim_{n\to\infty} \frac{n^{p}}{e^{n}}\]

Step 3: Use L'Hôpital's Rule

Since both the numerator and denominator functions grow without bound as \(n\) approaches infinity, we can use L'Hôpital's Rule, which states that if \(\lim_{n\to\infty} \frac{f(n)}{g(n)}\) is of the form \(\frac{\infty}{\infty}\), then: \[\lim_{n\to\infty} \frac{f(n)}{g(n)} = \lim_{n\to\infty} \frac{f'(n)}{g'(n)}\] where \(f'(n)\) and \(g'(n)\) are the derivatives of \(f(n)\) and \(g(n)\) with respect to \(n\). Taking the derivatives, we find: \[f'(n) = p n^{p-1}\] \[g'(n) = e^{n}\]

Step 4: Apply L'Hôpital's Rule

Applying L'Hôpital's Rule to the limit, we have: \[\lim_{n\to\infty} \frac{n^{p}}{e^{n}} = \lim_{n\to\infty} \frac{p n^{p-1}}{e^{n}}\] Now, we can apply L'Hôpital's Rule again since the limit is still of the form \(\frac{\infty}{\infty}\): \[f'(n) = p(p-1)n^{p-2}\] \[g'(n) = e^{n}\] \[\lim_{n\to\infty} \frac{pn^{p-1}}{e^n} = \lim_{n\to\infty} \frac{p(p-1)n^{p-2}}{e^n}\] This process will continue until the numerator takes the final derivative, which will have a constant term, while the denominator will always have the term \(e^n\).

Step 5: Final limit

After applying L'Hôpital's Rule \(p\) times, we get the final limit: \[\lim_{n\to\infty} \frac{p!}{e^n} = 0\] Since the limit exists and is equal to zero, the given sequence converges. The limit of the sequence is 0: \[\lim_{n\to\infty} \frac{n^{p}}{e^{n}} = 0\]

Key Concepts

L'Hôpital's Rule

When faced with limits that result in indeterminate forms, like \( \frac{\infty}{\infty} \) or \( \frac{0}{0} \), using L'Hôpital's Rule helps to find a tractable solution. L'Hôpital's Rule allows us to take the derivatives of the numerator and denominator, simplifying complex limits by analyzing their rate of growth.
This rule is especially helpful in calculus when evaluating functions as they approach infinity.

• Begin by confirming an indeterminate form exists.
• Take the derivative of the top function and the bottom function separately.
• Check the limit again; if it's still indeterminate, apply the rule repeatedly.
• Stop once you achieve a determinate form.

In this exercise, L'Hôpital's Rule was applied multiple times until the power in the polynomial term reduced fully, leaving a constant, ensuring a clear path to determine the sequence's convergence.

Limit Test

Limits are fundamental in evaluating the convergence or divergence of sequences. The limit test is straightforward: if the sequence's limit is finite and exists, it converges; if it doesn't exist, the sequence diverges. In this problem, our aim was to find:\[ \lim_{n\to\infty} \frac{n^{p}}{e^{n}} \]With sequences like this, considering the growth rates of both the numerator and the denominator helps determine the limit.
If we simplify correctly and find a solution, such as 0 in this case, the sequence converges. But, if problematic terms remain or approach infinity, the sequence diverges.For the given sequence, a limit result of 0 indicates convergence. This is confirmed by the exponential function's rapid growth, vastly overtaking the polynomial growth in the numerator as \( n \) tends toward infinity.

Exponential Functions

Exponential functions are essential in many mathematical contexts due to their unique growth properties. Comparing polynomial and exponential growth reveals that exponentials climb much faster. This exponential outpacing is straightforward when looking at limits like \( \lim_{n\to\infty} \frac{n^{p}}{e^{n}} \).

• Polynomials grow by increasing powers of \(n\).
• Exponential functions, such as \( e^n \), grow by multiplying themselves for each increment.
• Eventually, exponential terms dominate over polynomial ones in limits.

This balance shift makes exponential functions paramount in understanding sequence behavior, particularly when determining convergence or divergence in limit tests.

Calculus Techniques

Calculus provides essential tools for understanding complex mathematical ideas like sequence limits and convergence. Techniques like differentiation, integration, and limit evaluation help break down intricate problems. In the exercise, applying these calculus techniques allowed us to dismantle complex sequence behavior into manageable parts:

• Differentiation aided in using L'Hôpital's Rule successfully, by revealing growth rates.
• Limit evaluation gave insight into the ultimate behavior of the sequence, confirming convergence.

By grasping these foundational techniques, difficulties in assessing sequences become more navigable, particularly when approaching limits involving varying functions, such as polynomials and exponentials.