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Use the Ratio Test to determine the convergence or divergence of the given series. \(\sum_{n=1}^{\infty} \frac{n^{100}}{n !}\)

Short Answer

Expert verified
The series converges absolutely.

Step by step solution

01

Identify the Terms of the Series

The series given is \( \sum_{n=1}^{\infty} \frac{n^{100}}{n !} \). Here, the general term \( a_n \) is \( \frac{n^{100}}{n !} \).
02

Apply the Ratio Test

The Ratio Test states that for a series \( \sum a_n \), we consider \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). If \( L < 1 \), the series converges absolutely; if \( L > 1 \) or \( L = \infty \), the series diverges; if \( L = 1 \), the test is inconclusive.
03

Compute \( a_{n+1} \)

Find the next term in the sequence, \( a_{n+1} = \frac{(n+1)^{100}}{(n+1)!} \).
04

Find \( \frac{a_{n+1}}{a_n} \)

Calculate \[ \frac{a_{n+1}}{a_n} = \frac{(n+1)^{100}}{(n+1)!} \cdot \frac{n!}{n^{100}} = \frac{(n+1)^{100}}{n^{100} \cdot (n+1)} = \left( \frac{n+1}{n} \right)^{100} \cdot \frac{1}{n+1}. \]
05

Simplify the Expression

Simplify \[ \frac{(n+1)^{100}}{n^{100} \cdot (n+1)} = \left(\left( 1 + \frac{1}{n} \right)^{100} \right) \cdot \frac{1}{n+1}. \]
06

Evaluate the Limit as \( n \to \infty \)

Find \( L = \lim_{n \to \infty} \left(\left( 1 + \frac{1}{n} \right)^{100} \right) \cdot \frac{1}{n+1} \). The term \( \left( 1 + \frac{1}{n} \right)^{100} \to 1 \) as \( n \to \infty \), and \( \frac{1}{n+1} \to 0 \). So \( L = 1 \cdot 0 = 0 \).
07

Conclusion Based on the Ratio Test

Since \( L = 0 < 1 \), the series \( \sum_{n=1}^{\infty} \frac{n^{100}}{n !} \) converges absolutely by the Ratio Test.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Convergence: Understanding the Endpoint
In the world of mathematics, determining the "convergence" of a series helps us understand whether adding up its infinite terms brings us closer to a finite value. If a series converges, it means that as we continue to add more terms, we approach a specific number. A series that diverges, on the contrary, does not settle near any particular value, no matter how many terms we include.
Using tests like the Ratio Test, we can determine this behavior. For our series \( \sum_{n=1}^{\infty} \frac{n^{100}}{n !} \), applying the Ratio Test led to a limit calculation showing that \( L = 0 \), which is less than 1. This confirms that our series converges. Convergence plays a crucial role in determining the usefulness of a series in practical applications, such as in calculus and engineering.
Series: Unraveling the Sum
A "series" is simply the sum of terms of a sequence, allowing us to see what happens when we add up an infinite list of numbers. The series could either converge to a specific number or diverge entirely.
In practice, representing sums symbolically helps simplify calculations and understand the accumulation over a set range of terms. Our series, \( \sum_{n=1}^{\infty} \frac{n^{100}}{n !} \), has terms where \( n^n \) grows very quickly, but \( n! \) grows even faster, impacting the overall sum.
  • The sequence \( a_n = \frac{n^{100}}{n !} \) produces terms that heavily shrink as \( n \) grows because factorial growth is rapid.
  • This exponential versus factorial behavior is essential, forming the backbone of why the series converges as analyzed using the Ratio Test.
Factorial: A Game of Growth
The "factorial" of a number \( n \), denoted as \( n! \), means multiplying \( n \) by every positive integer below it. It grows astonishingly fast! This rapid growth can outperform polynomial growth like \( n^{100} \), where \( n \) is raised to a mere power.
For example:
  • 3! = 3 × 2 × 1 = 6
  • 5! = 5 × 4 × 3 × 2 × 1 = 120
  • 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040
This means that in our series, even as \( n^{100} \) increases, the factorial denominator increases at such a speed that it dwarfs the numerator, contributing fundamentally to the series' convergence as identified in our ratio test.
Limits: The Approach to Infinity
A "limit" is a value that a function or sequence "approaches" as the input or index approaches some value. Limits are especially valuable in calculus to define derivatives and integrals and in our case, to analyze convergence of series.
  • In the Ratio Test, we calculate the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).
  • For \( \sum_{n=1}^{\infty} \frac{n^{100}}{n !} \), we transitioned to the expression \( \left(\left( 1 + \frac{1}{n} \right)^{100} \right) \cdot \frac{1}{n+1} \). When we let \( n \) approach infinity, these terms reveal \( L = 0 \).
Thus, understanding limits helps us determine how expressions behave as they scale towards very large or very small figures, solidifying our conclusions about the behavior of series.

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