Question

Use the ratio test to determine whether \(\sum_{n=1}^{\infty} a_{n}\) converges, where \(a_{n}\) is given in the following problems. State if the ratio test is inconclusive. $$ \sum_{n=1}^{\infty} \frac{n !}{(n / e)^{n}} $$

Short answer

The series converges by the ratio test.

Step-by-step solution

Identify the Terms

Given the series \( \sum_{n=1}^{\infty} a_{n} \), where \( a_{n} = \frac{n!}{(n/e)^n} \). This is the term we'll analyze using the ratio test.

Apply the Ratio Test Formula

The ratio test involves evaluating \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_{n}} \right| \). We start by finding \( a_{n+1} = \frac{(n+1)!}{((n+1)/e)^{n+1}} \).

Substitute into the Ratio

Calculate \( \frac{a_{n+1}}{a_{n}} = \frac{(n+1)!}{((n+1)/e)^{n+1}} \times \frac{(n/e)^{n}}{n!} \). Simplify this expression.

Simplify the Expression

This leads to \( \frac{a_{n+1}}{a_{n}} = \frac{(n+1)(n/e)^n}{((n+1)/e)^{n+1}} \). Further simplify to \( \frac{(n+1) \cdot e^n}{(n+1)^{n+1}} \).

Limit Evaluation

Evaluate the limit \( \lim_{n \to \infty} \left( \frac{e^n}{(n+1)^n} \right) \). Use properties of limits and approximations for large \( n \).

Use Limit Properties

\( \left( \frac{e}{(n+1)} \right)^n \to 0 \) as \( n \to \infty \) because \( n+1 \) grows much faster than a constant base \( e \). The entire expression goes to 0.

Interpretation of Results

Since the limit is 0, which is less than 1, the series \( \sum_{n=1}^{\infty} \frac{n!}{(n/e)^n} \) converges according to the ratio test.

Key concepts

Convergence of Series

A series is a sum of terms, and we're often interested in knowing whether that sum converges to a specific value, or diverges (keeps growing or oscillating forever). To determine this,we use various tests. The ratio test is especially helpful when the terms of a series involve factorials or exponential expressions. It helps us determine if a series converges or divergesby looking at the limit of the ratio of consecutive terms.
In our exercise, we're given the series \( \sum_{n=1}^{\infty} \frac{n!}{(n/e)^n} \). The ratio test requires us to find the limit of \( \left| \frac{a_{n+1}}{a_{n}} \right| \) as \( n \) approaches infinity.If this limit is less than 1, the series converges. If it’s greater than 1, the series diverges. If it equals 1, the test is inconclusive, and we might need to try another method.

• The test is particularly effective for series where terms include factorials or powers.
• It's a quick way to check convergence without needing to guess or manually sum terms.
• Convergence means that the infinite sum results in a specific finite number.

Factorial in Sequences

Factorials are a mathematical operation often found in sequences or series, represented by a number followed by an exclamation mark, such as \( n! \). The notation \( n! \) means multiplying all whole numbers from \( n \) down to 1.
For example:

• \( 3! = 3 \times 2 \times 1 = 6 \)
• \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \)

In the original exercise, the sequence term \( a_{n} = \frac{n!}{(n/e)^n} \) includes factorials in the numerator. Factorials grow very quickly as \( n \) increases. They're often paired with exponential expressions in the denominatorin convergence problems like the one we're investigating.

• Using factorials helps evaluate the growth of sequence terms.
• Factorials increase much more rapidly than exponential terms.
• This rapid growth is crucial to determine how a series behaves as \( n \) becomes large.

Somewhat counterintuitively, even though factorial terms grow fast, they can still allow the series to converge when combined wisely with other factors that limit the growth, like larger powers in the denominator.

Limit Evaluation

When evaluating convergence using the ratio test, we often need to simplify expressions and compute limits involving sequences (or their ratios) as \( n \) approaches infinity.Limit evaluation involves using mathematical techniques to find what happens to a function or expression as the variable grows infinitely large.
In this problem, once we simplify the ratio \( \frac{a_{n+1}}{a_{n}} \), we're left with \( \frac{e^n}{(n+1)^n} \). As \( n \) grows, the exponential \( e^n \) and power terms both grow, but \( (n+1)^n \) grows faster because \( n \) is also raised to a power.

• The function \( \left( \frac{e}{n+1} \right)^n \) tends to zero because the denominator grows faster.
• We conclude that the limit is zero, which is less than one, implying convergence.
• Recognizing the growth rate differences is key in limit evaluation for convergence.

Understanding this allows us to see why the limit goes to zero and why the series converges in such cases. It's important in mathematics because it helps us predict the behavior of complex sequencesand series without direct computation of every term.