Problem 19 - Thomas Calculus

Question

Determining Convergence or Divergence In Exercises \(17-44,\) use any method to determine if the series converges or diverges. Give reasons for your answer. $$\sum_{n=1}^{\infty} n !(-e)^{-n}$$

Step-by-Step Solution

Answer

The series \( \sum_{n=1}^{\infty} n !(-e)^{-n} \) diverges by the Ratio Test.

Step 1: Understanding the Series

We need to determine the convergence of the series \( \sum_{n=1}^{\infty} n !(-e)^{-n} \). This series involves factorials and an exponential term, so we should consider using tests suitable for factorials or series with terms involving exponential functions.

Step 2: Applying the Ratio Test

The Ratio Test is effective for series involving factorials or exponential terms. We will apply it to determine convergence or divergence. The Ratio Test states that for \( a_n \), if \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L \), the series converges if \( L < 1 \), diverges if \( L > 1 \), and is inconclusive if \( L = 1 \).

Step 3: Setting Up the Ratio

For our series, \( a_n = n!(-e)^{-n} \). Now compute \( \frac{a_{n+1}}{a_n} = \frac{(n+1)! (-e)^{-(n+1)}}{n! (-e)^{-n}} = \frac{(n+1)!}{(-e)(n!)} = \frac{n+1}{e} \).

Step 4: Evaluating the Limit

Evaluate \( \lim_{n \to \infty} \frac{n+1}{e} = \lim_{n \to \infty} \frac{n}{e} + \frac{1}{e} = \infty \). Since this limit is greater than 1, the Ratio Test concludes that the series diverges.

Key Concepts

Ratio Test

The Ratio Test is a standard method used for determining the convergence or divergence of an infinite series, especially those involving factorials or exponential terms. When you apply the Ratio Test, you essentially look at the ratio of consecutive terms in the series. Here's how it works in simple terms:

• Suppose the series is given by \( \sum_{n=1}^{\infty} a_n \).
• Calculate the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).
• If \( L < 1 \), the series converges.
• If \( L > 1 \), the series diverges.
• If \( L = 1 \), the test is inconclusive.

In our case, for the series \( \sum_{n=1}^{\infty} n !(-e)^{-n} \), the Ratio Test helped us find that the limit \( L \) was infinity, leading us to conclude that the series diverges. This is because the terms do not sufficiently decrease in size to settle into a sum, which is why \( L > 1 \) indicates divergence.

Factorials in Series

Factorials often appear in series, especially those involving sequences like permutations and combinations, but they can also show up in mathematical expressions like power series. The factorial of a number \( n \), denoted as \( n! \), is the product of all positive integers up to \( n \). Factorials grow very rapidly, and this growth impacts whether a series converges or diverges.

• In the context of series, terms that include factorials tend to grow faster than polynomial terms.
• Due to this rapid growth, series with factorial terms often diverge unless controlled by significant exponential terms to counterbalance them.

For example, in the series \( \sum_{n=1}^{\infty} n !(-e)^{-n} \), the factorial term \( n! \) rapidly increases as \( n \) increases. The series attempted to counter this with the exponential decay \((-e)^{-n}\), but the exponential term \((-e)^{-n}\) was not enough to offset the growth of \( n! \), resulting in divergence.

Exponential Functions in Series

Exponential functions are crucial in the study of series, especially when determining convergence. These functions can have a dramatic impact on whether a series converges or diverges because exponential growth or decay can amplify or diminish the terms significantly.

• Exponential terms typically look like \( c^n \) where \( c \) is a constant.
• When involved in a series, exponential functions can lead to rapid growth (for \( c > 1 \)) or rapid decay (for \( 0 < c < 1 \)), both affecting convergence.

Consider our series \( \sum_{n=1}^{\infty} n !(-e)^{-n} \). Here, the exponential function \((-e)^{-n}\) was intended to cause decay as \( n \) increases, potentially allowing the series to converge. However, the factorial \( n! \) grows even faster than the exponential component decays. This imbalance caused the series to diverge, showing that not only the presence of exponential terms but their relationship with other factors, like factorials, determines convergence or divergence.