Question

Test the following series for convergence or divergence. Decide for yourself which test is easiest to use, but don't forget the preliminary test. Use the facts stated above when they apply. $$\sum_{n=1}^{\infty} \frac{(-1)^{n} n !}{10^{n}}$$

Short answer

The series diverges because the general term does not approach zero.

Step-by-step solution

Preliminary Test - Test for Divergence

Use the preliminary test for divergence. Compute the limit of the general term as n approaches infinity. \[ \lim_{{n \to \infty}} \frac{{(-1)^n n!}}{{10^n}} \]Notice that the factorial in the numerator grows much faster than the exponential term in the denominator. Hence, the limit will be infinite.

Evaluate the Limit

Calculate the limit from the previous step.\[ \lim_{{n \to \infty}} \frac{{(-1)^n n!}}{{10^n}} = \infty \]Since the limit does not equal zero, the series diverges according to the Divergence Test.

Key concepts

Understanding the Divergence Test

The Divergence Test, also known as the Test for Divergence, is one of the simplest ways to determine whether a series diverges. Here's how it works:

We start by examining the general term of the series, denoted as \(a_n\). We compute the limit of this term as \(n\) approaches infinity.

For the series \( \sum_{n=1}^{\infty} \frac{(-1)^{n} n !}{10^{n}} \), the general term is \( \frac{(-1)^n n!}{10^n} \). We evaluate the limit:
\[ \lim_{{n \to \infty}} \frac{(-1)^n n!}{10^n} \]
Notice that the factorial \( n! \) in the numerator grows significantly faster than the exponential term \(10^n\) in the denominator.

As a result, \(\frac{(-1)^n n!}{10^n}\rightarrow\infty\). Since this limit is not zero, the series diverges by the Divergence Test.

• The Divergence Test is particularly useful as a preliminary test, giving a quick indication if the series diverges.
• For a series to converge, the limit of its general term must be zero.

The Concept of the Limit of a Series

Evaluating the limit of a series is crucial in determining its convergence or divergence. To understand this more clearly, consider the following points:

A series is represented as the sum of terms: \ a_1 + a_2 + a_3 + ... \.
The general term of the series is denoted as \ a_n \.
We compute the limit of \ a_n \ as \ n \ approaches infinity:
\[ \lim_{{n \to \infty}} a_n \]

If this limit is not zero, the series diverges. This is because the individual terms do not get small enough to sum up to a finite value.

For the given series \( \sum_{n=1}^{\infty} \frac{(-1)^{n} n !}{10^{n}} \):
\[ \lim_{{n \to \infty}} \frac{(-1)^n n!}{10^n} = \infty \]

This means the terms grow without bound, leading to divergence.

• The limit helps in predicting the behavior of the entire series.
• Always remember, if the limit of the general term is not zero, the series diverges.

Exploring Factorial Growth

Factorials grow extraordinarily fast compared to other functions. Let's dive into what this means and how it affects series convergence:

The factorial of a number \ n \, denoted \ n! \,, is the product of all positive integers up to \ n \:
\[ n! = n \times (n-1) \times (n-2) ... \times 2 \times 1 \]

As \ n \ increases, \ n! \ becomes immensely large very quickly.

Consider the given series:
\[ \sum_{n=1}^{\infty} \frac{(-1)^{n} n !}{10^{n}} \]
The term \ n! \ in the numerator grows much faster than \ 10^n \ in the denominator.

This rapid growth of \ n! \ causes the general term \ \frac{(-1)^n n!}{10^n} \ to grow indefinitely as \ n \ approaches infinity:
\[ \lim_{{n \to \infty}} \frac{(-1)^n n!}{10^n} = \infty \]

• Factorial growth is much faster than exponential growth.
• This concept is vital in understanding why some series, especially those involving factorials, diverge.