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. $$\sum_{n=2}^{\infty}(-1)^{n} \frac{n}{n-1}$$

Short answer

The series diverges by the Divergence Test because the limit of the general term is \(\pm 1\) and not zero.

Step-by-step solution

- Preliminary Test: Divergence Test

First, apply the Divergence Test. Compute the limit of the general term \(a_n = (-1)^n \frac{n}{n-1}\) as \(n \to \infty \). \ \[ \lim_{n \to \infty} (-1)^n \frac{n}{n-1} \] \ This simplifies to: \ \[ \lim_{n \to \infty} (-1)^n \frac{n}{n-1} = \lim_{n \to \infty} (-1)^n \left(1 + \frac{1}{n-1}\right) = \pm 1 \] The limit does not approach zero, hence the series \(\sum_{n=2}^{\infty} (-1)^n \frac{n}{n-1}\) diverges by the Divergence Test.

Key concepts

Divergence Test

The Divergence Test is one of the simplest tools to check if a series diverges. This test is also known as the Test for Divergence or the n-th Term Test. To apply this test, we focus on the limit of the general term of the series as it approaches infinity.
If the limit of the general term, denoted as \(a_n\), does not equal zero, then the series diverges. However, if the limit is zero, the test is inconclusive, and we need to try other tests.
Let's look at an example:
For the series \( \sum_{n=2}^{\infty}(-1)^{n} \frac{n}{n-1} \), we start by computing the limit of the general term \(a_n \. \). Here, \( a_n = (-1)^n \frac{n}{n-1}\) and the limit is evaluated as \( n \to \infty \. \) This simplifies as follows:
\[ \lim_{n \to \infty} (-1)^n \frac{n}{n-1} = \pm 1 \]
Since the limit is neither zero nor non-existent, the series diverges by the Divergence Test.

Limit of General Term

Understanding the limit of the general term is crucial in determining the behavior of an infinite series. The general term is the expression we get for each term of the series as a function of \( n \. \)
The limit of the general term as \( n \to \infty \) tells us how the terms behave for very large \( n \. \)
In the above example, the general term is \(a_n = (-1)^n \frac{n}{n-1} \). To find the limit, we simplify it:
\[-1^n \frac{n}{n-1} = (-1)^n \left( 1 + \frac{1}{n-1} \right) \]
As \( n \to \infty \, \frac{1}{n-1} \) approaches zero, which leads to:
\[ \lim_{n \to \infty} (-1)^n \left( 1 + \frac{1}{n-1} \right) = \pm 1 \]
Since the limit does not equal zero, it indicates that the terms do not diminish to zero, and thus the series diverges. Always remember, when a limit of a general term is not zero, the series diverges.

Infinite Series

An infinite series is a sum of infinitely many terms. The series is written as \sum_{n=1}^{\to \infty} a_n \, where \( a_n \) is the general term. Understanding the convergence (whether the series adds up to a particular number) or divergence (whether the series doesn't approach a finite number) is key.
There are various tests to check the convergence or divergence of an infinite series, such as the Divergence Test, Root Test, Ratio Test, and others.
In our exercise, the test used is the Divergence Test because it is straightforward and quickly indicates divergence if the limit of the general term is non-zero. Always start checking for divergence first as it often simplifies the process.