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

Short answer

The series \( \sum_{n=2}^{\to \infty} \frac{(-1)^n}{n^2 - n} \) converges by the Alternating Series Test.

Step-by-step solution

- Preliminary Test

First, apply the preliminary test by checking the limit of the general term of the series as \( n \) approaches infinity. Consider $$ a_n = \frac{(-1)^n}{n^2 - n} $$ Calculate $$ \lim_{{n \to \infty}} \frac{(-1)^n}{n^2 - n} $$

- Evaluate the Limit

Evaluate the limit: As \( n \) approaches infinity, the denominator \( n^2 - n \) grows much faster than the numerator \( (-1)^n \). Therefore: $$ \lim_{{n \to \infty}} \frac{(-1)^n}{n^2 - n} = 0 $$ Since the limit is 0, we proceed with further tests.

- Alternating Series Test (Leibniz's Test)

Apply the Alternating Series Test. For a series \( \sum (-1)^n b_n \), if \( b_n \) is positive, decreasing, and \( \lim_{{n \to \infty}} b_n = 0 \), then the series converges. Here, let \( b_n = \frac{1}{n^2 - n} \). We have already shown that $$ \lim_{{n \to \infty}} b_n = 0 $$

- Check if \( b_n \) is Decreasing

Verify that \( b_n \) is decreasing. Consider $$ b_n = \frac{1}{n^2 - n} $$ Compute the derivative: $$ b_n' = \frac{d}{dn} \left( \frac{1}{n^2 - n} \right) = \frac{-2n + 1}{(n^2 - n)^2} $$ When \( n > 1 \), \( -2n + 1 < 0 \), so the derivative is negative, indicating \( b_n \) is decreasing for \( n \geq 2 \).

- Conclusion using Alternating Series Test

Since \( b_n = \frac{1}{n^2 - n} \) is positive, decreasing, and its limit as \( n \to \infty \) is 0, by the Alternating Series Test, the series $$ \sum_{n=2}^{\to \infty} \frac{(-1)^n}{n^2 - n} $$ converges.

Key concepts

preliminary test

Before you dive into any series convergence tests, it's vital to first apply the preliminary test, also known as the nth-term test for divergence. This initial step helps us quickly determine if a series diverges without needing more complex methods.

For a series to converge, the limit of its general term as it approaches infinity must be zero. In our exercise, the general term is \(a_n = \frac{(-1)^n}{n^2 - n}\). We need to evaluate this limit to apply the preliminary test:

• \(\lim_{n \to \infty} \frac{(-1)^n}{n^2 - n} \).

Because the denominator \(n^2 - n\) grows significantly faster than the numerator \((-1)^n\), the limit as \(n\) approaches infinity is 0:

• \(\lim_{n \to \infty} \frac{(-1)^n}{n^2 - n} = 0\).

If this limit had not been zero, the series would have diverged. Since it is zero, we move on to other tests to determine convergence.

alternating series test

One effective method for determining the convergence of an alternating series is the Alternating Series Test, also known as the Leibniz Test. This test is specifically useful if your series alternates in sign.

Our given series is \(\sum_{n=2}^{\infty} \frac{(-1)^n}{n^2 - n}\). This is an alternating series because the term \((-1)^n \) changes sign with each successive term. For the Leibniz Test to confirm convergence, three conditions must be met:

• The absolute value of the terms \(b_n\) (i.e. \(b_n = \frac{1}{n^2 - n}\)) must be positive.
• The sequence \(b_n\) must be decreasing.
• The limit of \(b_n \) as \(n\) approaches infinity must be zero.

We have already demonstrated that \(\lim_{n \to \infty} b_n = 0 \). Next, we need to check the remaining conditions.

limit evaluation

Evaluating the limit of the general term is a crucial part of the preliminary test and is also used in confirming conditions for other convergence tests.

For our series \(\sum_{n=2}^{\infty} \frac{(-1)^n}{n^2 - n}\), we previously determined:

• \(\lim_{n \to \infty} \frac{(-1)^n}{n^2 - n} = 0\).

This evaluation is significant because it satisfies the condition of the Alternating Series Test which requires the limit of \(b_n\) to be zero. Remember, a non-zero limit would indicate divergence. With the limit being zero, we proceed by verifying if the terms of the series decrease.

decreasing function

To use the Alternating Series Test, we must confirm that the sequence \(b_n\) is decreasing. This means each successive term is smaller than the one before it.

For our series, we need to check that \(b_n = \frac{1}{n^2 - n}\) is decreasing for \(n \geq 2 \). One way to verify this is to determine the derivative of \(b_n\):

• \(b_n' = \frac{d}{dn} \left( \frac{1}{n^2 - n} \right) = \frac{-2n + 1}{(n^2 - n)^2} \)

Since the derivative \(b_n'\) is negative for \(n > 1 \) (meaning \(-2n + 1 < 0\)), it indicates that \(b_n\) is indeed a decreasing sequence for \(n \geq 2 \).

By confirming that \(b_n\) is positive, decreasing, and its limit as \(n \to \infty\) is zero, we've met all the criteria of the Alternating Series Test. Thus, the series \(\sum_{n=2}^{\infty} \frac{(-1)^n}{n^2 - n}\) converges.