Question

Test the following series for convergence. $$\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{2}}$$

Short answer

The series converges by the Alternating Series Test.

Step-by-step solution

Identify the Series

Recognize the given series \(\frac{(-1)^n}{n^2}\) as an alternating series.

Apply the Alternating Series Test

Check if the series meets the criteria of the Alternating Series Test: \((-1)^n a_n\) where \(a_n = \frac{1}{n^2}\) 1) Determine if \(a_n\) is decreasing. 2) Determine if \(a_n \rightarrow 0\) as \(n \rightarrow \infty \).

Verify \(a_n\) is Decreasing

Since \(a_n = \frac{1}{n^2}\) is a positive decreasing sequence, it fulfills the first criterion: \(a_{n+1} < a_n\).

Check if \(a_n \rightarrow 0\)

Observe that \( \frac{1}{n^2} \rightarrow 0\) as \(n \rightarrow \infty \). Thus, this condition is also met.

Conclude the Convergence

Since both conditions of the Alternating Series Test are satisfied, the series converges.

Key concepts

alternating series test

To determine if an infinite series converges, one of the tests we can use is the Alternating Series Test. An alternating series is a series whose terms alternate in sign, such as \((-1)^n a_n\). The Alternating Series Test helps us determine if this type of series converges.

According to the test, a series of the form \sum (-1)^n a_n\ converges if two conditions are met:

• First, the sequence \(a_n\) must be positive and decreasing.
• Second, the sequence \(a_n\) should approach zero as \(n\) approaches infinity ( \(a_n \to 0 \) ).

These two conditions ensure that each term gets smaller and the positive-negative canceling effect eventually leads to convergence. By verifying these conditions, we can confirm the convergence of alternating series in different mathematical contexts.

decreasing sequence

A key condition in the Alternating Series Test is that the sequence \(a_n\) must be decreasing. A sequence is considered decreasing if each term is less than or equal to the previous term. Mathematically, this means that \(a_{n+1} \leq a_n\) for all \(n\).

In our series, \(a_n = \frac{1}{n^2}\), we need to show that each term gets smaller as \(n\) increases. Since \(n^2\) gets larger as \(n\) increases, each corresponding \( \frac{1}{n^2} \) gets smaller making the sequence decreasing.

This can be observed by comparing consecutive terms:

• For \ n = 1 \), \(a_1 = \frac{1}{1^2} = 1 \.
• For \( n = 2 \), \(a_2 = \frac{1}{2^2} = \frac{1}{4} \).
• For \( n = 3 \), \(a_3 = \frac{1}{3^2} = \frac{1}{9} \).

It's evident from these values that \(a_n \) is decreasing since \(1 > \frac{1}{4} > \frac{1}{9} \). This fulfills one of the necessary conditions of the Alternating Series Test.

limit to zero

The next key condition for the Alternating Series Test is that the sequence \(a_n\) must approach zero as \(n\) approaches infinity. In mathematical terms, this is written as \( \forall n \to \forall , a_n \to 0 \).

For our series, the terms are given by \(a_n = \frac{1}{n^2}\). To show that \ a_n \rightarrow 0 \, we observe that as \ n \ gets very large, \ n^2 \ also becomes very large and \ \frac{1}{n^2} \ gets closer and closer to zero.

For example:

• When \ n = 10 \, \(a_{10} = \frac{1}{10^2} = 0.01 \).
• When \ n = 100 \, \(a_{100} = \frac{1}{100^2} = 0.0001 \).
• When \ n = 1000 \, \(a_{1000} = \frac{1}{1000^2} = 0.000001 \).

Clearly, as \ n \ becomes larger, \ a_n \ gets smaller. Thus, the sequence \(a_n = \frac{1}{n^2} \) goes to zero as \ n \ approaches infinity.

This satisfies the second condition necessary for the Alternating Series Test.