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. $$\frac{1}{2^{2}}-\frac{1}{3^{2}}+\frac{1}{2^{3}}-\frac{1}{3^{3}}+\frac{1}{2^{4}}-\frac{1}{3^{4}}+\cdots$$

Short answer

The series converges because it is the difference of two convergent geometric series.

Step-by-step solution

Identify the general form of the series

The given series is: \[\frac{1}{2^{2}}-\frac{1}{3^{2}}+\frac{1}{2^{3}}-\frac{1}{3^{3}}+\frac{1}{2^{4}}-\frac{1}{3^{4}}+\text{ ... }\]. This indicates an alternating series where the general term can be written as: \[\frac{1}{2^n} - \frac{1}{3^n}\], for \(n\) starting from 2.

Separate the series into two simpler series

Express the given series as two separate series: \[\frac{1}{2^{2}}+\frac{1}{2^{3}}+\frac{1}{2^{4}}+\text{ ... } - (\frac{1}{3^{2}}+\frac{1}{3^{3}}+\frac{1}{3^{4}}+\text{ ... })\].

Evaluate the convergence of the series individually

First, consider the series \[\frac{1}{2^n} = \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \text{ ... }\]. This is a geometric series with common ratio \( \frac{1}{2} \), which converges because \( |r| < 1 \). Similarly, consider the series \[\frac{1}{3^n} = \frac{1}{3^2} + \frac{1}{3^3} + \frac{1}{3^4} + \text{ ... }\]. This is also a geometric series with common ratio \( \frac{1}{3} \), which converges because \( |r| < 1 \).

Combine the convergence results

Since both individual geometric series converge, the entire series \[\frac{1}{2^n} - \frac{1}{3^n}\] also converges by the difference of the two convergent series.

Key concepts

alternating series test

An alternating series is a series where the terms alternate in sign. This means one term is positive and the next term is negative.
To test if an alternating series converges, we can use the Alternating Series Test (Leibniz's test).
The test has two key conditions:

• The absolute value of the terms must be decreasing.
• The limit of the terms as n approaches infinity must be zero.

For example, in the given series, \(\frac{1}{2^2} - \frac{1}{3^2} + \frac{1}{2^3} - \frac{1}{3^3} + \frac{1}{2^4} - \frac{1}{3^4} + \text{...}\), if you look at the nth term: \(\frac{1}{2^n} - \frac{1}{3^n}\), each part follows the criteria.
The terms \(\frac{1}{2^n}\) and \(\frac{1}{3^n}\) individually decrease and approach zero. Therefore, the series converges.

geometric series

A geometric series is a series with a constant ratio between successive terms. The general form of a geometric series is \(\text{a} + \text{ar} + \text{ar}^2 + \text{ar}^3 + \text{ ...}\).
The common ratio is denoted as r.
We can determine the convergence of a geometric series based on the common ratio, r:

• If \(|r| < 1\), the series converges.
• If \(|r| \geq 1\), the series diverges.

When we separated the given series into two simpler geometric series: \(\frac{1}{2^n} \) and \(\frac{1}{3^n}\), each had common ratios of \(\frac{1}{2}\) and \(\frac{1}{3}\) respectively.
Because both \(\frac{1}{2}\) and \(\frac{1}{3}\) are less than 1, each individual series converges. Therefore, their combination in the original series also converges.

convergence and divergence

In the context of series, convergence and divergence are key concepts.
A series converges if the sum of its terms approaches a finite value as more terms are added.
Conversely, a series diverges if the sum does not approach a finite value and instead grows indefinitely or oscillates without settling.
There are various tests to determine convergence or divergence, such as the Geometric Series Test, Alternating Series Test, and others like the Ratio Test and Integral Test.
In our example, we used the criteria for a geometric series to determine convergence.
Because both component series \(\frac{1}{2^n}\) and \(\frac{1}{3^n}\) converged individually, we concluded that the entire series converges.

• Recognize the type of series.
• Apply the appropriate test.
• Ensure that all conditions are met for the test used.

By understanding convergence and divergence, and how to test for it, students can more easily tackle various series-related problems.