Problem 32 - Thomas Calculus

Question

Which of the series Converge absolutely, which converge, and which diverge? Give reasons for your answers. $$ \sum_{n=1}^{\infty}(-5)^{-n} $$

Step-by-Step Solution

Answer

The series \( \sum_{n=1}^{\infty}(-5)^{-n} \) converges absolutely.

Step 1: Identify the Series Type

The given series is \( \sum_{n=1}^{\infty}(-5)^{-n} \) which is a geometric series. A geometric series has the form \( \sum_{n=0}^{\infty} ar^n \). In this case, the series starts at \( n=1 \) and the first term to consider would be \((-5)^{-1}\). The common ratio \( r \) here is \((-5)^{-1}\) i.e. \( \frac{1}{-5} \).

Step 2: Check for Absolute Convergence

To check whether the series converges absolutely, consider the series formed by the absolute values of the terms: \( \sum_{n=1}^{\infty} |(-5)^{-n}| = \sum_{n=1}^{\infty} \frac{1}{5^n} \). This is another geometric series with first term \( \frac{1}{5} \) and a common ratio \( \frac{1}{5} \).

Step 3: Apply the Geometric Series Test for Absolute Convergence

A geometric series \( \sum_{n=0}^{\infty} ar^n \) converges if \( |r| < 1 \). Here \( r = \frac{1}{5} \), so \(|r| = 0.2\). Since \( 0.2 < 1 \), the series \( \sum_{n=1}^{\infty} \frac{1}{5^n} \) converges. Thus, \( \sum_{n=1}^{\infty} (-5)^{-n} \) converges absolutely.

Step 4: Determine Overall Convergence

Since the series converges absolutely, by definition it also converges. There is no need to test for conditional convergence separately as absolute convergence ensures overall convergence.

Step 5: Conclude with Divergence

Given that our series converges absolutely, it certainly does not diverge. A series either converges or diverges, and here we have established convergence.

Key Concepts

Geometric Series

A geometric series is a special type of series characterized by each term being a constant multiple of the previous one. You can spot a geometric series by its typical form: \( \sum_{n=0}^{\infty} ar^n \), where \( a \) is the first term and \( r \) is the common ratio. Think of it like a snowball rolling down a hill, each roll adding a consistent proportion more snow.

In our example, we see the series \( \sum_{n=1}^{\infty}(-5)^{-n} \) fits this profile. Here, each term is created by multiplying the previous term by \((-\frac{1}{5})\). The common ratio \((-\frac{1}{5})\) tells us how each term relates to the one before it, dictating the series' behavior.

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

Identifying the series as geometric helps us apply the right tests for convergence or divergence.

Absolute Convergence

Absolute convergence is an important concept that guarantees a stronger form of convergence. If a series is found to converge absolutely, it means that the series remains converged even when we take the absolute values of each term.

For the series \( \sum_{n=1}^{\infty}(-5)^{-n} \), the absolute values form the series \( \sum_{n=1}^{\infty} \left|(-5)^{-n}\right| \), which simplifies to \( \sum_{n=1}^{\infty} \frac{1}{5^n} \). By working with these positive terms only, it guarantees that fluctuations in signs won't affect convergence.
The criteria for absolute convergence in geometric series is that the common ratio \( |r| \) must be strictly less than 1. With \(|r| = 0.2\) (since \( \frac{1}{5} = 0.2\)), we meet the condition for absolute convergence. Once we've secured this, we can conclusively say the original series converges as well.

Series Divergence

A series is divergent if it fails to approach a stable value, often growing without bound or oscillating indefinitely. Unlike convergence, divergence means the "sum" does not settle.
The given series \( \sum_{n=1}^{\infty}(-5)^{-n} \) does not diverge because it meets the criteria for absolute convergence. Typically, for a geometric series to diverge, the common ratio's absolute value would be 1 or more, yet here it's a fraction less than 1. This secures convergence, ruling out any form of divergence.

Should a series diverge, it implies that no finite limit exists. However, in cases like this one, where absolute convergence is demonstrated, both divergence and conditional convergence are neatly ruled out.