Problem 7 - Calculus Early Transcendentals: Pearson New International Edition

Question

Show that each series converges absolutely. $$ \sum_{n=1}^{\infty}\left(-\frac{3}{4}\right)^{n} $$

Step-by-Step Solution

Answer

The series converges absolutely because the absolute series is a convergent geometric series with a ratio \( r = \frac{3}{4} < 1 \).

Step 1: Consider the Absolute Value of the Series

To show that the series converges absolutely, we need to consider the absolute value of each term. The given series is \( \sum_{n=1}^{\infty} \left( -\frac{3}{4} \right)^{n} \). Taking the absolute value, we have \( \sum_{n=1}^{\infty} \left| -\frac{3}{4} \right|^{n} = \sum_{n=1}^{\infty} \left( \frac{3}{4} \right)^{n} \).

Step 2: Identify the Series

The series \( \sum_{n=1}^{\infty} \left( \frac{3}{4} \right)^{n} \) is a geometric series with the first term \( a = \frac{3}{4} \) and common ratio \( r = \frac{3}{4} \).

Step 3: Determine Convergence of the Geometric Series

A geometric series \( \sum_{n=1}^{\infty} ar^{n} \) converges if the absolute value of the common ratio \( |r| < 1 \). Here, \( \left| \frac{3}{4} \right| = \frac{3}{4} < 1 \), so the series \( \sum_{n=1}^{\infty} \left( \frac{3}{4} \right)^{n} \) converges.

Step 4: Conclusion on Absolute Convergence

Since the series \( \sum_{n=1}^{\infty} \left( \frac{3}{4} \right)^{n} \) converges, the original series \( \sum_{n=1}^{\infty} \left( -\frac{3}{4} \right)^{n} \) converges absolutely as well.

Key Concepts

Geometric Series

A geometric series is a series of numbers in which each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. This kind of series can be written as:

• First term: \( a \)
• Common ratio: \( r \)
• Series: \( a + ar + ar^2 + ar^3 + \ldots \)

The general form of a geometric series is \( \sum_{n=0}^{\infty} ar^{n} \). This series helps us understand how values grow or decay over iterations, depending on whether the common ratio is greater or less than one. In the specified problem, the series given is \( \sum_{n=1}^{\infty} \left( -\frac{3}{4} \right)^{n} \), which transforms into a geometric series of \( \sum_{n=1}^{\infty} \left( \frac{3}{4} \right)^{n} \) when considering absolute convergence. Understanding geometric series is essential in examining how these repeated multiplications affect the series’ behavior.

Absolute Convergence

Absolute convergence is a concept where a series \( \sum a_n \) converges absolutely if the series of absolute values \( \sum |a_n| \) converges. It provides a stricter criterion than simple convergence. This means if a series converges absolutely, it also converges in the regular sense.
For the series \( \sum_{n=1}^{\infty} \left( -\frac{3}{4} \right)^{n} \), we examine its absolute convergence by looking at \( \sum_{n=1}^{\infty} \left( \frac{3}{4} \right)^{n} \). Since \( \sum_{n=1}^{\infty} \left( \frac{3}{4} \right)^{n} \) is a geometric series with a common ratio of \( \frac{3}{4} \), which is less than 1, it converges absolutely.
Absolute convergence assures that rearranging terms in a series will not affect the sum. This is particularly useful in mathematical analysis and solving complex series-related problems.

Common Ratio

The common ratio in a geometric series determines the factor by which we multiply each term to get the next term in the series.
Denoted as \( r \), it plays a critical role in defining the series’ behavior. When \( |r| < 1 \), the series converges because the terms get smaller and approach zero as you progress.

• If \( |r| = 1 \), the series is either constant, repeating, or divergent.
• If \( |r| > 1 \), the series diverges because the terms grow larger with each iteration.

In our exercise, the common ratio is \( \frac{3}{4} \), perfectly fitting the condition for convergence. Thus, knowing and identifying the common ratio allows us to quickly evaluate if a geometric series will converge or diverge, making it a crucial element in series analysis.