Problem 41 - Essential Calculus

Question

Find the sum of the convergent series. $$ 3-1+\frac{1}{3}-\frac{1}{9}+\cdots $$

Step-by-Step Solution

Answer

The sum of the convergent geometric series \(3-1+1/3-1/9+...\) is 2.25.

Step 1: Identify the initial term and common ratio

The first term \(a\) of the series is 3, and the common ratio \(r\) can be found by dividing the second term by the first term or the third term by the second term. In this case, \(r\) is \(-1/3\).

Step 2: Check for Convergence

Before finding the sum of the geometric series, it's important to check if the series is convergent. A geometric series \(a + ar + ar^2 + ar^3 + ... \) is convergent if -1 < r < 1, which is true for our series because \(-1 < -1/3 < 1\). So, this series is convergent.

Step 3: Find the Sum of the Series

The sum \(S\) of an infinite geometric series with first term \(a\) and common ratio \(r\) can be found by the formula \(S = a / (1 - r)\). Substituting our values, we have \(S = 3 / (1 - (-1/3)) = 3 / (1 + 1/3) = 3 / (4/3) = 9/4 = 2.25\).

Key Concepts

Understanding Geometric Series

Geometric series are a powerful mathematical concept used to add up terms in a sequence where each term is a constant multiple, called the common ratio, of the previous term.

Understanding the structure of a geometric series helps in identifying its characteristics efficiently. A typical geometric series might be written as:

• A first term, often represented as \( a \), which starts the series.
• Each successive term is the previous term multiplied by a constant called the common ratio, \( r \).
• The series continues infinitely, represented as \( a + ar + ar^2 + ar^3 + \, \ldots \)

For example, the series given in the problem starts with 3, then -1, \( \frac{1}{3} \), \( -\frac{1}{9} \), and so on. Here, each term is being progressively modified by multiplying the previous term by the common ratio \( -\frac{1}{3} \).

Deciphering the Common Ratio

The common ratio is the backbone of a geometric series, dictating how each term transitions into the next. Recognizing and computing this ratio is fundamental to understanding the behavior of the series.

Here's how you can determine the common ratio, \( r \):

• Select a term in the series and the term immediately following it.
• Divide the subsequent term by the previous one to find \( r \).
• In our exercise, the series moves from 3 to -1, so \( r = -\frac{1}{3} \) because \( -1 \div 3 = -\frac{1}{3} \).

It's important to note the sign and magnitude of \( r \). The sign affects whether subsequent terms will alternate between positive and negative, and the magnitude affects whether the terms grow or shrink in size.

Delving into Infinite Series and Convergence

Infinite series extend the concept of a series by continuing indefinitely. A critical aspect of infinite series, like our geometric series, is whether they converge or not.

Convergence in the context of a geometric series means that if you sum up all the infinitely many terms, the sum approaches a finite value. Here's how convergence is determined:

• The series converges only if the absolute value of the common ratio \( r \) is less than 1, that is \( -1 < r < 1 \).
• For our series, \( r \) is \(-\frac{1}{3} \), which falls within the convergence criteria, \(-1 < -\frac{1}{3} < 1 \).

Once a series is confirmed to be convergent, its sum can be found using the formula:
\[ S = \frac{a}{1 - r} \] substituting the given values, \( S = \frac{3}{1 - (-\frac{1}{3})} = \frac{9}{4} = 2.25 \).

Convergence ensures the total sum of the series doesn't spiral off to infinity, making the concept invaluable in both pure and applied mathematics.