Problem 63 - Algebra and Trigonometry

Question

Determine whether each infinite geometric series converges or diverges. If it converges, find its sum. $$ \sum_{k=1}^{\infty} \frac{1}{2} \cdot 3^{k-1} $$

Step-by-Step Solution

Answer

The series diverges since \(|r| = 3 > 1\).

Step 1: Identify the First Term (a)

In the given infinite geometric series, identify the first term, denoted as \(a\). The series is $$\frac{1}{2} \times 3^{k-1}$$. When \(k=1\), the first term is: $$a = \frac{1}{2} \times 3^{1-1} = \frac{1}{2} \times 3^0 = \frac{1}{2}.$$

Step 2: Identify the Common Ratio (r)

Next, identify the common ratio, denoted as \(r\). The common ratio can be found by dividing the second term by the first term. The second term occurs when \(k=2\): $$r = \frac{3^{1}}{3^{0}} = 3.$$

Step 3: Test for Convergence

Determine if the series converges by checking the value of \(r\). An infinite geometric series converges if \(|r| < 1\) and diverges otherwise. Here, \(r = 3\): $$|r| = 3 > 1.$$ Therefore, the series diverges.

Key Concepts

Convergence and Divergence

Infinite geometric series can either converge or diverge based on the value of their common ratio, denoted as \(r\).
**Convergence**: An infinite geometric series converges if the absolute value of the common ratio, |r|, is less than 1. When a series converges, the terms keep getting smaller and approach zero, allowing the series to sum to a finite value.
**Divergence**: If the common ratio's absolute value, |r|, is greater than or equal to 1, the series diverges. This means the terms do not approach zero and the series does not sum to a finite value.
In our example, the series \sum_{k=1}^{\text{∞}} \frac{1}{2} \times 3^{k-1}\ has a common ratio \(r = 3 \). Since \( |3| > 1 \), the series diverges.

Common Ratio

The common ratio, often denoted by \(r\), is a crucial part of identifying whether an infinite geometric series converges or diverges.
To find the common ratio, divide any term in the series by the preceding term.
In the given series, \sum_{k=1}^{\text{∞}} \frac{1}{2} \times 3^{k-1}\, the first term is \( \frac{1}{2} \). When \( k = 2 \), the second term is \( \frac{1}{2} \times 3^{2-1} = \frac{1}{2} \times 3 \).
Thus, the common ratio \(r\) is calculated as:
\[ r = \frac{ \frac{1}{2} \times 3} { \frac{1}{2}} = 3 \]
Knowing \(r\) helps in determining whether the geometric series will converge or diverge and is essential for finding the sum if it converges.

Sum of a Series

One major aspect of infinite geometric series, if they converge, is the possibility to find their sum.
For a convergent infinite geometric series where \( |r| < 1 \), the sum can be determined using the formula:
\[ S = \frac{a}{1 - r} \] where \( S \) is the sum, \( a \) is the first term, and \( r \) is the common ratio.
In our problem, we found that the series diverges because \( |r| = 3 > 1 \). Therefore, the sum does not exist for this series. When dealing with problems like these, it's crucial first to determine whether the series converges or diverges, before attempting to calculate the sum.