Problem 67 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

The series converges and its sum is 6.

Step 1: Identify the first term and common ratio

For the geometric series \sum_{k=1}^{\textbackslash infty} 3\textbackslash left(\textbackslash frac{2}{3}\textbackslash right)^{k}\, identify the first term \( a \) and the common ratio \( r \). The first term \( a \) is the coefficient of the term when \( k = 1 \), and it can be found as: \[ a = 3 \textbackslash left(\frac{2}{3}\textbackslash right)^{1} = 3 \left( \textbackslash frac{2}{3} \right) = 2. \] The common ratio is \( r = \textbackslash frac{2}{3} \).

Step 2: Determine if the series converges

A geometric series converges if the absolute value of the common ratio \( |r| < 1 \). Here, \( |\frac{2}{3}| = \frac{2}{3} \), which is less than 1. Therefore, the series converges.

Step 3: Find the sum of the series

Since the series converges, use the formula for the sum of an infinite geometric series: \[ S = \frac{a}{1 - r} \] where \( a = 2 \) and \( r = \frac{2}{3} \). Substituting these values in, calculate the sum as follows: \[ S = \frac{2}{1 - \frac{2}{3}} = \frac{2}{\frac{1}{3}} = 2 \times 3 = 6. \]

Key Concepts

Geometric Series Convergence

When dealing with infinite geometric series, one key concept is convergence. For an infinite geometric series \(\textstyle \sum_{k=1}^{\text{∞}} ar^{k}\), the series converges if the absolute value of the common ratio is less than 1. This means \(|r| < 1\). Convergence here means that as we keep adding terms, the sum gets closer and closer to a specific value instead of growing indefinitely.
In our original exercise, the series is \(\textstyle \sum_{k=1}^{\text{∞}} 3 \left( \frac{2}{3} \right)^{k}\). By identifying the common ratio \(\textstyle r = \frac{2}{3} \), we see that \(|r| = \frac{2}{3}|\) which is less than 1. Hence, the series converges.
It's crucial to always check the value of \( r? \) to determine whether our series will sum to a finite value or diverge. If \(|r| \geq 1\), the series would diverge, meaning the sum would go to infinity and would not settle to a specific number.

Common Ratio

The common ratio \( r \) is a key term in any geometric series. It represents the constant factor between consecutive terms of the series. For a geometric series expressed as \(\text{a, ar, ar}^2, ar^3, \ldots\), the ratio between any term and the one before it is always \( r \).
In our problem \(\textstyle \sum_{k=1}^{\text{∞}} 3 \left( \frac{2}{3} \right)^{k}\), we determined that the common ratio is \( r = \frac{2}{3}\). This is found by identifying the factor being exponentiated by \( k \) in each term.
Understanding the common ratio helps in determining the behavior of the series. If \(| r | < 1\), we know the terms are getting smaller and the series is likely to converge. Conversely, if \(| r | \geq 1\), the terms either stay the same size or grow, leading to divergence.

Sum of Infinite Series

An essential formula for geometric series is the sum of an infinite series. When \( | r | < 1 \), the sum \( S \) of an infinite geometric series can be computed using:
\[ S = \frac{a}{1 - r} \]
Here, \( a\) is the first term of the series, and \(- r\) is the common ratio.
Applying this to our exercise \(\textstyle \sum_{k=1}^{\text{∞}} 3 \left( \frac{2}{3} \right)^{k}\), we've previously identified \(a = 2\) (because the first term when \(k = 1\) is 2) and \(r = \frac{2}{3}\). Plugging these into the formula:

\[ S = \frac{2}{1 - \frac{2}{3}} = \frac{2}{\frac{1}{3}} = 2 \times 3 = 6 \]
Thus, the sum of our infinite geometric series is 6. This final value tells us where the series converges.
Remember, this formula is only valid when \(| r | < 1\) because otherwise, the series does not settle to a finite sum.