Problem 62 - Algebra and Trigonometry

Question

Determine whether each infinite geometric series converges or diverges. If it converges, find its sum. $$ \sum_{k=1}^{\infty} 8\left(\frac{1}{3}\right)^{k-1} $$

Step-by-Step Solution

Answer

\( S = 12 \)

Step 1 - Identify the first term and common ratio

In an infinite geometric series, the general form is given by \( a r^{k-1}\) , where \(a\) is the first term and \(r\) is the common ratio. Here, the first term \(a = 8\) and the common ratio \(r = \frac{1}{3}\).

Step 2 - Determine Convergence or Divergence

An infinite geometric series converges if and only if the absolute value of the common ratio is less than 1, that is \(|r| < 1\). In this case, \(r = \frac{1}{3}\), and since \(\frac{1}{3} < 1\), the series converges.

Step 3 - Use the Sum Formula for Convergent Series

When a geometric series converges, its sum can be found using the formula for the sum of an infinite geometric series, \(S = \frac{a}{1-r}\). Substituting the known values, \(a = 8\) and \(r = \frac{1}{3}\), the sum is \( S = \frac{8}{1 - \frac{1}{3}} \) .

Step 4 - Simplify the Sum

Now, simplify the equation \( S = \frac{8}{1 - \frac{1}{3}}\). This simplifies to \( S = \frac{8}{\frac{2}{3}} \) , which further simplifies to \( S = 8 \times \frac{3}{2} = 12\). Therefore, the sum of the series is 12.

Key Concepts

infinite geometric series

An infinite geometric series is a type of series where you keep adding terms indefinitely. Each term in the series is obtained by multiplying the previous term by a constant factor known as the common ratio. The series can be written in the form: