Problem 6 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Determine whether the series converges or diverges. If it converges, \(n d\) its sum. $$ \frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\frac{1}{32}+\cdots $$

Step-by-Step Solution

Answer

The series converges and its sum is \( \frac{1}{6} \)

Step 1: Identification of First Term and Ratio

The first term \(a\) is \(\frac{1}{4}\) and the common ratio \(r\) can be found by dividing the second term by the first or the third term by the second, giving \(r=-\frac{1}{2}\). This is less than one, hence the series converges.

Step 2: Apply the Geometric Series Sum Formula

We apply the formula for the sum of an infinite geometric series, which is \(S = \frac{a}{1 - r}\). Substituting our values in, we get \(S = \frac{\frac{1}{4}}{1 - (-\frac{1}{2})}\)

Step 3: Simplify the Result

By performing the operation in the formula, we get \(S = \frac{\frac{1}{4}}{1.5} = \frac{1}{6}\)

Key Concepts

Convergence of Series

When we talk about the convergence of a series, we refer to whether the sum totals up to a specific finite number. In the context of a geometric series, convergence is specifically determined by the common ratio, which is a constant that each term of the series is multiplied by to get the next term. A geometric series converges if the absolute value of the common ratio is less than 1, meaning it shrinks each subsequent term, leading the series towards a limit. Conversely, if the absolute value of the common ratio is equal to or greater than 1, the series will diverge, meaning it won't settle towards a single sum. In our exercise, with a common ratio of \(-\frac{1}{2}\), observe how it lies between \(-1\) and \(1\). Therefore, this indicates convergence, implying the series will approach a particular sum rather than growing indefinitely.

Sum of Infinite Series

The sum of an infinite series refers to the total value obtained by adding all terms of a series, stretching towards infinity. For a geometric series that converges, we can calculate this sum using the formula: \[ S = \frac{a}{1 - r} \] where \(a\) is the first term of the series, and \(r\) is the common ratio. This formula gives us a shortcut to find the sum without manually adding up all the terms, which isn't possible for an infinite series. In our exercise, we have the first term \(a = \frac{1}{4}\) and the common ratio \(r = -\frac{1}{2}\). By plugging these numbers into the formula:

• Calculate the denominator: \(1 - (-\frac{1}{2}) = 1.5\)
• Apply the formula: \(S = \frac{\frac{1}{4}}{1.5}\)
• Simplify the fraction: \(S = \frac{1}{6}\)

This result indicates that the series converges to a sum of \(\frac{1}{6}\).

Common Ratio

At the heart of geometric series, the common ratio is crucial as it determines the pattern through which the series progresses. In simple terms, it is the fixed factor by which we multiply one term to obtain the next term in the series. For example, in our series, starting from the first term \(\frac{1}{4}\):

• The second term is derived by multiplying the first term by \(-\frac{1}{2}\), yielding \(-\frac{1}{8}\).
• Repeating this multiplication derives subsequent terms like \(\frac{1}{16}\), \(-\frac{1}{32}\), and so forth.

A negative common ratio, such as \(-\frac{1}{2}\), not only reduces the terms but also flips their signs, alternating their values to positive and negative throughout the series. Identifying this ratio correctly is essential because it dictates the series' behavior and ultimately assists in determining its convergence or divergence. Whether the series converges or diverges boils down to the range of the common ratio, reaffirming the significant role it plays in evaluating the limitless bounds of a geometric series.