Problem 3 - Thomas Calculus

Question

In Exercises \(1-6,\) find a formula for the \(n\) th partial sum of each series and use it to find the series' sum if the series converges. $$1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\cdots+(-1)^{n-1} \frac{1}{2^{n-1}}+\cdots$$

Step-by-Step Solution

Answer

The sum of the series is \(\frac{2}{3}\).

Step 1: Identify the Series Pattern

The series given is an alternating series with terms of the form \((-1)^{n-1} \frac{1}{2^{n-1}}\). The pattern suggests a geometric series with a first term \(a = 1\) and a common ratio \(r = -\frac{1}{2}\).

Step 2: Write the General Formula for Partial Sum of a Geometric Series

The sum of the first \(n\) terms of a geometric series is given by the formula: \(S_n = a \frac{1 - r^n}{1 - r}\). Substituting the values \(a = 1\) and \(r = -\frac{1}{2}\), the formula becomes \(S_n = \frac{1 - \left(-\frac{1}{2}\right)^n}{1 - (-\frac{1}{2})}\).

Step 3: Simplify the Formula for the Partial Sum

Simplify \(S_n = \frac{1 - (-\frac{1}{2})^n}{1 + \frac{1}{2}} = \frac{1 - (-\frac{1}{2})^n}{\frac{3}{2}}\). This further simplifies to \(S_n = \frac{2}{3} (1 - (-\frac{1}{2})^n)\).

Step 4: Determine if the Series Converges

A geometric series converges if the common ratio \(|r| < 1\). Here, \(|r| = |\frac{1}{2}| = \frac{1}{2} < 1\), so the series converges.

Step 5: Find the Sum of the Converging Series

For a converging series with \(|r| < 1\), the sum is \(S = \frac{a}{1 - r}\). Substituting \(a = 1\) and \(r = -\frac{1}{2}\), we find \(S = \frac{1}{1 - (-\frac{1}{2})} = \frac{1}{\frac{3}{2}} = \frac{2}{3}\).

Key Concepts

Partial Sum

In mathematics, especially when dealing with infinite series, it is often useful to consider the partial sum. A partial sum is the sum of a given number of first terms of a sequence. This concept is crucial as it helps us study and understand the behavior of series, whether they reach a finite value or continue to grow without limits. For a geometric series, like the one presented in the original problem with alternating signs, the formula for the partial sum of the first \(n\) terms is given by:

• \(S_n = a \frac{1 - r^n}{1 - r}\)

where \(a\) is the first term and \(r\) is the common ratio of the series. In this sequence, \(a = 1\) and \(r = -\frac{1}{2}\), so the partial sum formula becomes:

• \(S_n = \frac{2}{3} (1 - (-\frac{1}{2})^n)\)

By using this formula, we can calculate the cumulative value of these first \(n\) terms, which gives insight into the progression of the series.

Series Convergence

Series convergence is a fundamental concept when dealing with infinite series. To determine if a series converges, we check if the sum of its infinite terms approaches a finite limit. In geometrical terms, for a series with a common ratio \(r\), convergence occurs if \(|r| < 1\). This ensures the terms shrink towards zero, preventing the sum from growing infinitely. In this particular series, the common ratio is \(-\frac{1}{2}\) and we ascertain convergence because \(|-\frac{1}{2}| = \frac{1}{2} < 1\). This finding is an essential step, as it confirms that this series does indeed converge to a specific value, allowing further calculations.

Alternating Series

An alternating series is a series in which the terms change sign each time—think of positive, negative, positive, and so on. This pattern is crucial because it affects the way the series approaches its sum. The given series:

• \(1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \cdots\)

is an alternating geometric series. Each term is of the form:\((-1)^{n-1} \frac{1}{2^{n-1}}\). The difference in sign from one term to the next can lead to more rapid convergence as the positive and negative terms partially cancel each other out. This results in a "zig-zag" behavior towards a specific limit, making alternating series an interesting study of series convergence. In this example, the alternating nature plays a critical role in the convergence and final sum, ultimately leading to a sum of \(\frac{2}{3}\).