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

Step 1: Understand the series pattern

The series given is a geometric series where the first term \(a = 1\), and the common ratio \(r = -\frac{1}{2}\). The terms alternate in sign due to the \((-1)^{n-1}\) factor.

Step 2: Determine the formula for the n-th partial sum

The formula for the sum of the first \(n\) terms of a geometric series is \[S_n = a \frac{1-r^n}{1-r}\]. Substituting the values \(a = 1\) and \(r = -\frac{1}{2}\), we get \[S_n = \frac{1 - \left(-\frac{1}{2}\right)^n}{1 + \frac{1}{2}} = \frac{1 - \left(-\frac{1}{2}\right)^n}{\frac{3}{2}} = \frac{2}{3} \left(1 - \left(-\frac{1}{2}\right)^n\right)\].

Step 3: Check for convergence of the series

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

Step 4: Calculate the limit of the series as n approaches infinity

Since the series converges, we find the sum by taking the limit of the partial sum formula as \(n\) approaches infinity: \[\sum = \lim_{n \to \infty} S_n = \frac{2}{3} \left(1 - \lim_{n \to \infty}\left(-\frac{1}{2}\right)^n\right) = \frac{2}{3} (1 - 0) = \frac{2}{3}\].

Key Concepts

Partial Sum

In mathematics, especially when dealing with series, the concept of a partial sum is very important. A partial sum represents the sum of the first few terms of a series. This helps us to understand how the series behaves as we add more terms.
For a geometric series, like the one in our original problem, the partial sum of the first \(n\) terms is found using a specific formula.
If a series starts with the first term \(a\) and has a common ratio \(r\), the formula to find the \(n\)th partial sum \(S_n\) is:

• \( S_n = a \frac{1 - r^n}{1-r} \)

This formula allows us to calculate the sum of the terms up to any desired point \(n\). It's particularly useful because it provides a direct way to sum up the series without having to add each term individually.
Understanding partial sums is a key step toward analyzing the overall behavior of series, especially in determining whether the series converges or diverges.

Series Convergence

A central concept in the study of series is convergence. When we talk about series convergence, we're interested in whether summing all terms of an infinite series results in a finite number.
For geometric series, there's a straightforward rule: If the absolute value of the common ratio \(|r| < 1\), then the series converges. Otherwise, it diverges and does not result in a finite sum.
In the exercise provided, we found that the common ratio \(r\) was \(-\frac{1}{2}\). Because \(|-\frac{1}{2}| = \frac{1}{2} < 1\), we can conclude that the series converges.
Convergence is crucial in many areas of mathematics as it helps determine the applicability and stability of various methods, particularly in calculus and analysis. It tells us that even though the series has an infinite number of terms, they effectively "sum up" to a specific value.

Infinite Series

An infinite series is essentially the sum of infinitely many terms. This sounds a bit paradoxical because we typically think of sums as being finite. However, in mathematics, we often stretch our definitions to include infinite processes.
When dealing with geometric series, like the one in the exercise, we focus on whether this endless sum leads to a meaningful result, usually by assessing convergence.
If a geometric series converges, the sum of all its infinite terms can be calculated using the formula for the sum of an infinite series:

• \( S = \frac{a}{1-r} \)

This formula works when \(|r| < 1\), ensuring that the series converges. In such a case, the terms become smaller and smaller as we keep adding them, ultimately approaching a particular value.
Infinite series are foundational in mathematical analysis, physics, and many applied fields, allowing us to solve complex problems by approximating seemingly unmanageable sums.