Problem 81 - Thomas Calculus

Question

Make up an infinite series of nonzero terms whose sum is a. 1\(\quad\) b. \(-3 \quad\) c. 0

Step-by-Step Solution

Answer

a. Use \(\sum_{n=1}^{\infty} \frac{1}{2^n}\). b. Use \(\sum_{n=1}^{\infty} \frac{-3}{2^n}\). c. Use \(\sum_{n=1}^{\infty} \left(\frac{(-1)^{n+1}}{n}\right)\).

Step 1: Understanding Infinite Series

An infinite series is the sum of the terms of an infinite sequence. Our task is to create infinite series that converge to given sums. We will use known series as a base for each part.

Step 1: Sum of Series is 1

Use the geometric series formula \[ \sum_{n=0}^{\infty} ar^n = \frac{a}{1-r} \] where \( |r| < 1 \). We know if \( a = 1 \) and \( r = \frac{1}{2} \), the sum is \[ \sum_{n=0}^{\infty} \left(\frac{1}{2}\right)^n = \frac{1}{1-\frac{1}{2}} = 2. \] We need to adjust terms to make the sum 1.

Step 3: Create Series for Part a

To have a series sum to 1, take \[ \sum_{n=1}^{\infty} \frac{1}{2^n}. \] This is the series with terms such as \( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots \) and its sum is realized as \[ 1 - \left(\sum_{n=0}^{\infty} \left(\frac{1}{2}\right)^n - 1\right) = 1. \]

Step 2: Sum of Series is -3

Use a series similar to \[ a + ar + ar^2 + ar^3 + \ldots \]where \[ a = -3 \text{ and } r = \frac{1}{2}. \] Sum becomes \[ -3 \times \frac{1}{1-\frac{1}{2}} = -6. \] Use a different base series for \(-3\) by setting \( a = -3 \), ensuring convergence to \(-3\), not \(-6\).

Step 5: Create Series for Part b

Insert a known geometric property: \[ \sum_{n=1}^{\infty} -3\left(\frac{1}{2}\right)^{n} = -3 \times \left( \sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^n \right) = -3. \] This series features terms such as \( - \frac{3}{2} - \frac{3}{4} - \frac{3}{8} + \ldots \).

Step 3: Sum of Series is 0

A straightforward series with alterations alternates signs; famously: \[ 1 - 1 + 1 - 1 + 1 - \ldots \] takes a form leading to an undefined behavior but forms alike can sum to zero in the limit.

Step 7: Create Series for Part c

Adapt the alternating sum by using \[ \sum_{n=1}^{\infty} \left(\frac{(-1)^{n+1}}{n}\right). \] This method maintains convergence whether a combination respectively sums to zero over periods. It features an alternating series with compensating \( n \) in the denominator for convergence.

Key Concepts

Geometric Series

Understanding a geometric series is key to grasping the concept of infinite series, especially when we want to determine specific sums. A geometric series has the form:

• \( a + ar + ar^2 + ar^3 + \ldots \)
• Where \( a \) is the first term and \( r \) is the common ratio

For the series to be geometric, the ratio between successive terms must be constant. When dealing with an infinite geometric series, it can converge to a specific sum if the absolute value of the common ratio \( |r| \) is less than one. The formula used to find the sum of an infinite geometric series is:\[\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r}\]To solve for specific sums using geometric series, adapt the values of \( a \) and \( r \) accordingly. For instance, to achieve a sum of 1 with an infinite series, choosing \( a = 1 \) and \( r = \frac{1}{2} \), then altering the series approach further led us to the correct sum. Adjusting the series led to:\[\sum_{n=1}^{\infty} \frac{1}{2^n} = 1\]This demonstrates the power of geometric series in modeling infinite sums.

Convergence

Convergence is a fundamental concept when working with infinite series. It addresses whether the series approaches a finite value as the number of terms increases indefinitely. For a series to be considered convergent, its terms must approach zero in a specific manner.Convergence of a geometric series depends heavily on the common ratio \( r \). If \(|r| < 1\), the geometric series will converge to a finite sum, calculated using the formula:\[\text{Sum} = \frac{a}{1-r}\]However, if \(|r| \geq 1\), the series will diverge, meaning it doesn’t settle at any finite value.For example, when we found a series with a sum of -3, we carefully chose the first term and the common ratio:

• First term \( a = -3 \)
• Common ratio \( r = \frac{1}{2} \)

This led to:\[\sum_{n=1}^{\infty} -3 \left(\frac{1}{2}\right)^n = -3\]Understanding convergence is crucial when manipulating infinite series towards desired sums.

Alternating Series

An alternating series is notable for its terms changing sign between positive and negative. This sees terms follow a pattern such as \( 1, -1, 1, -1, \ldots \), contributing much to the possibility of constructing series with specific sums.To form a series with a sum of zero, we can use an alternating pattern. A classical form is:

• \( 1 - 1 + 1 - 1 + \ldots \)

While such a sequence does not converge in the typical sense, one can construct related series that do by carefully managing the terms.A more refined version could be:\[\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\]This series converges thanks to its alternating nature and thus can sum to zero. Alternating series open pathways to tailor sums through convergence techniques and are instrumental for series manipulation.