Problem 10 - Thomas Calculus

Question

In Exercises \(7-14,\) write out the first eight terms of each series to show how the series starts. Then find the sum of the series or show that it diverges. $$ \sum_{n=0}^{\infty}(-1)^{n} \frac{5}{4^{n}} $$

Step-by-Step Solution

Answer

The series converges and its sum is 4.

Step 1: Identify the series

The given series is \( \sum_{n=0}^{\infty}(-1)^{n} \frac{5}{4^{n}} \). This is an alternating series, as indicated by the \((-1)^n\) term.

Step 2: Write the general term for the series

The general term of the series is \( a_n = (-1)^n \frac{5}{4^n} \).

Step 3: Generate the first eight terms of the series

Compute each term for \( n = 0 \) to \( n = 7 \): - For \( n = 0 \), \( a_0 = (-1)^0 \frac{5}{4^0} = 5 \) - For \( n = 1 \), \( a_1 = (-1)^1 \frac{5}{4^1} = -\frac{5}{4} \) - For \( n = 2 \), \( a_2 = (-1)^2 \frac{5}{4^2} = \frac{5}{16} \) - For \( n = 3 \), \( a_3 = (-1)^3 \frac{5}{4^3} = -\frac{5}{64} \) - For \( n = 4 \), \( a_4 = (-1)^4 \frac{5}{4^4} = \frac{5}{256} \) - For \( n = 5 \), \( a_5 = (-1)^5 \frac{5}{4^5} = -\frac{5}{1024} \) - For \( n = 6 \), \( a_6 = (-1)^6 \frac{5}{4^6} = \frac{5}{4096} \) - For \( n = 7 \), \( a_7 = (-1)^7 \frac{5}{4^7} = -\frac{5}{16384} \)

Step 4: Determine if the series converges and find the sum

This is an infinite geometric series with first term \( a = 5 \) and common ratio \( r = -\frac{1}{4} \). Since the absolute value of the common ratio \(|r| < 1\), the series converges. The sum of the series can be calculated using the formula for the sum of an infinite geometric series: \[ S = \frac{a}{1 - r} = \frac{5}{1 - (-\frac{1}{4})} = \frac{5}{\frac{5}{4}} = 4. \]

Key Concepts

Understanding Alternating Series

An alternating series is one where the signs of the terms alternate, meaning they switch between positive and negative as the series progresses. This change in sign is typically indicated by a factor like \((-1)^n\), which ensures that even terms are positive and odd terms are negative. Alternating series are common in mathematics because they are often used to approximate functions or represent them more simply.

When analyzing an alternating series, one important aspect is checking how quickly the terms decrease in size and whether they converge to a small number or zero. In many cases, such as in our example series \( \sum_{n=0}^{\infty}(-1)^{n} \frac{5}{4^{n}} \), the terms \((-1)^n \frac{5}{4^n}\) decrease in magnitude as \(-1/4\) is a fraction less than one, indicating the alternating series will likely converge.

This alternating behavior is not only important for convergence, but it can also offer advantages in numerical calculations and stability in approximation models.

Convergence of a Series

Convergence in the context of a series refers to whether the sum of infinitely many numbers reaches a finite limit. Some series will converge, others won't. A powerful way to determine convergence is through the geometric series test, specifically when dealing with series like \( \sum_{n=0}^{\infty} (-1)^{n} \frac{5}{4^n} \).

To use this test, we observe the common ratio \( r \) derived from the series terms. For a series to converge, the absolute value of this ratio \( |r| \) must be less than one. This condition is met in our series with \( r = -\frac{1}{4} \), hence \(|r| < 1\). When these conditions are satisfied, the series converges to a particular sum.

Understanding whether a series converges and how quickly it does can help determine whether the series is useful for approximation. Converging series provide reliable sum estimates, making them handy in practical calculations. Hence, convergence is a crucial property when evaluating series.

Calculating the Sum of the Series

The sum of an infinite geometric series can be determined using a specific formula, provided the series converges. For the infinite series \( \sum_{n=0}^{\infty} a_n \) to converge, the absolute value of the common ratio \( |r| \) must be less than one. Once convergence is established, the sum \( S \) of such a series can be calculated using:

• \( S = \frac{a}{1 - r} \)

where \( a \) represents the first term of the series and \( r \) represents the common ratio. In our specific series, \( a = 5 \) and \( r = -\frac{1}{4} \).

Plugging in these values into the formula, we can compute the sum:

• \( S = \frac{5}{1 - (-\frac{1}{4})} = \frac{5}{\frac{5}{4}} \)
• \( S = 4 \)

This indicates that the entire series simplifies into the value 4. This procedure is helpful in many mathematical and scientific applications where understanding the total impact of adding infinitely many diminishing quantities is desired. The ability to sum all terms precisely enhances the certainty and accuracy in various fields such as engineering and physics.