Problem 13 - 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. $$ \sum_{n=0}^{\infty}\left(\frac{1}{2^{n}}+\frac{(-1)^{n}}{5^{n}}\right) $$

Step-by-Step Solution

Answer

The series converges with a sum of \( \frac{17}{6} \).

Step 1: Identify the Series

We are given the infinite series \( \sum_{n=0}^{\infty}\left(\frac{1}{2^{n}}+\frac{(-1)^{n}}{5^{n}}\right) \). This is a combination of two separate series: a geometric series \( \frac{1}{2^{n}} \) and an alternating geometric series \( \frac{(-1)^{n}}{5^{n}} \). We will write out the first eight terms of this series.

Step 2: Calculate the First Eight Terms

For each term in the series: - When \( n=0 \), the term is \( 1 + 1 = 2 \).- When \( n=1 \), the term is \( \frac{1}{2} - \frac{1}{5} = \frac{5}{10} - \frac{2}{10} = \frac{3}{10} \).- When \( n=2 \), the term is \( \frac{1}{4} + \frac{1}{25} = \frac{25}{100} + \frac{4}{100} = \frac{29}{100} \).- When \( n=3 \), the term is \( \frac{1}{8} - \frac{1}{125} \). To find a common denominator: - \( \frac{1}{8} = \frac{125}{1000} \) - \( \frac{1}{125} = \frac{8}{1000} \) - Thus, the term is \( \frac{125}{1000} - \frac{8}{1000} = \frac{117}{1000} \).Continue this calculation up to \( n=7 \), listing the terms in a similar manner.

Step 3: Identify the Nature of Each Component Series

The series \( \sum_{n=0}^{\infty} \frac{1}{2^n} \) is a geometric series with \( a = 1 \) and \( r = \frac{1}{2} \). It converges because \(|r| < 1\). The sum is \( \frac{1}{1 - \frac{1}{2}} = 2 \).The series \( \sum_{n=0}^{\infty} \frac{(-1)^n}{5^n} \) is an alternating geometric series with \( a = 1 \) and \( r = -\frac{1}{5} \). It converges since \(|r| < 1\). The sum is \( \frac{1}{1 - (-\frac{1}{5})} = \frac{1}{1 + \frac{1}{5}} = \frac{1}{\frac{6}{5}} = \frac{5}{6} \).

Step 4: Find the Total Sum of the Series

Since both components of the original series converge, the whole series converges as well. The sum of the series is the sum of the individual series: \( 2 + \frac{5}{6} = 2 + 0.8333... = 2.8333... = \frac{17}{6} \).

Step 5: Conclusion

The original series \( \sum_{n=0}^{\infty}\left(\frac{1}{2^{n}}+\frac{(-1)^{n}}{5^{n}}\right) \) converges with a total sum of \( \frac{17}{6} \).

Key Concepts

Geometric Series

A geometric series is a type of series where each term is obtained by multiplying the previous term by a constant factor, known as the common ratio. You can write a geometric series as:

• \[ a + ar + ar^2 + ar^3 + \dots \]

Here, \( a \) is the first term and \( r \) is the common ratio. A crucial property of geometric series is that they can either converge or diverge based on the value of \( r \):

• If \(|r| < 1\), the series converges, meaning its sum approaches a finite number.
• If \(|r| \geq 1\), the series diverges, and the sum reaches infinity.

For convergent geometric series, the sum can be calculated by the formula:

• \[S = \frac{a}{1 - r},\quad \text{where } |r| < 1\]

In the original exercise, the component series \( \sum_{n=0}^{\infty} \frac{1}{2^n} \) is geometric with \( a = 1 \) and \( r = \frac{1}{2} \), so it converges with a sum of 2.

Alternating Series

An alternating series is a series whose terms alternate in sign. It can be expressed as:

• \[ a_1 - a_2 + a_3 - a_4 + \ldots \]

Alternating series often involve terms like \((-1)^n\) or \((-1)^{n+1}\), ensuring the series alternates between positive and negative. The key concept for checking the convergence of an alternating series is the Alternating Series Test:

• If the absolute value of the terms decreases steadily to zero, the series will converge.

Let's consider the alternating geometric series from the exercise: \( \sum_{n=0}^{\infty} \frac{(-1)^n}{5^n} \). This series has:

• \( a = 1 \) and \( r = -\frac{1}{5} \)

Given that \(|r| = \frac{1}{5} < 1\), this alternating series converges. The sum of such a series can be evaluated similarly to non-alternating geometric series:

• \[S = \frac{a}{1 - r}\]
• The sum for this series is \( \frac{1}{1 + \frac{1}{5}} = \frac{5}{6} \).

Infinite Series

Infinite series are sums that have an infinite number of terms. They can present complex behavior, either converging to a specific finite value or diverging, having no limit. Understanding infinite series is crucial in many areas of calculus and mathematical analysis.
There are several types of infinite series, including arithmetic series, geometric series, harmonic series, and many more. Our focus is on geometric and alternating series, as seen in the original exercise.
The exercise involved an infinite series built by summing the terms of a geometric series and an alternating series together:

• \[\sum_{n=0}^{\infty}\left(\frac{1}{2^{n}}+\frac{(-1)^{n}}{5^{n}}\right)\]

To find whether an infinite series converges or diverges, it is essential to understand its components. Since both component series in our exercise converge, the entire infinite series also converges.

• The total sum is derived by adding the sums of the individual convergent series: \( 2 + \frac{5}{6} = \frac{17}{6} \).

In summary, understanding infinite series involves recognizing the nature of its components. When you determine each part's behavior, you can conclude the series's overall behavior.