Problem 17 - Thomas Calculus

Question

In Exercises \(15-22\) , determine if the geometric series converges or diverges. If a series converges, find its sum. $$\left(\frac{1}{8}\right)+\left(\frac{1}{8}\right)^{2}+\left(\frac{1}{8}\right)^{3}+\left(\frac{1}{8}\right)^{4}+\left(\frac{1}{8}\right)^{5}+\cdots$$

Step-by-Step Solution

Answer

The series converges with a sum of \( \frac{1}{7} \).

Step 1: Identify Series Type

The given series is \( \left(\frac{1}{8}\right)+\left(\frac{1}{8}\right)^{2}+\left(\frac{1}{8}\right)^{3}+\left(\frac{1}{8}\right)^{4}+\cdots \). This is a geometric series where the first term \( a = \frac{1}{8} \) and the common ratio \( r = \frac{1}{8} \).

Step 2: Test for Convergence

A geometric series \( a + ar + ar^2 + ar^3 + \cdots \) converges if the common ratio \( |r| < 1 \). Here, \( r = \frac{1}{8} \), and since \(|\frac{1}{8}| < 1\), the series converges.

Step 3: Calculate the Sum of the Series

The sum \( S \) of a converging geometric series can be found using the formula \( S = \frac{a}{1-r} \) where \( a \) is the first term and \( r \) is the common ratio. Here, \( a = \frac{1}{8} \) and \( r = \frac{1}{8} \), so \[ S = \frac{\frac{1}{8}}{1 - \frac{1}{8}} = \frac{\frac{1}{8}}{\frac{7}{8}} = \frac{1}{7}. \]

Key Concepts

Convergence of Series

In mathematical analysis, the convergence of a series is crucial to understanding whether the series approaches a finite value as the number of terms increases indefinitely. A series converges if the sum of its terms approaches a specific number, rather than growing without bound.
For geometric series, like the one in our exercise, a key factor in determining convergence is the *common ratio* \( r \). If the absolute value of the common ratio \( |r| \) is less than 1, the series converges. Conversely, if \( |r| \geq 1 \), the series diverges, meaning it does not settle towards any particular value.
In the example given, we see that the common ratio is \( \frac{1}{8} \). Since the absolute value \( |\frac{1}{8}| < 1 \), the series indeed converges.

Sum of Series

Once it is established that a geometric series converges, the next step is to calculate its sum. The sum of an infinite converging geometric series can be determined using the formula:

• \[ S = \frac{a}{1-r} \], where \( a \) is the first term, and \( r \) the common ratio.

Let's apply this to our given series. With \( a = \frac{1}{8} \) and \( r = \frac{1}{8} \), substituting these into the formula gives:

• \[ S = \frac{\frac{1}{8}}{1 - \frac{1}{8}} \]
• This simplifies to \[ \frac{\frac{1}{8}}{\frac{7}{8}} \]
• Finally, this results in \( S = \frac{1}{7} \).

Thus, the sum of this series as the terms extend infinitely is \( \frac{1}{7} \).

Common Ratio

The common ratio is a critical element in understanding and working with geometric series. It is defined as the ratio between successive terms in the series.
In our example, the series \( \left(\frac{1}{8}\right), \left(\frac{1}{8}\right)^2, \left(\frac{1}{8}\right)^3, \ldots \) has a common ratio \( r = \frac{1}{8} \). This means each term is \( \frac{1}{8} \) of the previous term.

• To find the common ratio, divide any term by the previous term.
• Formula: \( r = \frac{a_{n}}{a_{n-1}} \)
• Example: \( \frac{\left(\frac{1}{8}\right)^3}{\left(\frac{1}{8}\right)^2} = \frac{1}{8} \)

The common ratio determines not only the convergence but also the nature and behavior of the series.

Infinite Series

An infinite series, as the term suggests, is a sum of infinitely many terms. In mathematics, it brings about the challenge of finding if such a series has a finite sum or diverges to infinity.
When dealing with geometric series like the one in the exercise, the infinite aspect implies continuing the pattern established by the series terms indefinitely.
Important points to consider:

• An infinite series may or may not converge.
• Convergence depends largely on the common ratio, especially in geometric series.
• For convergence, it is essential that each subsequent term becomes negligibly small, essentially approaching zero as the series progresses.

In our exercise, despite infinitely many terms, the series converges to a sum because the common ratio ensures the diminishing value of each term.