Problem 32 - Thomas Calculus

Question

Which of the series in Exercises 13 46 converge, and which diverge? Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series' convergence or divergence.) $$ \sum_{n=1}^{\infty} \frac{1}{(\ln 3)^{n}} $$

Step-by-Step Solution

Answer

The series converges because it is a geometric series with a common ratio less than 1.

Step 1: Identify the Series Type

The series given is \( \sum_{n=1}^{\infty} \frac{1}{(\ln 3)^{n}} \). This is a geometric series, where the general term can be identified as \( a_n = \left( \frac{1}{\ln 3} \right)^n \).

Step 2: Determine the Common Ratio

For a geometric series \( \sum_{n=1}^{\infty} ar^n \), the common ratio \( r \) can be identified in \( a_n = ar^n \). Here, \( r = \frac{1}{\ln 3} \).

Step 3: Evaluate the Common Ratio

For a geometric series to converge, the absolute value of the common ratio \( |r| \) must be less than 1. Let's evaluate \( |r| = \left| \frac{1}{\ln 3} \right| \). Since \( \ln 3 > 1 \), \( \frac{1}{\ln 3} < 1 \).

Step 4: Conclusion on Convergence

Since the common ratio \( r < 1\), the series converges. A geometric series with \( |r| < 1 \) converges by the geometric series test.

Key Concepts

series convergence

Convergence in series is like determining whether a series "settles down" to a specific value or continues to grow indefinitely. In simple terms, when we talk about a series converging, it means that as we keep adding more and more terms of the series, the total sum approaches a fixed number. For a series to converge, it must satisfy certain criteria, which depend on the type of series we are dealing with.
Understanding convergence is crucial because it helps determine the fate of the infinite sums.

• A convergent series stabilizes to a certain number.
• A divergent series keeps increasing or decreasing without settling on a particular value.

In the case of geometric series, which we are dealing with here, the common ratio plays a central role in determining convergence.

common ratio

The common ratio in a geometric series is a key factor that helps us determine how each term in the series relates to the previous one. In a geometric series, each term is obtained by multiplying the previous term by a fixed number known as the common ratio.
The formula for a geometric series looks like this:

• ext{General Term:} \, a_n = ar^n

Here, \( r \) is the common ratio. Understanding this ratio is vital because it influences whether the series will converge or diverge.
To find the common ratio, you simply divide any term by the previous term. In our example, the given series is \( \sum_{n=1}^{\infty} \left(\frac{1}{\ln 3}\right)^n \). Here, the common ratio \( r \) is \( \frac{1}{\ln 3} \). Knowing \( r \) allows us to use the geometric series test to check for convergence.

geometric series test

The geometric series test is a simple yet powerful tool for deciding if a geometric series converges or diverges. Once you have identified that you are dealing with a geometric series, you apply the geometric series test to check the behavior of the series.
This test involves evaluating the absolute value of the common ratio \( |r| \).

• If \( |r| < 1 \), the series converges.
• If \( |r| \geq 1 \), the series diverges.

Let's apply this test to our example. We identified the common ratio as \( r = \frac{1}{\ln 3} \). Since \( \ln 3 > 1 \), it follows that \( \|r\| = \left| \frac{1}{\ln 3} \right| < 1 \). Hence, the series converges. This reflects the underlying principle of the geometric series test, which emphasizes the significance of the common ratio being less than one for convergence to occur.