Problem 7 - Calculus: Early Transcendentals

Question

Determine whether the series converges or diverges. \( \displaystyle \sum_{n = 1}^{\infty} \frac {9^n}{3 + 10^n} \)

Step-by-Step Solution

Answer

The series converges.

Step 1: Recognize the Series Type

Identify that the series is \[ \sum_{n=1}^{\infty} \frac{9^n}{3 + 10^n} \]and it resembles a geometric series form but requires an alternate test due to the more complex denominator.

Step 2: Examine the Terms for Large n

For large values of \( n \), the term \( 10^n \) in the denominator \( 3 + 10^n \) dominates over the constant term 3. Thus, the terms of the series \( \frac{9^n}{3 + 10^n} \) can be approximated as \( \frac{9^n}{10^n} = \left( \frac{9}{10} \right)^n \).

Step 3: Apply the Limit Comparison Test

Apply the Limit Comparison Test with the geometric series \( \sum_{n=1}^{\infty} \left( \frac{9}{10} \right)^n \), which is known to converge. Compute the limit:\[ \lim_{{n \to \infty}} \frac{\frac{9^n}{3 + 10^n}}{\left(\frac{9}{10}\right)^n} = \lim_{{n \to \infty}} \frac{1}{3 + 10^n}\cdot 10^n \approx \lim_{{n \to \infty}} \frac{10^n}{10^n} = 1 \].

Step 4: Conclude on Convergence

Since the limit from the Limit Comparison Test established equality (not 0), and we compared with a convergent geometric series, the original series converges by the Limit Comparison Test.

Key Concepts

Limit Comparison Test

When analyzing an infinite series, especially one with complex terms, the Limit Comparison Test (LCT) is a useful convergence test. The LCT helps determine the convergence or divergence of a series by comparing it to another series with known behavior. The general form of using LCT involves:

• Selecting a second series, \( \sum b_n \), which is simple and has known convergence properties.
• Computing the limit: \( \lim_{{n \to \infty}} \frac{a_n}{b_n} \) for each term \( a_n \) of the original series and \( b_n \) of the comparison series.
• If the limit is a finite, non-zero constant, then both series converge or diverge together.

In the given exercise, we began by recognizing the terms \( a_n = \frac{9^n}{3+10^n} \) and compared it with a geometric series \( \left(\frac{9}{10}\right)^n \). The limit calculated \( \frac{10^n}{3 + 10^n} \) approaches 1, indicating similar convergence behavior.

Geometric Series

Geometric series are one of the simplest types of series, characterized by a constant ratio between successive terms. A geometric series can be represented as:\[ \sum_{n=0}^{\infty} ar^n = a + ar + ar^2 + ar^3 + \ldots \]where \( a \) is the first term and \( r \) is the common ratio. The convergence of a geometric series depends primarily on the value of \( r \):

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

In our exercise, the series \( \left( \frac{9}{10} \right)^n \) is a geometric series with \( r = \frac{9}{10} \) which is less than 1, indicating convergence. This became a basis for our comparison in the Limit Comparison Test.

Infinite Series

Infinite series involve summing an infinite sequence of terms and can result in either a convergent or divergent sum. To determine the behavior of an infinite series, a variety of tests can be used, including but not limited to:

• Geometric Series Test - for series with a constant ratio.
• Limit Comparison Test - for series that are complex but share a similar growth rate to a simpler series.
• Ratio Test - evaluates the limit of the ratio between successive terms.

In determining whether a series is an infinite convergent series, we seek a finite sum as the number of terms goes to infinity. During the evaluation of the series \( \sum_{n=1}^{\infty} \frac{9^n}{3 + 10^n} \), a combination of tests like the Limit Comparison Test helps confirm the series' convergence, giving us clear insight that as \( n \to \infty \), the series approaches a finite limit.