Problem 66 - Thomas Calculus

Question

Assume that \(b_{n}\) is a sequence of positive numbers converging to 1\(/ 3 .\) Determine if the following series converge or diverge. $$ \\\\{ b. }\quad sum_{n=1}^{\infty} \frac{b_{n+1} b_{n}}{n 4^{n}} \quad \text { b. } \sum_{n=1}^{\infty} \frac{n^{n}}{n ! b_{1}^{2} b_{2}^{2} \cdots b_{n}^{2}} $$

Step-by-Step Solution

Answer

Series A converges; Series B diverges.

Step 1: Analyze Series A

Consider the series \( \sum_{n=1}^{\infty} \frac{b_{n+1} b_{n}}{n 4^{n}} \). We want to apply the Limit Comparison Test with a geometric series. Since \(b_n \to \frac{1}{3}\), for large \(n\), \(b_n \approx \frac{1}{3}\). Let's compare with \(\sum \frac{1}{n 4^n}\), a known convergent series.

Step 2: Apply Limit Comparison Test to Series A

Using the Limit Comparison Test, compute \(\lim_{n \to \infty} \frac{\frac{b_{n+1}b_n}{n4^n}}{\frac{1}{n4^n}} = \lim_{n \to \infty} b_{n+1}b_n = \left(\frac{1}{3}\right)^2 = \frac{1}{9}. \) Since this is a positive finite number, and \(\sum \frac{1}{n 4^n}\) converges, \( \sum_{n=1}^{\infty} \frac{b_{n+1} b_{n}}{n 4^{n}} \) also converges.

Step 3: Analyze Series B

Consider the series \( \sum_{n=1}^{\infty} \frac{n^{n}}{n ! b_{1}^{2} b_{2}^{2} \cdots b_{n}^{2}} \). We will use the Ratio Test to determine its convergence.

Step 4: Apply Ratio Test to Series B

Determine the nth term \( a_n = \frac{n^{n}}{n! b_{1}^{2} b_{2}^{2} \cdots b_{n}^{2}} \). Use the Ratio Test by examining \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \frac{(n+1)^{n+1}}{(n+1)! b_1^2 b_2^2 \cdots b_{n+1}^2} \cdot \frac{n! b_1^2 b_2^2 \cdots b_n^2}{n^n} \). Simplifying yields \( \lim_{n \to \infty} \frac{(n+1)^{n+1} n! n^n}{(n+1)! n^n \cdot b_{n+1}^2} = \lim_{n \to \infty} \frac{(n+1)^n}{b_{n+1}^2} \cdot \frac{n+1}{n}\).

Step 5: Conclude Ratio Test for Series B

Further simplify to \( \lim_{n \to \infty} (1 + \frac{1}{n})^n \cdot \frac{1}{b_{n+1}^2} = e \cdot \frac{1}{\frac{1}{9}} = 9e. \) Since this limit is greater than 1, the series \( \sum_{n=1}^{\infty} \frac{n^{n}}{n ! b_{1}^{2} b_{2}^{2} \cdots b_{n}^{2}} \) diverges by the Ratio Test.

Key Concepts

Limit Comparison Test

The Limit Comparison Test is a powerful tool to determine whether a series converges or diverges by comparing it to another series whose behavior is known. It is particularly useful when dealing with series that include complex expressions, like the ones involving terms converging to a constant.Here's how it works:

• You find a series \sum a_n with known convergence properties.
• Look at the series \sum b_n you're interested in. Compute \( \lim_{n \to \infty} \frac{b_n}{a_n} \).
• If the limit is a positive finite number \( c \), then either both series converge or both diverge.

In our original problem, we used the limit comparison test with \( \sum \frac{1}{n 4^n} \), a convergent series. Since the sequence \( b_n \) converges to \( \frac{1}{3} \), for large \( n \), \( b_n \approx \frac{1}{3} \). We set \( a_n = \frac{1}{n 4^n} \), calculated the limit to be \( \frac{1}{9} \), a positive finite number. Hence, the test confirmed both series converge.

Ratio Test

The Ratio Test is a common method for determining the convergence or divergence of a series, especially when terms involve factorials or exponential elements. It uses the ratio of consecutive terms to assess the behavior of the series.To apply the Ratio Test:

• Consider the series \( \sum a_n \).
• Compute the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).
• If \( L < 1 \), the series converges. If \( L > 1 \) or if \( L = \infty \), the series diverges. If \( L = 1 \), the test is inconclusive.

For Series B in the original exercise, applying the Ratio Test involved examining terms of the form \( a_n = \frac{n^n}{n! b_1^2 b_2^2 \cdots b_n^2} \). After simplifying, it led to \( L = 9e \), which is greater than 1, indicating that the series diverges. The factorials and exponential growth dominated the terms, increasing the ratio and confirming divergence.

Series Convergence and Divergence

Understanding convergence and divergence is crucial for analyzing infinite series. Convergence means that the series approaches a certain finite number, while divergence implies that it doesn't settle to any specific limit as more terms are added.Key factors in determining convergence or divergence include:

• The behavior of terms as \( n \) becomes very large.
• The form of the series itself, whether it's geometric, p-series, or involves exponential elements.
• The results from tests like the Limit Comparison and Ratio Tests.

In our exercise, series convergence was decided by examining known series and applying the relevant tests effectively:- **For Series A**, we identified a similar geometric series, employed the Limit Comparison Test, and demonstrated convergence.- **For Series B**, the Ratio Test revealed divergence due to growing exponential factors and factorial terms.
Thus, these tests help in making meaningful decisions about the series in question by comparing or analyzing their growth and decay.