Problem 63 - Thomas Calculus

Question

Which of the series in Exercises \(57-64\) converge, and which diverge? Give reasons for your answers. $$ \sum_{n=1}^{\infty} \frac{1 \cdot 3 \cdot \cdots \cdot(2 n-1)}{4^{n} 2^{n} n !} $$

Step-by-Step Solution

Answer

The series converges by the Root Test.

Step 1: Identify the General Term

The general term of the given series is \( a_n = \frac{1 \cdot 3 \cdot \cdots \cdot(2n-1)}{4^n 2^n n!} \). The numerator, \(1 \cdot 3 \cdot \cdots \cdot (2n-1)\), represents the product of the first \(n\) odd numbers.

Step 2: Simplify the Numerator

The numerator \(1 \cdot 3 \cdot \cdots \cdot (2n-1)\) is the double factorial \((2n-1)!!\). The formula for the product of the first \(n\) odd numbers is \((2n-1)!! = \frac{(2n)!}{2^n n!}\).

Step 3: Substitute and Simplify

Substituting the double factorial expression into \(a_n\), we get:\[a_n = \frac{(2n)!}{2^n n! \cdot 4^n 2^n n!} = \frac{(2n)!}{(2^{2n})(n!)^2}.\]

Step 4: Analyze with the Ratio Test

To apply the Ratio Test, consider the ratio \( \frac{a_{n+1}}{a_n} \). Calculate it:\[\frac{a_{n+1}}{a_n} = \frac{((2(n+1))!)}{(2^{2(n+1)})(n+1)!^2} \times \frac{(2^n n!)^2}{(2n)!}\]This simplifies to:\[\frac{(2n+2)(2n+1)}{4(n+1)^2} \to \frac{4n^2 + 6n + 2}{4n^2 + 8n + 4}.\]

Step 5: Find the Limit of the Ratio

Simplify the fraction by dividing by \(n^2\):\[\lim_{n \to \infty} \frac{4n^2 + 6n + 2}{4n^2 + 8n + 4} = \frac{4}{4} = 1.\]

Step 6: Conclude Using the Ratio Test

The Ratio Test is inconclusive when the limit is 1. Therefore, the Ratio Test does not tell us if the series converges or diverges.

Step 7: Stirling's Approximation

When the Ratio Test is inconclusive, use Stirling's approximation for \(n!\): \( n! \sim \sqrt{2\pi n} \left(\frac{n}{e}\right)^n\). Apply this to \((2n)!\) and \((n!)^2\) in \(a_n\): \[(2n)! \sim \sqrt{4\pi n} \left(\frac{2n}{e}\right)^{2n}, \quad (n!)^2 \sim (2\pi n) \left(\frac{n}{e}\right)^{2n}.\]Substituting these approximations results in:\[a_n \sim \frac{1}{\sqrt{\pi n}} \times \frac{1}{4^n} \to 0 \text{ as } n \to \infty.\]

Step 8: Apply the Root Test

Consider the root test by finding the nth root of the nth term:\[\lim_{n \to \infty} \sqrt[n]{|a_n|} = \lim_{n \to \infty} \frac{1}{\sqrt[n]{\pi n} 4} \to \frac{1}{4} < 1.\]Since this limit is less than 1, the series converges by the Root Test.

Key Concepts

Ratio Test

The Ratio Test is a popular method in determining the convergence or divergence of a series. To apply it, you consider the ratio of successive terms in the series, specifically \[ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|. \]

• If \(L < 1\), the series converges absolutely.
• If \(L > 1\), the series diverges.
• If \(L = 1\), the test is inconclusive.

In the given problem, applying the Ratio Test resulted in \( L = 1 \). This means that the Ratio Test could not decisively tell us if the series converges or diverges. When the test is inconclusive, other methods, such as the Root Test or using approximations, can be useful.

Root Test

The Root Test, also known as Cauchy's Root Test, helps determine the behavior of a series. For the series with terms \(a_n\), you calculate \[ \lim_{n \to \infty} \sqrt[n]{|a_n|}. \]

• If the limit is less than 1, the series converges absolutely.
• If the limit is greater than 1, it diverges.
• If the limit equals 1, the test is again inconclusive.

For the given series, substituting simplified terms and using the Root Test led to a limit of \(\frac{1}{4} < 1\). Hence, the Root Test confirms that this series converges, providing a conclusive result where the Ratio Test could not.

Double Factorial

The concept of a double factorial is essential for the analysis of some series. It involves multiplying every second integer up to a specified number. For \((2n-1)!!\), you multiply all odd numbers up to \(2n-1\), noted as \[(2n-1)!! = 1 \cdot 3 \cdot 5 \cdots (2n-1).\]Alternatively, the double factorial can be expressed using regular factorials as \[(2n-1)!! = \frac{(2n)!}{2^n \cdot n!}.\]In the exercise, recognizing and rewriting the double factorial using factorials allowed the solution to effectively utilize tools like Stirling’s Approximation.

Stirling's Approximation

Stirling's Approximation offers an efficient way to estimate factorials for large numbers. It states that \[ n! \approx \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n. \]This approximation becomes especially useful in complex series and limits analysis, where direct computation of large factorials is impractical. For the given series problem, Stirling's Approximation aided in simplifying \((2n)!\) and \((n!)^2\), helping to demonstrate that the series terms tend towards zero.By confirming the behavior of these terms, it further informed the decision that the series converges, supporting the conclusions drawn from the Root Test.