Problem 38 - Thomas Calculus

Question

In Exercises \(37-40,\) find the series' radius of convergence. $$ \sum_{n=1}^{\infty}\left(\frac{2 \cdot 4 \cdot 6 \cdot \cdot \cdot(2 n)}{2 \cdot 5 \cdot 8 \cdot \cdot(3 n-1)}\right)^{2} x^{n} $$

Step-by-Step Solution

Answer

The series' radius of convergence is \( \frac{9}{4} \).

Step 1: Understand the Series

The series given is \( \sum_{n=1}^{\infty} \left(\frac{2 \cdot 4 \cdot 6 \cdot \cdots \cdot(2n)}{2 \cdot 5 \cdot 8 \cdot \cdots \cdot(3n-1)}\right)^{2} x^{n} \). This series is in the form of \( \sum a_n x^n \), where \( a_n = \left(\frac{2 \cdot 4 \cdot 6 \cdot \cdots \cdot(2n)}{2 \cdot 5 \cdot 8 \cdot \cdots \cdot(3n-1)}\right)^2 \). We need to find the radius of convergence.

Step 2: Analyze Factorial-like Terms

The numerator \( 2 \cdot 4 \cdot 6 \cdot \cdots \cdot(2n) \) is equal to \( 2^n n! \), since it is the product of the first \( n \) even numbers. The denominator \( 2 \cdot 5 \cdot 8 \cdot \cdots \cdot(3n-1) \) doesn't follow a factorial pattern, but observe that it involves arithmetic terms with increments of 3. We need to extract the pattern for both numerator and denominator terms.

Step 3: Simplify the Terms

Focus on simplifying the expression \( a_n \). Write \( a_n = \left(\frac{2^n n!}{2 \cdot 5 \cdot 8 \cdot \cdots \cdot(3n-1)} \right)^2 \). We notice the denominator is a product of terms forming an arithmetic sequence \( a_k = 3k - 1 \) from \( k=1 \) to \( n \). The expression for each \( a_n \) simplifies to \( a_n \approx \left(\frac{2^n \cdot n!}{\prod_{k=1}^{n} (3k-1)} \right)^2 \).

Step 4: Use Ratio Test

To find the radius of convergence, apply the ratio test. Compute \( \lim_{n \to \infty} \left| \frac{a_{n+1}x^{n+1}}{a_n x^n} \right| = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} x \right| \). This focuses on the terms involving \( a_n \) and \( a_{n+1} \). This can be expressed as:\[ \frac{a_{n+1}}{a_n} = \left(\frac{(2(n+1))}{(3(n+1)-1)}\right)^2 \approx \left(\frac{2(n+1)}{(3n + 2)}\right)^2. \]

Step 5: Solve for Limit

Calculate the limit: \(\lim_{n \to \infty} \left(\frac{2(n+1)}{(3n + 2)}\right)^2 = \lim_{n \to \infty} \left(\frac{2}{3}\right)^2 = \left(\frac{2}{3}\right)^2 = \frac{4}{9}. \) This simplifies the ratio test's root analysis with terms involving \( x \).

Step 6: Determine Radius of Convergence

For the series to converge, this limit must be less than one:\[ \left| x \right| \cdot \frac{4}{9} < 1, \]\[ \left| x \right| < \frac{9}{4}. \]Hence, the radius of convergence \( R \) is \( \frac{9}{4} \).

Key Concepts

Ratio Test

The ratio test is a powerful tool for determining the convergence of an infinite series. It's particularly useful when the series involves terms with factorial-like patterns or exponential growth. To apply the ratio test, we examine the limit of the ratio of successive terms in the series.

For a series defined as \( \sum a_n x^n \), the test involves computing:\[\lim_{n \to \infty} \left| \frac{a_{n+1} x^{n+1}}{a_n x^n} \right| = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} x \right|\]Key steps include:

• Calculate the ratio \( \frac{a_{n+1}}{a_n} \) for the sequence elements.
• Simplify the expression.
• Take the limit as \( n \to \infty \).

After finding this limit, the series converges absolutely if this limit is less than one. This directly informs us about the radius of convergence of the series.

Factorial-like Terms

Understanding factorial-like terms is crucial when working with series expressions containing products of sequences, especially when they resemble patterns similar to factorials.

In the exercise, the expression for \( a_n \) includes terms such as \( 2 \cdot 4 \cdot 6 \cdot \cdots \cdot(2n) \), which can be rewritten in terms of factorials:

• The sequence of even numbers multiplied together equals \( 2^n n! \).

On the denominator side, we observe arithmetic sequences with an increment. This can be condensed into a pattern but doesn't fit the standard factorial mold of consecutive numbers.

When dealing with these terms, it helps to focus on extracting and simplifying patterns to match or approximate known functions like factorials.

Infinite Series

Infinite series are sums of infinite sequences that extend indefinitely. Understanding their behavior helps determine whether they converge (approach a finite limit) or diverge (grow without bound).

A key form is \( \sum a_n x^n \), where \( x \) is a variable and \( a_n \) are coefficients depending on \( n \).

In our task, each term in the series involves a complicated expression of \( a_n \), squared even and arithmetic terms, multiplied by powers of \( x \).

Analysis tools like convergence tests, particularly for radius of convergence, guide us to conclude the behavior of such infinite sums.

• If the series converges for \( |x| < R \), \( R \) is termed the radius of convergence.

This concept is vital to ascertain the domain of \( x \) values for which the series behaves nicely by converging.

Convergence Criteria

Convergence criteria let us determine the conditions under which a given series will converge.

These include techniques like:

• Comparison tests
• Alternating series test
• Integral test
• Root test

The ratio test, utilized in our exercise, serves as a convergence criterion that involves examining the limit of ratios of successive terms in the series.

When the limit evaluated is less than one, absolute convergence is ensured, pinpointing the range of \( x \) values where the series converges for certain, as mentioned in our radius of convergence formula: \[\left| x \right| < \frac{9}{4}.\]Understanding and systematically applying convergence criteria are key aspects of analyzing infinite series.