Chapter 4: Problem 49
In the following exercises, use an appropriate test to determine whether the series converges. $$ a_{k}=2^{k} /\left(\begin{array}{l} 3 k \\ k \end{array}\right) $$
Short Answer
Expert verified
The series converges, as shown by the ratio test with limit \( L = \frac{4}{27} < 1 \).
Step by step solution
01
Understand the Problem
We need to determine whether the series defined by the term \( a_k = \frac{2^k}{\binom{3k}{k}} \) converges. This involves considering the behavior of the series as \( k \to \infty \).
02
Simplify the Binomial Coefficient
The term can be rewritten using the binomial coefficient formula: \( \binom{3k}{k} = \frac{(3k)!}{k! (2k)!} \). So: \[ a_k = \frac{2^k}{\frac{(3k)!}{k! (2k)!}} = \frac{2^k \cdot k!(2k)!}{(3k)!} \]
03
Use Stirling's Approximation
For large \( n \), Stirling's approximation \( n! \approx \sqrt{2\pi n} \left(\frac{n}{e}\right)^n \) simplifies factorials. Applying this, \( (3k)! = \sqrt{2\pi 3k} \left( \frac{3k}{e} \right)^{3k} \) and similarly for \( k! \) and \( (2k)! \).
04
Estimate the Factorial
Using Stirling's approximation for each factorial, substitute back into the formula to estimate \( a_k \). The factorial terms grow rapidly, and substituting leads to: \[ a_k \approx \frac{2^k \cdot k^{k+1/2} \cdot (2k)^{2k+1/2}}{(3k)^{3k+1/2}} \]
05
Analyze the Resulting Expression
Simplify the expression: the ratio effectively concerns the exponential terms. Notably, \( \left(\frac{2k}{3k}\right)^{3k} \) dominates for large \( k \), showing \( a_k \sim \left( \frac{4}{3^3} \right)^k = \left( \frac{4}{27} \right)^k \), which tends to zero.
06
Apply the Ratio Test
The ratio test for convergence uses \( \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| \). Compute the limit, finding \( L = \frac{4}{27} < 1 \), indicating convergence of the series.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Stirling's Approximation
Stirling's approximation is a powerful tool used to estimate the factorial of large numbers. It's particularly helpful when dealing with complex expressions involving factorial terms. The formula is: Stirling's approximation: \[ n! \approx \sqrt{2\pi n} \left(\frac{n}{e}\right)^n \] This means that for a large integer \( n \), the factorial \( n! \) can be closely approximated using this expression. Stirling's approximation simplifies the computation since factorials grow rapidly and can be cumbersome to compute directly for large \( n \). In the context of our exercise, by substituting the factorials \( (3k)! \), \( k! \), and \( (2k)! \) with Stirling's approximation, we greatly simplify the term \( a_k \) of the series, leading us to an easier analysis and ultimately towards determining convergence.
Ratio Test
The Ratio Test is a method for determining whether an infinite series converges. If you have a series with terms \( a_k \), the Ratio Test checks the limit of the ratio of successive terms. This test is especially useful when terms involve factorials or exponential functions. _Here's how it works:_
- Calculate \( \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| \), denoted as \( L \).
- If \( L < 1 \), the series converges absolutely.
- If \( L > 1 \), the series diverges.
- If \( L = 1 \), the test is inconclusive.
Binomial Coefficients
Binomial coefficients are fundamental in combinatorics and appear in various mathematical contexts, such as series and expansions. Expressed as \( \binom{n}{k} \), they represent the number of ways to choose \( k \) elements from \( n \) elements without caring about the order. The formula is:\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \] In our exercise, we used this formula to express the binomial coefficient \( \binom{3k}{k} \). We expanded it as \( \frac{(3k)!}{k!(2k)!} \), which allowed us to integrate Stirling’s approximation to simplify \( a_k \), giving us a clearer path to apply the Ratio Test. Understanding how to manipulate binomial coefficients is crucial in simplifying series expressions and determining convergence efficiently. By breaking down these expressions, we can further explore the behavior of complex series.