Question

In the following exercises, use either the ratio test or the root test as appropriate to determine whether the series \(\sum a_{k}\) with given terms \(a_{k}\) converges, or state if the test is inconclusive. $$ a_{k}=\frac{k !}{1-3 \cdot 5 \cdots(2 k-1)} $$

Short answer

The series converges by the ratio test.

Step-by-step solution

Identify the series term

Examine the given term of the series: \(a_k = \frac{k!}{1 \cdot 3 \cdot 5 \cdots (2k-1)}\). This term consists of a factorial in the numerator and a product of odd numbers in the denominator.

Simplify the denominator

The denominator is a product of all odd numbers up to \((2k-1)\). This can be expressed more compactly using a double factorial notation: \((2k-1)!!\). Hence, the term simplifies to \(a_k = \frac{k!}{(2k-1)!!}\).

Choose a Series Test

Considering the form of \(a_k\), the ratio test could be a suitable approach since factorials are involved. The ratio test examines the limit of the absolute value of the ratio of consecutive terms: \(L = \lim_{{k \to \infty}} \left| \frac{a_{k+1}}{a_k} \right|\).

Apply the Ratio Test

Calculate \(\frac{a_{k+1}}{a_k}\): \( a_{k+1} = \frac{(k+1)!}{(1 \cdot 3 \cdot 5 \cdots (2k+1))} \), \( a_k = \frac{k!}{(1 \cdot 3 \cdot 5 \cdots (2k-1))} \). Therefore, \[ \frac{a_{k+1}}{a_k} = \frac{(k+1)!}{(2k+1)(k!)} \times \frac{(1 \cdot 3 \cdot 5 \cdots (2k-1))}{k!} = \frac{k+1}{2k+1} \].

Evaluate the limit

Find the limit of the ratio as \(k\) approaches infinity:\[ L = \lim_{{k \to \infty}} \frac{k+1}{2k+1} = \lim_{{k \to \infty}} \frac{1 + \frac{1}{k}}{2 + \frac{1}{k}} = \frac{1}{2} \].

Interpret the result

Since \(L = \frac{1}{2} < 1\), the ratio test states that the series \(\sum a_k\) converges.

Key concepts

Ratio Test

The Ratio Test is a handy tool for determining if a series converges or diverges. It’s especially useful when factorials or exponential terms are present. The main idea behind the ratio test is to examine the limit of the absolute value of the ratio of consecutive terms of a series.

Here's how it works:

• Calculate the ratio of the \(k+1\) term to the \(k\) term of the series, \(\frac{a_{k+1}}{a_k}\).
• Determine the limit \(L\) as \(k o \infty\) of the absolute value of this ratio, \(|\lim_{{k \to \infty}} \frac{a_{k+1}}{a_k}|)\).

If \(L < 1\), the series converges. If \(L > 1\), the series diverges. If \(L = 1\), the test is inconclusive and another convergence test is needed. This test is powerful because it simplifies working with series that might otherwise be complex.

Series Convergence

Understanding whether a series converges is crucial in calculus. Convergence tells us if the series sums to a finite value. If a series converges, the sum of its infinite terms is limited to a certain number instead of being infinite.

Series convergence can be determined using:

• The Ratio Test, for series with factorial or exponential terms.
• The Root Test, for series with powers, especially roots.
• Other tests like the Integral Test, Direct Comparison Test, and Alternating Series Test, depending on the situation.

For example, calculating the limit of the ratios, \(L \), helps establish if a given series \(\sum a_k\) converges. It illustrates how adding terms infinitely can still sum to a finite value, reinforcing the concept of series in calculus.

Factorial Notation

Factorial notation, represented by an exclamation point (!), is used to denote the product of all positive integers up to a certain number. The factorial of a non-negative integer \(n\), written as \(n!\), is the product \((n \cdot (n-1) \cdot (n-2) \cdots 2 \cdot 1))\).

Factorials grow rapidly because each term multiplies by the next integer in line. A related concept is the double factorial, written as \((2k-1)!!\), which specifically applies to multiplying all odd numbers up to \(2k-1\). Factorials appear frequently in problems involving permutations, combinations, and in series like the one we examined.

Grasping factorial notation is essential as it simplifies and breaks down expressions, making complex series more manageable.