Problem 22 - Calculus: One and Several Variables

Question

Determine whether the series converges or diverges. $$\sum \frac{(k !)^{2}}{(2 k) !}$$

Step-by-Step Solution

Answer

The series \(\sum \frac{(k !)^{2}}{(2 k) !}\) converges, based on the ratio test, as the limit of the ratio of consecutive terms is \(\frac{1}{4}\), which is less than 1.

Step 1: Consider the Ratio Test

In this case, we will consider the ratio test, which states that if the limit as n goes to infinity of the absolute value of the ratio of consecutive terms of the series is L, then the series converges if L < 1, diverges if L > 1, and is inconclusive if L = 1. We will first compute the ratio of consecutive terms and then find the limit of this ratio as n goes to infinity: \[\lim_{k \to \infty} \frac{\frac{((k+1)!)^2}{(2(k+1))!}}{\frac{(k!)^2}{(2k)!}}\]

Step 2: Simplify the Ratio

Next, we will simplify this ratio, which will help us compute the limit. \[\lim_{k \to \infty} \frac{\frac{((k+1)!)^2}{(2(k+1))!}}{\frac{(k!)^2}{(2k)!}} = \lim_{k \to \infty} \frac{((k+1)!)^2 (2k)!}{(k!)^2 (2(k+1))!}\] Now, we will simplify the factorial terms: \[\lim_{k \to \infty} \frac{((k+1)!)^2 (2k)!}{(k!)^2 (2(k+1))!} = \lim_{k \to \infty} \frac{((k+1)!)^2 (2k)!}{(k!)^2 (2k+2)(2k+1)(2k)!}\] Now cancel terms to get: \[\lim_{k \to \infty} \frac{((k+1)!)^2}{(k!)^2 (2k+2)(2k+1)}\]

Step 3: Compute the Limit

Compute the limit by factoring and dividing terms: \[\lim_{k \to \infty} \frac{((k+1)(k!))^2}{(k!)^2 (2k+2)(2k+1)} = \lim_{k \to \infty} \frac{(k+1)^2}{(2k+2)(2k+1)}\] Now, divide the terms by the highest power of k: \[\lim_{k \to \infty} \frac{\frac{(k+1)^2}{k^2}}{\frac{(2k+2)(2k+1)}{k^2}} = \lim_{k \to \infty} \frac{(1+\frac{1}{k})^2}{(2+\frac{2}{k})(2+\frac{1}{k})}\]

Step 4: Determine Convergence or Divergence

By calculating the limit, we can now determine whether the series converges or diverges: \[\lim_{k \to \infty} \frac{(1+\frac{1}{k})^2}{(2+\frac{2}{k})(2+\frac{1}{k})} = \frac{1^2}{(2)(2)} = \frac{1}{4}\] Since the limit is less than 1, the ratio test tells us that the series converges. Therefore, the given series \(\sum \frac{(k !)^{2}}{(2 k) !}\) converges.

Key Concepts

Ratio Test

The Ratio Test is a powerful tool to check if an infinite series converges or diverges. It involves the calculation of a limit based on the ratios of successive terms in a series. Here's how it works:

• Start by finding the ratio of the (n+1)th term to the nth term of the series.
• Take the absolute value of that ratio.
• Calculate the limit of this ratio as n approaches infinity.

The key part is the result of this limit, denoted by \( L \). If \( L \) is less than 1, the series converges, meaning it sums to a finite value. If \( L \) is greater than 1, the series diverges, implying the series sums to infinity.
If \( L \) equals 1, the test is inconclusive, and other methods might be needed.
For example, in the series \( \sum \frac{((k !)^{2}}{(2 k) !} \), the ratio of successive terms is simplified, and the limit is computed to be \( \frac{1}{4} \), which is less than 1. Thus, the series converges.

Factorial Notation

Factorial notation, denoted by an exclamation mark (!), is a function that multiplies a series of descending natural numbers. For any positive integer \( n \), \( n! \) is defined as:

• \( n! = n \times (n-1) \times (n-2) \times ... \times 2 \times 1 \)

Factorial notation grows very quickly with increasing \( n \). For example, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \). The factorial of zero is defined as 1: \( 0! = 1 \). This definition is useful in various mathematical concepts, including permutations, combinations, and especially in series as shown above.
In our exercise, factorials appear in both the numerator and the denominator, and simplifying them helps in calculating the limit via the Ratio Test.

Limit of a Sequence

The limit of a sequence is the value that the terms of the sequence approach as the term number becomes very large. Calculating limits often involves simplification and sometimes breaking down complex terms. For sequences, if the limit exists, the sequence is said to converge to that limit value.
When applying the Ratio Test to find convergence, computing the limit of simplified terms is crucial. Factorials can make these calculations challenging, but simplifying terms by canceling common factorials can help. For example, when terms like \( (k + 1)(k)\) are involved, it's handy to divide by the highest power present to find a clearly defined limit. Simplifying givens a consistent path to the limit as \( k \to \infty \).
In the exercise, simplifying the sequence and computing its limit reveals \( \lim_{k \to \infty} \frac{(1+\frac{1}{k})^2}{(2+\frac{2}{k})(2+\frac{1}{k})} = \frac{1}{4} \), showing how the terms behave at infinity, confirming convergence.

Infinite Series

An infinite series is the sum of an infinite sequence of terms. It can be represented as \( \sum_{n=1}^{\infty} a_n \), where \( a_n \) is the nth term. The nature of these series depends on the behavior of their sums as more and more terms are added. A series converges if the sum of its terms approaches a specific value as more terms are added.
Several tests exist to determine if an infinite series converges or diverges, including the Ratio Test used in this exercise. Infinite series are foundational in calculus and analysis, appearing in contexts such as power series and Fourier series, with applications extending to physics and engineering.
In this series \( \sum \frac{(k!)^2}{(2k)!} \), the application of the Ratio Test and the resulting limit \( \frac{1}{4} \) assures us that adding more terms brings the series closer to a finite value, hence proving its convergence.