Question

Stirling's formula Complete the following steps to find the values of \(p>0\) for which the series \(\sum_{k=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots(2 k-1)}{p^{k} k !}\) converges. a. Use the Ratio Test to show that \(\sum_{k=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots(2 k-1)}{p^{k} k !}\) converges for \(p>2\). b. Use Stirling's formula, \(k !=\sqrt{2 \pi k} k^{k} e^{-k}\) for large \(k,\) to determine whether the series converges when \(p=2\). (Hint: \(1 \cdot 3 \cdot 5 \cdots(2 k-1)=\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdots(2 k-1) 2 k}{2 \cdot 4 \cdot 6 \cdots 2 k}\) (See the Guided Project Stirling's formula and \(n\) ? for more on this topic.)

Short answer

Answer: The series converges for any value of \(p > 0\), where \(p \geq 2\).

Step-by-step solution

Ratio Test for Convergence

We will perform the Ratio Test, which requires computing the limit of the ratio between consecutive terms as \(k \to \infty\) and determining if this limit is less than 1. Let \(a_k=\frac{1 \cdot 3 \cdot 5 \cdots(2 k-1)}{p^{k} k!}\). We will find now require \(\lim_{k\to\infty}\frac{a_{k+1}}{a_k}\): \(\lim_{k\to\infty} \frac{\frac{1\cdot3\cdot5\cdots(2k+1)}{p^{k+1}(k+1)!}}{\frac{1\cdot3\cdot5\cdots(2k-1)}{p^k k!}}\) \(=\lim_{k\to\infty} \frac{p^k k!}{p^{k+1}(k+1)!} (2k + 1)\) \(=\lim_{k\to\infty} \frac{1}{p} \frac{k!}{(k+1)!}(2k + 1)\) \(=\lim_{k\to\infty} \frac{1}{p} \cdot \frac{1}{k+1}(2k + 1)\). For this limit to be less than 1, we need: \(\lim_{k\to\infty} \frac{1}{p} \cdot \frac{1}{k+1}(2k + 1) < 1\) \(\lim_{k\to\infty} \frac{2k+1}{p(k+1)} < 1\). As \(k \to \infty\), the limit above will be less than 1 if \(p > 2\). Therefore, the series converges for \(p > 2\).

Use Stirling's formula to determine convergence when \(p = 2\)

Using Stirling's formula, we can approximate the factorials and determine if the series converges when \(p = 2\). We are given a hint to rewrite the terms in the series as: \(1 \cdot 3 \cdot 5 \cdots(2 k-1)=\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdots(2 k-1) 2 k}{2 \cdot 4 \cdot 6 \cdots 2 k}\). For large values of \(k\), we can approximate the numerator Factorial using Stirling's Formula, \(k! \approx \sqrt{2 \pi k}(k^k e^{-k})\). We can approximate the denominator by noting \({(2k)!} \approx \sqrt{4\pi k}(2^k k^k e^{-2k})\). Now, let's look at the \(k\)-th term in the series with \(p=2\): \(a_k=\frac{(2k)!}{2^k k!} \cdot \frac{1}{2^k k!} = \frac{(2k)!}{(2^{2k} (k!)^2)}\) Using Stirling's formula, and assuming large \(k\), we get: \(a_k \approx \frac{\sqrt{4\pi k}(2^k k^k e^{-2k})}{(2^{2k})(\sqrt{2 \pi k} k^k e^{-k})(\sqrt{2 \pi k} k^k e^{-k})}\) Further simplifying, we can cancel terms: \(a_k \approx \frac{\sqrt{4\pi k}(2^k k^k e^{-2k})}{(2^{2k})(2\pi k) k^{2k} e^{-2k}}\) \(a_k \approx \frac{1}{(2^{2k-1])(2 \pi k) k^k}\). As \(k \to \infty\), the term \(\frac{1}{(2^{2k-1})(2 \pi k) k^k}\) approaches 0. Hence, the series converges when \(p = 2\) as well. In conclusion, the series \(\sum_{k=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots(2 k-1)}{p^{k} k !}\) converges for any value of \(p > 0\), where \(p \geq 2\).

Key concepts

Ratio Test

Understanding the Ratio Test is crucial when determining series convergence, especially for series involving factorials or powers. The Ratio Test involves looking at the ratio of successive terms in a series: if the limit of this ratio (as the term number goes to infinity) is less than one, the series converges.

Apply the Ratio Test by taking the limit of \( \frac{a_{k+1}}{a_k} \) as \( k \) grows larger and larger. If this limit is less than 1, the series converges; if it is greater than 1, it diverges. A result of exactly 1 is inconclusive with the Ratio Test needing another method to determine convergence.

In the problem, by applying the Ratio Test to the series \( \sum_{k=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots(2k-1)}{p^{k} k!} \), we find that the series converges for \( p > 2 \) since the limit of the ratio of consecutive terms is less than 1.

Series Convergence

A fundamental topic in calculus and analysis is the convergence of series. A series is simply the sum of the terms of a sequence, and 'convergence' refers to the idea that as you add more and more terms, the sum approaches a certain finite value.

An infinite series may or may not have a finite sum, depending on whether the individual terms eventually shrink towards zero as more terms are added and they do not grow too rapidly. In our exercise, the convergence of the series depends on the parameter \( p \). By exploring different values of \( p \) and applying tests like the Ratio Test, we can determine the conditions under which the sum will not shoot off to infinity but instead settle down to a finite number.

Factorial Approximation

Factorial approximation, particularly Stirling's formula, is a powerful tool for estimating the factorial of a large number, which plays a considerable role in mathematical applications such as probability and series convergence.

Stirling's formula, \( k! \approx \sqrt{2 \pi k}(k^k e^{-k}) \) for large \( k \), provides a way to simplify factorials in terms of more elementary functions. This approximation becomes very accurate as \( k \) becomes large. In the context of our exercise, Stirling's formula is used to rewrite the terms of the series to determine if the series converges when \( p = 2 \).

This approach allows us to evaluate the behavior of the series when traditional methods may be cumbersome or impossible to carry out exactly due to the complex nature of factorials.

Infinite Series

An infinite series is a sum of infinitely many terms. Not all infinite series converge; whether a series does so depends on the properties of its terms. Convergence can sometimes be counterintuitive: even if individual terms become tiny, their sum does not necessarily form a finite number.

In the exercise, we're dealing with an infinite series in which each term contains a product of odd numbers and powers of a variable \( p \) and factorials. The challenge with infinite series is determining if, when you keep adding more terms, the total sum approaches a limit (converges) or grows without bound (diverges). Tools like the Ratio Test help in this analysis. Even though the series in this problem has complex terms, we are able to reach a conclusion about convergence by harnessing such mathematical tests and approximations like Stirling's formula.