# Calculating limit involving factorials.

I want to show that $\lim\limits_{k\to\infty} \frac{\pi^kk!}{(2k+1)!} = 0$. I've been trying to use the squeeze theorem, but am having a hard time finding some expression $P$ involving $k$ that is greater and $P\to 0$ as $k\to \infty$.

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You can use Stirling's formula if you know it, but there is a simpler line of reasonning here.

Notice that $$\frac{\pi^kk!}{(2k+1)!} = \frac{\pi^k}{(k+1)(k+2)\cdots(2k+1)} = \frac{\pi}{k+1} \frac{\pi}{k+2}\dots \frac{\pi}{2k} \frac{1}{2k+1}$$

For $k\geq 2\pi$, each term is bounded by $\dfrac{1}{2}$, yielding $$0 \leq \frac{\pi^k k!}{(2k+1)!} \leq \frac{1}{2^{k+1}}\qquad\dots$$

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That is very nice. Thank you for your help. – user59083 Apr 5 '13 at 8:46

Hint: \begin{align} \frac{\pi^kk!}{(2k+1)!} &=\frac{\pi^k}{\underbrace{(k+1)(k+2)(k+3)\cdots(2k+1)}_{k+1\text{ terms}}}\\ &\le\frac{\pi^k}{(k+1)^{k+1}}\\ &=\frac1\pi\left(\frac\pi{k+1}\right)^{k+1} \end{align}

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Use the fact that

$$k! < e \sqrt{k} \left ( \frac{k}{e}\right )^k$$

$$(2 k+1)! > \sqrt{2 \pi (2 k+1)} \left ( \frac{2 k+1}{e}\right )^{2 k+1}$$

Then

$$\frac{\pi^k k!}{(2 k+1)!} < \frac{e^2}{\sqrt{2 \pi}} \sqrt{\frac{k}{(2 k+1)^3}} \left (\frac{\pi e k}{(2 k+1)^2} \right )^k$$

Show that the RHS goes to zero as $k \rightarrow \infty$.

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Where did you get the first two inequalities from? I have never seen those before. – user59083 Apr 5 '13 at 11:13
The text passage contains a mathematical question about calculating a limit involving factorials, and multiple answers or hints to solve the problem are provided.
single_question_with_answer
Given a text passage, determine:
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    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
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- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# Calculating limit involving factorials.

I want to show that $\lim\limits_{k\to\infty} \frac{\pi^kk!}{(2k+1)!} = 0$. I've been trying to use the squeeze theorem, but am having a hard time finding some expression $P$ involving $k$ that is greater and $P\to 0$ as $k\to \infty$.

-

You can use Stirling's formula if you know it, but there is a simpler line of reasonning here.

Notice that $$\frac{\pi^kk!}{(2k+1)!} = \frac{\pi^k}{(k+1)(k+2)\cdots(2k+1)} = \frac{\pi}{k+1} \frac{\pi}{k+2}\dots \frac{\pi}{2k} \frac{1}{2k+1}$$

For $k\geq 2\pi$, each term is bounded by $\dfrac{1}{2}$, yielding $$0 \leq \frac{\pi^k k!}{(2k+1)!} \leq \frac{1}{2^{k+1}}\qquad\dots$$

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That is very nice. Thank you for your help. – user59083 Apr 5 '13 at 8:46

Hint: \begin{align} \frac{\pi^kk!}{(2k+1)!} &=\frac{\pi^k}{\underbrace{(k+1)(k+2)(k+3)\cdots(2k+1)}_{k+1\text{ terms}}}\\ &\le\frac{\pi^k}{(k+1)^{k+1}}\\ &=\frac1\pi\left(\frac\pi{k+1}\right)^{k+1} \end{align}

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Use the fact that

$$k! < e \sqrt{k} \left ( \frac{k}{e}\right )^k$$

$$(2 k+1)! > \sqrt{2 \pi (2 k+1)} \left ( \frac{2 k+1}{e}\right )^{2 k+1}$$

Then

$$\frac{\pi^k k!}{(2 k+1)!} < \frac{e^2}{\sqrt{2 \pi}} \sqrt{\frac{k}{(2 k+1)^3}} \left (\frac{\pi e k}{(2 k+1)^2} \right )^k$$

Show that the RHS goes to zero as $k \rightarrow \infty$.

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Where did you get the first two inequalities from? I have never seen those before. – user59083 Apr 5 '13 at 11:13
user
The text passage contains a mathematical question about calculating a limit involving factorials, and multiple answers or hints to solve the problem are provided.

TYPE: SINGLE_QUESTION_WITH_ANSWER