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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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3 Answers 3

up vote 3 down vote accepted

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. –  5space 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. –  5space Apr 5 '13 at 11:13
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