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Find $a,r>0$ such that

$\lim_{n\to \infty} n^r \cdot \frac12 \cdot \frac34 \cdots \frac{2n-1}{2n}=a$

I don't have any idea to solve it. How can I solve it?

  • 2
    BTW: 1. please avoid `$` in titles. 2. When writing a title, don't make it entirely in \LaTeX$. Add some words!2011-09-26

3 Answers 3

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Hints: (1) Write the product of rational numbers as a single rational number, using only powers of $2$ and factorials. (2) Use Stirling's formula to compute simple equivalents of the numerator and the denominator. The ratio of these should be your $an^{-r}$.

(To help you check your computations, I mention that $r=\frac12$.)

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I'll start out from a celebre limit, namely Wallis product that states that:

$ \frac{\pi}{2} = \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \frac{8}{7} \cdot \frac{8}{9} \cdot \cdot \cdot $

Without loss of generality, we consider an even factors number of the limit excepting ${n^r}$, and then by applying Wallis product we get that:

$ \lim_{n\to \infty}\frac{n^r}{\sqrt{2n+1}} \frac{\sqrt{2}}{\sqrt{\pi}}$ that obviously gives us $L =\frac{1}{\sqrt{\pi}}$ for $r=\frac{1}{2}$

The proof is complete.

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Here are the details of @did's answer. Write $ \frac12 \cdot \frac34 \cdots \frac{2n-1}{2n} = \frac{(2n)!}{(2^n n!)^2}=\frac{1}{4^{n}}{2n \choose n} $ We have the following asymptotics for the central binomial coefficient: $ {2n \choose n} \sim \frac{4^n}{\sqrt{\pi n}}\text{ as }n\rightarrow\infty $ Hence $ \frac12 \cdot \frac34 \cdots \frac{2n-1}{2n} \sim \frac{1}{\sqrt{\pi n}} $ and so $ n^{1/2} \frac12 \cdot \frac34 \cdots \frac{2n-1}{2n} \sim \frac{1}{\sqrt{\pi}} $