# Finding $a$ and $r$ such that $\lim\limits_{n\to \infty} n^r \cdot \frac12 \cdot \frac34 \cdots \frac{2n-1}{2n}=a$

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?

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BTW: 1. please avoid $$ in titles. 2. When writing a title, don't make it entirely in \LaTeX. Add some words! – Guess who it is. Sep 26 '11 at 2:36 ## 3 Answers Hints: (1) Write the product of rational numbers as a single rational number, using only powers of 2 and factorials. (2) Use Stirling's formula to compute simple equivalents of the numerator and the denominator. The ratio of these should be your an^{-r}. (To help you check your computations, I mention that r=\frac12.) - Here are the details of @did's answer. Write$$ \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}} $$- I'll start out from a celebre limit, namely Wallis product that states that:$$ \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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