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What is the following product in terms of $n$?

$$\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdot \frac{7}{8} \cdot \cdots \cdot \frac{2n-1}{2n}$$

Thank you.

Ivo Terek
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Sahana
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3 Answers3

7

$$\frac{1}{2}\frac{3}{4}\frac{5}{6} \dotsb \frac{2n-1}{2n}=\frac{1\mathbf{2}3\mathbf{4}5 \dotsb (2n-1)\mathbf{2n}}{(2 \cdot 4 \cdot 6 \dotsb 2n)^2}=\frac{(2n)!}{2^{2n}(n!)^2}$$

Anurag A
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As an alternative answer that has its uses, this product of fractions can be written as a definite integral:

$$\begin{align} \frac12\frac34\frac56\frac78\dots\frac{2n-1}{2n}&=\frac{(2n-1)!!}{(2n)!!}\\ &=\frac{2}{\pi}\int_{0}^{\pi/2}\sin^{2n}{x}\,\mathrm{d}x \end{align}$$

David H
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    This solution shows that $ \displaystyle \lim_{n \to \infty} \prod_{k = 1}^{n} \frac{2 k - 1}{2 k} = 0 $. As $ {\sin^{2 n}}(x) $ converges to $ 0 $ for all $ x \in \left[ 0,\dfrac{\pi}{2} \right) $, and as $ |\sin^{2 n}| $ is dominated by $ 1 $ for all $ n \in \mathbb{N} $, it follows from the Lebesgue Dominated Convergence Theorem that \begin{align} \lim_{n \to \infty} \prod_{k = 1}^{n} \frac{2 k - 1}{2 k} & = \lim_{n \to \infty} \frac{2}{\pi} \int_{0}^{\pi / 2} {\sin^{2 n}}(x) ~ \mathrm{d}{x} \ & = \frac{2}{\pi} \int_{0}^{\pi / 2} 0 ~ \mathrm{d}{x} \ & = 0. \end{align} – Berrick Caleb Fillmore Jun 21 '14 at 20:14
1

Given

$$ (1/2)(3/4)(5/6)(7/8)\cdots([2n-1]/2n) $$

Thus

$$ \frac{1}{2} \frac{3}{4} \frac{5}{6} \cdots \frac{2n-1}{2n} = \frac{1 \cdot 2 \cdot 3 \cdot 4 \cdots [2n-1] \cdot 2n}{2^{2n} \cdot (1 \cdot 2 \cdot 3 \cdot 4 \cdots 2n)^2 } = \frac{[2n]!}{2^n (n!)^2} $$

johannesvalks
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