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I came across to this: evaluate $$\prod_{k = 1}^{n} \frac {2k - 1} {2k}.$$

I brought it to a closed-form expression in asymptotics but I suspect my asymptotics is bad and I also want to see different solutions, hence the question.

My original solution: Write

$$\begin {eqnarray} \prod_{k = 1}^{n} \frac {2k - 1} {2k} & = & \exp \left (\sum_{k = 1}^{n} \log \left (1 - \frac {1} {2k}\right) \right) \nonumber \\ & = & \exp \left (- \sum_{k = 1}^{n} \sum_{m = 1}^{\infty} \frac {1} {(2k)^m m}\right). \end {eqnarray}$$

Let $$S = \sum_{m = 1}^{\infty} \frac {1} {(2k)^m m} \tag {1}.$$ Multiply both sides of $(1)$ by $2k$ to get $2k S < S + 1$, and by $4k$ to get $4kS > S + 2$. Hence, $$\frac {2} {4k - 1} < S < \frac {1} {2k - 1}.$$ Then, $$\frac {1} {2} \left( \log (4n - 1) - \log 3 \right) - 2 + O \left(\frac {1} {n^2}\right) < \sum_{k = 1}^{n} \sum_{m = 1}^{\infty} \frac {1} {(2k)^m m} < \frac {1} {2} \log (2n - 1) - 1 + O \left(\frac {1} {n^2}\right).$$ Hence, $$\sqrt {\frac {c} {2n - 1}} < \prod_{k = 1}^{n} \frac {2k - 1} {2k} < \sqrt {\frac {3 c^2} {4n - 1}},$$ where $c$ is a constant. Any better approach?

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    this product should be $${\frac { \left( n-1/2 \right) !}{n!,\sqrt {\pi }}}$$ – Dr. Sonnhard Graubner Dec 23 '15 at 15:28
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    Be neither angry nor offended: you can expand your question showing your attempt and where you're stuck or uncertain. Or just where you'd like to see improvements to your argument. By the way, I didn't downvote. – egreg Dec 23 '15 at 15:54
  • (-1) for the rant. If you expand your question to include the asymptotics you came up with, or to include any questions you have about the answers in the duplicated thread, I'm sure the question wouldn't have been downvoted at the start. –  Dec 23 '15 at 16:02
  • @user What? The answers in the "duplicated" thread are just plain answers, not solutions. I also know that you can express it in terms of digamma function, which is non-elementary, but I just wished to see which interesting ways the master user here will come. I am here to learn, so it's very natural that I ask. –  Dec 23 '15 at 16:22

2 Answers2

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$$\prod_{k=1}^{n}\frac{2k-1}{2k}=\frac{2-1}{2}\cdot\frac{4-1}{4}\cdot\frac{6-1}{6}\cdot\dots\cdot\frac{2n-1}{2n}=$$ $$\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\dots\cdot\frac{2n-1}{2n}=\frac{1\cdot3\cdot5\dots\cdot(2n-1)}{2\cdot4\cdot6\dots\cdot2n}=\frac{\frac{2^n\Gamma\left(n+\frac{1}{2}\right)}{\sqrt{\pi}}}{2^n\Gamma\left(n+1\right)}=\frac{\Gamma\left(n+\frac{1}{2}\right)}{n!\sqrt{\pi}}$$

Jan Eerland
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  • This I know, man! I want a naive and elementary solution, an outcome in terms of elementary functions. Actually, from your result, also, one can make use of Stirling's formula to arrive at an approximation in terms of (finitely many) elementary functions, but then again, one can't even bother expanding that. –  Dec 23 '15 at 17:24
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I tried a different and definitely a far more elementary approach and, surprisingly, it gave a better estimate. We have obviously (which I had overlooked!)

$$\prod_{k = 1}^{n} \frac {2k - 1} {2k} = \frac {(2n)!} {4^n (n!)^2}$$

which, by Stirling formula, is

$$=\frac {1} {\sqrt {\pi n}} \exp \left (\frac {1} {192 n^3} - \frac {1} {8n}\right).$$

This estimate is indeed very precise, since for the very small value $n = 5$ the actual value is $0.24609375$ and the estimated value is $0.2460938695063\cdots$.