Evaluate $\prod \frac {2k - 1} {2k}$ 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?

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 $0.2460938695063\cdots$.
