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?

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