How to do this approximation

Question

I am asked to prove that when $k$ is sufficiently large $$ \frac 1 {2k}\frac{1}{2(k-1)} \prod_{n=1}^{k-2} \left(1+\frac 1 {2n}\right)\approx \frac 1 {2\sqrt{\pi}k^{3/2}} $$

Anyone got some ideas?

Answer 1

If you write out your product it is $$\frac {3\cdot 5 \cdot 7 \ldots \cdot (2k-1)}{2\cdot 4 \cdot 6 \ldots \cdot (2k-2)} =\frac {(2k-1)!}{(2^{k-1}(k-1)!)^2}$$ Now use Stirling's approximation and you should get there