find the limits : $\lim_{n}( \frac{1\cdot3\cdot5\cdot \cdot \cdot(2n-1)}{2\cdot4\cdot6 \cdot \cdot \cdot (2n)} ) =?$ Find the limits (without Stirling's approximation ) :
$$\lim_{n}( \frac{1\cdot3\cdot5\cdot \cdot \cdot(2n-1)}{2\cdot4\cdot6 \cdot \cdot \cdot (2n)} ) =?$$


My Try :
$$\frac{1\times 3\times 5\cdots \times(2n-1)}{2\times 4\times 6\cdots \times(2n)}=\frac{(2n)!}{4^n(n!)^2}$$
now what ?
This is
$$\prod_{n=1}^\infty\left(1-\frac1{2n}\right).$$
Its logarithm is
$$\sum_{n=1}^\infty\left(-\frac1{2n}+O(n^{-2})\right).$$
Does that sum converge/diverge/what?
maybe it will help you. the answer in the image is a partial result



Since the root of k is necessary for it to converge to a value, I must assume that the limit of the production tends to zero, and that its rate of growth is equal to that of the square root of k