Find the sum of the infinite series: ∑n=1∞1⋅3⋅5⋯(2n−1)22nn!\sum_{n=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2^{2n} n!}∑n=1∞22nn!1⋅3⋅5⋯(2n−1)
2−22 - \sqrt{2}2−2
2\sqrt{2}2
12−1\frac{1}{\sqrt{2}} - 121−1
2−1\sqrt{2} - 12−1