Do either of the following series converge: $\sum_{n = 1}^\infty \frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots{2n}}$ Does $$\sum_{n = 1}^\infty \frac{1\cdot3\cdot5\cdots{2n-1}}{2\cdot4\cdot6\cdots{2n}}$$  converge and additionally does the following converge $$\sum_{n = 1}^\infty \frac{1^2\cdot3^2\cdot5^2\cdots{(2n-1)}^2}{2^2\cdot4^2\cdot6^2\cdots{(2n)}^2}$$
 A: Hint:
$$ \frac{1\cdot3\cdot5\dots{2n-1}}{2\cdot4\cdot6\dots{2n-2}}= \frac{3}{2}  \frac{5}{4}... \frac{2n-1}{2n-2} >1$$
A: $\newcommand{\+}{^{\dagger}}
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$\ds{\sum_{n = 1}^{\infty}
     {1\cdot 3\cdot 5 \cdots \pars{2n-1} \over 2\cdot4\cdot6\cdots\pars{2n}}:\
     {\large ?}\,,\qquad\sum_{n = 1}^{\infty}
     \bracks{1\cdot 3\cdot 5 \cdots \pars{2n-1}
     \over 2\cdot4\cdot6\cdots\pars{2n}}^{2}:\ {\large ?}}$

Note that
  $$
{1\cdot 3\cdot 5 \cdots \pars{2n-1} \over 2\cdot4\cdot6\cdots\pars{2n}}
={1\cdot 2\cdot 3\cdot 4\cdot 5 \cdots \pars{2n-1}\pars{2n}
  \over \bracks{2\cdot4\cdot6\cdots\pars{2n}}^{2}}
={1 \over 4}\,{\pars{2n}! \over n!^{2}} = {1 \over 4}\,{2n \choose n}
$$

When $\ds{n \gg 1}$:
\begin{align}
{1 \over 4}\,{2n \choose n}&\approx
{1 \over 4}\,{\root{2\pi}\pars{2n}^{2n + 1/2}\expo{-2n}
\over \bracks{\root{2\pi}n^{n + 1/2}\expo{-n}}^{2}}=
{2^{2n - 2} \over \root{\pi}}\,{1 \over n^{1/2}} \stackrel{n \to \infty}{\Large \to} +\infty
\\[3mm]
{1 \over 16}\,{2n \choose n}^{2}&\approx
{2^{4n - 4} \over \pi}\,{1 \over n} \stackrel{n \to \infty}{\Large \to} +\infty
\end{align}
