Ratio Test Series - How to solve: $\sum\limits_{n=1}^\infty \frac{1\cdot3 \dots (2n-1)}{4^n 2^n n!}$? The problem is:
$$\sum_{n=1}^\infty  \frac{1\cdot3 \cdot ... \cdot (2n-1)}{4^n 2^n n!}$$
Could some help me how to solve it?
 A: Let $$p=\lim_{n\to \infty}\left\lvert\frac{a_{n+1}}{a_n}\right\rvert$$ which is the Ratio Test.
$$a_{n+1}=\frac{1*3*5*....*(2n-1)(2n+1)}{2^{n+1}4^{n+1}(n+1)!}$$
Note that $(2n+1)$ is obtained from substituting the $n$ in $(2n-1)$ with $(n+1)$
$$a_n=\frac{1*3*5*....*(2n-1)}{2^{n}4^{n}(n)!}$$
We will then substitute them inside the Ratio Test formula:
$$p=\lim_{n\to \infty}\left\lvert\frac{\frac{1*3*5*....*(2n-1)(2n+1)}{2^{n+1}4^{n+1}(n+1)!}}{\frac{1*3*5*....*(2n-1)}{2^{n}4^{n}(n)!}}\right\rvert$$
We can then see that the fraction can be simplified into:
$$\frac{1*3*5*....*(2n-1)(2n+1)}{2^{n+1}4^{n+1}(n+1)!}*\frac{2^{n}4^{n}(n)!}{1*3*5*....*(2n-1)}$$
Cancelling terms will give us:
$$\frac{2n+1}{8(n+1)}$$
Essentially:
$$p=\lim_{n\to \infty}\left\lvert\frac{2n+1}{8(n+1)}\right\rvert$$
Could you solve this limit? Remember that if $p<1$, the series converges absolutely.
A: If the problem is to determine convergence or divergence of the given series, observe: 
$$1\cdot 3 \cdot 5 \cdots (2n-1) < 2\cdot 4 \cdot 6 \cdots 2n = 2^n\cdot n!.$$
Thus the $n$th term in our series, which is positive, is less than $1/4^n.$ Of course $\sum_n 1/4^n$ converges, being a geometric series. Therefore our series converges by the comparison text. 
A: Hint. Using the generalized binomial theorem, one has
$$\begin{eqnarray*}
\sum_{n=1}^\infty{2n\choose n}u^{n}
&=& \frac{1}{\sqrt{1-4u}} -1, \qquad |u|<1/4.
\end{eqnarray*}$$
A: The Ratio Test is not the only method that works for this series.We have $2^{2n}=(1+1)^{2n}=\sum_{j=0}^{j=2n}\binom {2n}{j}\geq$ $ \binom {2n}{n}.$ Therefore $$\frac {1\cdot 3\cdot ...\cdot (2n-1)}{4^n2^n n!}=\frac {(2n)!}{(2^nn!)(4^n2^nn!)}=\binom {2n}{n}2^{-4n}\leq 2^{2n}2^{-4n}=2^{-2n}.$$ For the Ratio Test, the term in $(n+1)$ divided by the term in $n$ is  $(2n+1)/8(n+1)<(2n+2)/8(n+1)=1/8.$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\sum_{n = 1}^{\infty}{1 \times 3 \cdots \pars{2n - 1} \over 4^{n}\,2^{n}\,n!} & =
\sum_{n = 1}^{\infty}{\pars{2n}!/\pars{2^{n}\,n!} \over 2^{3n}\,n!} =
\sum_{n = 1}^{\infty}{\Gamma\pars{2\bracks{n + 1/2}} \over 2^{4n}\pars{n!}^{2}}
\\[5mm] & =
\sum_{n = 1}^{\infty}{\pars{2\pi}^{-1/2}\,2^{2n + 1/2}\,\Gamma\pars{n + 1/2}
\Gamma\pars{n + 1} \over 2^{4n}\pars{n!}^{2}} =
{1 \over \root{\pi}}\sum_{n = 1}^{\infty}{\Gamma\pars{n + 1/2} \over 2^{2n}\,n!}
\\[5mm] & =
{1 \over \root{\pi}}\,\pars{-\,{1 \over 2}}!\sum_{n = 1}^{\infty}{\pars{n - 1/2}! \over n!\pars{-1/2}!}\,\pars{1 \over 4}^{n}
\\[5mm] & =
{1 \over \root{\pi}}\,\Gamma\pars{1 \over 2}\sum_{n = 1}^{\infty}{n - 1/2 \choose n}\pars{1 \over 4}^{n} =
\sum_{n = 1}^{\infty}\bracks{{-1/2 \choose n}\pars{-1}^{n}}\pars{1 \over 4}^{n}
\\[5mm] & = 
\sum_{n = 1}^{\infty}{-1/2 \choose n}\pars{-\,{1 \over 4}}^{n} =
\bracks{1+\pars{-\,{1 \over 4}}}^{-1/2} - 1 = \bbx{\ds{2\root{3} \over 3} - 1}
\end{align}
