Problem solving a series with convergence test: $\sum_{k=1}^{\infty}\frac{k!}{(2k)!}$ Good morning, I have a big problem solving this:
$\sum_{k=1}^{\infty}\frac{k!}{(2k)!}\:$
I'm trying solving this limit with test of D'Alembert, but I have a problem solving the limit.
$\lim_{k\rightarrow\infty}\frac{(k+1)!k!}{(2k+2)!2k!}=(?)$
please, help me.
 A: We have that
$$
\lim_{k\to\infty}\frac{(k+1)!(2k)!}{k!(2(k+1))!}=\lim_{k\to\infty}\frac{(k+1)(2k)!}{(2k+2)!}=\lim_{k\to\infty}\frac{k+1}{(2k+2)(2k+1)}=\lim_{k\to\infty}\frac1{2(2k+1)}=0.
$$
Hence, the series converges by the ratio test.
A: Note that 
$$
\frac{(k+1)!}{(2k+2)!}\cdot \frac{(2k)!}{k!}=\frac{k+1}{(2k+2)(2k+1)}=\frac{1}{2(2k+1)}\to 0 \text{ as }k \to\infty.
$$
Thus, by the ratio test, the series converges.
A: $$S=\sum_{k=1}^{+\infty}\frac{k!}{(2k)!}=\sum_{k\geq 1}\frac{1}{(2k)!}\int_{0}^{+\infty} x^k e^{-x}\,dx =\int_{0}^{+\infty}e^{-x}\sum_{k\geq 1}\frac{x^k}{(2k)!}\,dx\tag{1}$$
hence:
$$ S = 2\int_{0}^{+\infty} z e^{-z^2} \sum_{k\geq 1}\frac{z^{2k}}{(2k)!}\,dz = \int_{0}^{+\infty} 2z\,e^{-z^2}\left(\cosh(z)-1\right)\,dz \tag{2}$$
or:
$$ S = \int_{0}^{+\infty} e^{-z^2}\sinh(z)\,dz = \color{red}{\frac{\sqrt{\pi}}{2} e^{1/4}\,\text{erf}\left(\frac{1}{2}\right)}\approx 0.5922965.\tag{3}$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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 \newcommand{\half}{{1 \over 2}}
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 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,{\rm Li}_{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
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\begin{align}
\color{#f00}{\sum_{k = 1}^{\infty}{k! \over \pars{2k}!}} & =
\sum_{k = 1}^{\infty}{1 \over \pars{k - 1}!}\,
{\Gamma\pars{k + 1}\Gamma\pars{k} \over \Gamma\pars{2k + 1}} =
\sum_{k = 1}^{\infty}{1 \over \pars{k - 1}!}
\int_{0}^{1}t^{k}\pars{1 - t}^{k - 1}\,\dd t
\\[3mm] & =
\int_{0}^{1}t\sum_{k = 0}^{\infty}{\bracks{t\pars{1 - t}}^{k} \over k!}\,\dd t =
\int_{0}^{1}t\expo{t\pars{1 - t}}\,\dd t =
\int_{-1/2}^{1/2}\pars{t + \half}\expo{1/4 - t^{2}}\,\dd t
\\[3mm] & =
{\root{\pi} \over 2}\expo{1/4}\
\overbrace{\bracks{{2 \over \root{\pi}}\int_{0}^{1/2}\expo{-t^{2}}\,\dd t}}
^{\ds{\textrm{erf}\pars{1/2}}}\ =\
\color{#f00}{{\root{\pi} \over 2}\expo{1/4}\textrm{erf}\pars{\half}} \approx
0.5923
\end{align}
