$$\newcommand{a}[0]{\alpha}
\newcommand{b}[0]{\beta}
\newcommand{beq}[0]{\begin{eqnarray*}}
\newcommand{C}[0]{\mathbb{C}}
\newcommand{ds}[0]{\displaystyle}
\newcommand{eeq}[0]{\end{eqnarray*}}
\newcommand{G}[1]{\Gamma({#1})}
\newcommand{g}[0]{\gamma}
\newcommand{hp}[0]{\frac{\pi}2}
\newcommand{limn}[0]{\lim_{n\to\infty}}
\newcommand{N}[0]{\mathbb{N}}
\newcommand{Q}[0]{\mathbb{Q}}
\newcommand{R}[0]{\mathbb{R}}
\newcommand{sumk}[0]{\sum_{k=1}^n}
\newcommand{sumn}[1]{\sum_{n={#1}}^\infty}
\newcommand{t}[0]{\theta}
\newcommand{tc}[0]{\TextCenter}
\newcommand{Z}[0]{\mathbb{Z}}
$$
${}$
最近少し話題の, 手書きの数式を$\LaTeX$に変換してくれるサイトさん (
https://webdemo.myscript.com/views/math/index.html
) とペンタブを使って, 手軽に数式が打てそうだと思ったので, お気に入りの級数のまとめを作ってみました.
${}$
$$\sum ^{\infty }_{n=0}\dfrac{n!^{2}}{\left( 2n\right) !}=\dfrac{4}{3}+\dfrac{2\pi }{9\sqrt{3}}$$
$$\sum ^{\infty }_{n=0}\dfrac{2^{n}n!^{2}}{\left( 2n\right) !}=2+\dfrac{\pi }{2}$$
$$\sum ^{\infty }_{n=0}\dfrac{\left( -1\right) ^{n}n!^{2}}{\left( 2n\right) !}=\dfrac{4}{5}\left( 1-\dfrac{\log \varphi }{\sqrt{5}}\right)$$
${}$
$$\sum ^{\infty }_{n=0}\dfrac{\left( 5n+3\right) n!\left( 2n\right) !}{2^{n}\left( 3n+2\right) !}=\dfrac{\pi }{2}$$
$$\sum ^{\infty }_{n=0}\dfrac{\left( 5n+4\right) n!\left( 2n\right) !}{2^{n}\left( 3n+2\right) !}=3\log 2$$
${}$
$$\sum ^{\infty }_{n=0}\dfrac{n!\left( 2n\right) !}{2^{n}\left( 3n\right) !}=\dfrac{3}{5^{3}}\left( \dfrac{11}{6}\pi -2\log 2+45\right)$$
${}$$$\sum ^{\infty }_{n=1}\dfrac{1}{2n-1}\left( \dfrac{\left( 2n\right) !}{2^{2n}n!^{2}}\right) =1$$
$$\sum ^{\infty }_{n=1}\dfrac{1}{2n-1}\left( \dfrac{\left( 2n\right) !}{2^{2n}n!^{2}}\right) ^{2}=1-\dfrac{2}{\pi }$$
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$$\sum ^{\infty }_{n=1}\dfrac{H_{n}H_{n+1}}{n\left( n+1\right) }=\zeta(2) +2\zeta(3)
$$
$$\sum ^{\infty }_{n=1}\dfrac{H_{n}H_{n+1}H_{n+2}}{n\left( n+1\right) \left( n+2\right) }=-\dfrac{1}{2}\zeta(2) +\dfrac{5}{4}\zeta(3) +\dfrac{5}{8}\zeta(4)$$
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$$\sum ^{\infty }_{n=1}\dfrac{\left( 2n\right) !}{2^{2n}n!^{2}}\dfrac{H_{n}}{n+1}=4\log 2$$
$$\sum ^{\infty }_{n=1}\dfrac{\left( 2n\right) !}{2^{2n}n!^{2}}\dfrac{H_{n}^{2}}{n+1}=\pi ^{2}+4\log ^{2}2$$
${}$