中央2項係数を含む通常型母関数

中央2項係数を含む通常型母関数

(1)通常型母関数の部分和

\[ \sum_{k=0}^{n}\frac{1}{k+1}C\left(2k,k\right)z^{k}=\frac{1}{2z}\left\{ 1-\left(1-4z\right)^{\frac{1}{2}}\right\} -\frac{C\left(2\left(n+1\right),n+1\right)}{n+2}z^{n+1}F\left(1,n+\frac{3}{2};n+3;4z\right) \]

(2)通常型母関数

\[ \sum_{k=0}^{\infty}\frac{1}{k+1}C\left(2k,k\right)z^{k}=\frac{1}{2z}\left\{ 1-\left(1-4z\right)^{\frac{1}{2}}\right\} \]

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\(F\)は超幾何関数。

(1)

\begin{align*} \sum_{k=0}^{n}\frac{1}{k+1}C\left(2k,k\right)z^{k} & =\sum_{k=0}^{n}\frac{\left(2k\right)!}{\left(k+1\right)!k!}z^{k}\\ & =\sum_{k=0}^{n}\frac{\left(2k\right)!!\left(2k-1\right)!!}{\left(k+1\right)!k!}z^{k}\\ & =\sum_{k=0}^{n}\frac{2^{k}k!2^{k}\left(k-\frac{1}{2}\right)!}{\left(k+1\right)!k!\Gamma\left(\frac{1}{2}\right)}z^{k}\\ & =\sum_{k=0}^{n}\frac{4^{k}Q\left(\frac{1}{2},k\right)}{\left(k+1\right)!}z^{k}\\ & =\sum_{k=0}^{n}\frac{Q\left(\frac{1}{2},k\right)}{\left(k+1\right)!}\left(4z\right)^{k}\\ & =\sum_{k=0}^{\infty}\frac{Q\left(\frac{1}{2},k\right)}{\left(k+1\right)!}\left(4z\right)^{k}-\sum_{k=n+1}^{\infty}\frac{Q\left(\frac{1}{2},k\right)}{\left(k+1\right)!}\left(4z\right)^{k}\\ & =\sum_{k=0}^{\infty}\frac{\left(-1\right)^{k}P\left(-\frac{1}{2},k\right)}{\left(k+1\right)!}\left(4z\right)^{k}-\sum_{k=n+1}^{\infty}\frac{Q\left(\frac{1}{2},k\right)}{Q\left(2,k\right)}\left(4z\right)^{k}\\ & =\sum_{k=0}^{\infty}\frac{P\left(-\frac{1}{2},k\right)}{\left(k+1\right)!}\left(-4z\right)^{k}-\sum_{k=0}^{\infty}\frac{Q\left(\frac{1}{2},n+1+k\right)}{Q\left(2,n+1+k\right)}\left(4z\right)^{n+1+k}\\ & =-\frac{1}{4z}\sum_{k=0}^{\infty}\frac{P\left(-\frac{1}{2},k\right)}{\left(k+1\right)!}\left(-4z\right)^{k+1}-\frac{Q\left(\frac{1}{2},n+1\right)}{Q\left(2,n+1\right)}\left(4z\right)^{n+1}\sum_{k=0}^{\infty}\frac{Q\left(\frac{1}{2}+n+1,k\right)}{Q\left(2+n+1,k\right)}\left(4z\right)^{k}\\ & =-\frac{1}{2z}\sum_{k=0}^{\infty}\frac{P\left(\frac{1}{2},k+1\right)}{\left(k+1\right)!}\left(-4z\right)^{k+1}-\frac{Q\left(\frac{1}{2},n+1\right)}{Q\left(2,n+1\right)}\left(4z\right)^{n+1}\sum_{k=0}^{\infty}\frac{Q\left(n+\frac{3}{2},k\right)}{Q\left(n+3,k\right)}\left(4z\right)^{k}\\ & =-\frac{1}{2z}\sum_{k=0}^{\infty}C\left(\frac{1}{2},k+1\right)\left(-4z\right)^{k+1}-\frac{\left(2n+1\right)!}{2^{2n+1}n!\left(n+2\right)!}\left(4z\right)^{n+1}F\left(1,n+\frac{3}{2};n+3;4z\right)\\ & =-\frac{1}{2z}\sum_{k=1}^{\infty}C\left(\frac{1}{2},k\right)\left(-4z\right)^{k}-2\frac{C\left(2n+1,n\right)}{n+2}z^{n+1}F\left(1,n+\frac{3}{2};n+3;4z\right)\\ & =-\frac{1}{2z}\left\{ \sum_{k=0}^{\infty}C\left(\frac{1}{2},k\right)\left(-4z\right)^{k}-C\left(\frac{1}{2},0\right)\left(-4z\right)^{0}\right\} -\frac{C\left(2\left(n+1\right),n+1\right)}{n+2}z^{n+1}F\left(1,n+\frac{3}{2};n+3;4z\right)\\ & =\frac{1}{2z}\left\{ 1-\left(1-4z\right)^{\frac{1}{2}}\right\} -\frac{C\left(2\left(n+1\right),n+1\right)}{n+2}z^{n+1}F\left(1,n+\frac{3}{2};n+3;4z\right) \end{align*}

(2)

\begin{align*} \sum_{k=0}^{\infty}\frac{1}{k+1}C\left(2k,k\right)z^{k} & =\lim_{n\rightarrow\infty}\sum_{k=0}^{n}\frac{1}{k+1}C\left(2k,k\right)z^{k}\\ & =\lim_{n\rightarrow\infty}\left\{ \frac{1}{2z}\left\{ 1-\left(1-4z\right)^{\frac{1}{2}}\right\} -\frac{C\left(2\left(n+1\right),n+1\right)}{n+2}z^{n+1}F\left(1,n+\frac{3}{2};n+3;4z\right)\right\} \\ & =\frac{1}{2z}\left\{ 1-\left(1-4z\right)^{\frac{1}{2}}\right\} \end{align*}

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中央2項係数を含む通常型母関数

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