2項係数の総和その他

2項係数の総和その他
2項係数の総和について次が成り立つ。

(1)

\(n\in\mathbb{N}\)とする。
\[ \sum_{k=1}^{n-1}\frac{C\left(k-n,k\right)}{k}=-H_{n-1} \]

(2)

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

(3)

\[ \sum_{k=z}^{\infty}\frac{C\left(k,z\right)}{k^{2}}=\frac{\pi}{z\sin\left(\pi z\right)} \]

(4)

\[ \sum_{k=z}^{\infty}\frac{C\left(k,z\right)}{k}=\frac{F\left(z;;1\right)}{z} \]

-

\(H_{n}\)は調和数
\(F\)は一般化超幾何関数

(1)

\begin{align*} \sum_{k=1}^{n-1}\frac{C\left(k-n,k\right)}{k} & =\sum_{k=1}^{n-1}\frac{\left(-1\right)^{k}}{k}C\left(n-1,k\right)\\ & =\sum_{k=1}^{n-1}\frac{\left(-1\right)^{k}}{k}\left(C\left(n-2,k\right)+C\left(n-2,k-1\right)\right)\\ & =\sum_{k=1}^{n-1}\left(\frac{\left(-1\right)^{k}}{k}C\left(n-2,k\right)+\frac{\left(-1\right)^{k}}{n-1}C\left(n-1,k\right)\right)\\ & =\sum_{k=1}^{n-1}\frac{\left(-1\right)^{k}}{k}C\left(n-2,k\right)+\frac{1}{n-1}\sum_{k=0}^{n-1}\left(-1\right)^{k}C\left(n-1,k\right)-\frac{C\left(n-1,0\right)}{n-1}\\ & =\sum_{k=1}^{n-2}\frac{\left(-1\right)^{k}}{k}C\left(n-2,k\right)+\frac{1}{n-1}\delta_{0,n-1}-\frac{1}{n-1}\\ & =\sum_{k=1}^{n-2}\frac{C\left(k-n+1,k\right)}{k}+\frac{1}{n-1}\delta_{n,1}-\frac{1}{n-1}\\ & =\sum_{j=2}^{n}\left(\sum_{k=1}^{j-1}\frac{C\left(k-j,k\right)}{k}-\sum_{k=1}^{j-2}\frac{C\left(k-j+1,k\right)}{k}\right)\\ & =\sum_{j=2}^{n}\left(\frac{1}{j-1}\delta_{j,1}-\frac{1}{n-1}\right)\\ & =-\sum_{j=2}^{n}\left(\frac{1}{n-1}\right)\\ & =-\sum_{j=1}^{n-1}\left(\frac{1}{j}\right)\\ & =-H_{n-1} \end{align*}

(2)

\begin{align*} \sum_{k=z}^{\infty}\frac{C\left(k,z\right)}{k+1} & =\sum_{k=0}^{\infty}\frac{C\left(k+z,z\right)}{k+z+1}\\ & =\sum_{k=0}^{\infty}\frac{C\left(k+z,k\right)}{k+z+1}\\ & =B\left(-z,z+1\right)\cmt{\because B_{z}\left(\alpha,\beta\right)=z^{\alpha}\sum_{k=0}^{\infty}\frac{C\left(k-\beta,k\right)}{\alpha+k}z^{k}}\\ & =\frac{\Gamma\left(-z\right)\Gamma\left(z+1\right)}{\Gamma\left(-z+z+1\right)}\\ & =\Gamma\left(-z\right)\Gamma\left(z+1\right)\\ & =\Gamma\left(-z\right)\Gamma\left(1-\left(-z\right)\right)\\ & =\frac{\pi}{\sin\left(-\pi z\right)}\\ & =-\frac{\pi}{\sin\left(\pi z\right)} \end{align*}

(3)

\begin{align*} \sum_{k=z}^{\infty}\frac{C\left(k,z\right)}{k^{2}} & =\sum_{k=0}^{\infty}\frac{C\left(k+z,z\right)}{\left(k+z\right)^{2}}\\ & =\sum_{k=0}^{\infty}\frac{C\left(k+z,k\right)}{\left(k+z\right)^{2}}\\ & =\frac{1}{z}\sum_{k=0}^{\infty}\frac{C\left(k+z-1,k\right)}{k+z}\\ & =\frac{1}{z}B\left(z,1-z\right)\cmt{\because B_{z}\left(\alpha,\beta\right)=z^{\alpha}\sum_{k=0}^{\infty}\frac{C\left(k-\beta,k\right)}{\alpha+k}z^{k}}\\ & =\frac{1}{z}\cdot\frac{\Gamma\left(z\right)\Gamma\left(1-z\right)}{\Gamma\left(z+1-z\right)}\\ & =\frac{1}{z}\cdot\Gamma\left(z\right)\Gamma\left(1-z\right)\\ & =\frac{\pi}{z\sin\left(\pi z\right)} \end{align*}

(4)

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