簡単な望遠鏡和で導出するApery Like級数

[1]
\begin{eqnarray} (n;k)&=&n\prod_{m=1}^{k}(n^{2}-m^{2})\\ &=&n(n-1)(n-2)\cdots(n-k)(n+1)(n+2)\cdots (n+k)\\ &=&\frac{n!(n+k)!}{(n-k-1)!n!}\\ &=&\frac{(n+k)!}{(n-k-1)!} \end{eqnarray}
[2]
\begin{eqnarray} \frac{1}{(n;k)}&=&\frac{n-k-1}{n+k}\frac{(n-k-2)!}{(n+k-1)!}\\ &=&\frac{n-k-1}{n+k}\frac{1}{(n-1;k)}\\ &=&\frac{1}{(n-1;k)}-\frac{2k+1}{n+k}\frac{1}{(n-1;k)} \end{eqnarray}
\begin{eqnarray} \frac{1}{(n-1;k)}-\frac{1}{(n;k)}&=&\frac{2k+1}{n+k}\frac{1}{(n-1;k)} \end{eqnarray}
\begin{eqnarray} \frac{n+k-1}{(n-1;k)}-\frac{n+k}{(n;k)}&=&\frac{2k}{(n-1;k)} \end{eqnarray}
ゆえに
\begin{eqnarray} \sum_{n=k+2}^{\infty}\frac{1}{(n-1;k)}&=&\frac{1}{2k}\sum_{n=k+2}^{\infty}\{ \frac{n+k-1}{(n-1;k)}-\frac{n+k}{(n;k)}\}\\ &=&\frac{2k+1}{2k(k+1;k)}\\ &=&\frac{1}{2k(2k)!} \end{eqnarray}
[3]
\begin{eqnarray} \frac{1}{(n;k-1)}&=&\frac{n^{2}-k^{2}}{(n;k)} \end{eqnarray}

\begin{eqnarray} \varphi_{k-1}(x)&=&\sum_{n=k}^{\infty}\frac{1}{(n^{2}-x^{2})(n;k-1)}\\ &=&\frac{1}{(k^{2}-x^{2})(k;k-1)}+\sum_{n=k+1}^{\infty}\frac{1}{(n^{2}-x^{2})(n;k-1)}\\ &=&\frac{1}{(k^{2}-x^{2})(2k-1)!}+\sum_{n=k+1}^{\infty}\frac{n^{2}-k^{2}}{(n^{2}-x^{2})(n;k)}\\ &=&\frac{1}{(k^{2}-x^{2})(2k-1)!}+\sum_{n=k+1}^{\infty}\frac{1}{(n;k)}+(x^{2}-k^{2})\sum_{n=k+1}^{\infty}\frac{1}{(n^{2}-x^{2})(n;k)}\\ &=&\frac{1}{(k^{2}-x^{2})(2k-1)!}+\frac{1}{2k(2k)!}+(x^{2}-k^{2})\varphi_{k}(x)\\ &=&\frac{5k^{2}-x^{2}}{2k(k^{2}-x^{2})(2k)!}+(x^{2}-k^{2})\varphi_{k}(x)\\ &=&\frac{1}{2k(2k)!}\frac{5k^{2}-x^{2}}{k^{2}-x^{2}}+(x^{2}-k^{2})\varphi_{k}(x) \end{eqnarray}

\begin{eqnarray} \varphi_{k-1}(x)&=&\frac{1}{2k(2k)!}\frac{5k^{2}-x^{2}}{k^{2}-x^{2}}+(x^{2}-k^{2})\varphi_{k}(x)\\ &=&\frac{1}{2k(2k)!}\frac{5k^{2}-x^{2}}{k^{2}-x^{2}}-\frac{1}{2(k+1)\{2(k+1)\}!}\frac{5(k+1)^{2}-x^{2}}{(k+1)^{2}-x^{2}}(k^{2}-x^{2})\\ &+&\frac{1}{2(k+2)\{2(k+2)\}!}\frac{5(k+2)^{2}-x^{2}}{(k+1)^{2}-x^{2}}\{k^{2}-x^{2}\}\{(k+1)^{2}-x^{2}\}+\cdots\\ &=&\sum_{n=k}^{\infty}(-1)^{n-1}\frac{1}{2n(2n)!}\frac{5n^{2}-x^{2}}{n^{2}-x^{2}}\prod_{m=1}^{n-k}\{(k+m-1)^{2}-x^{2}\} \end{eqnarray}
特に$k=1$として
\begin{eqnarray} \sum_{n=1}^{\infty}\frac{1}{n(n^{2}-x^{2})}&=&\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{2n(2n)!}\frac{5n^{2}-x^{2}}{n^{2}-x^{2}}\prod_{m=1}^{n-1}(m^{2}-x^{2})\\ &=&\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{2n(2n)!}\frac{5n^{2}-x^{2}}{n^{2}-x^{2}}(n!)^{2}\frac{1}{n^{2}}\prod_{m=1}^{n-1}(1-\frac{x^{2}}{m^{2}})\\ &=&\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{2n^{3}\begin{pmatrix}2n\\n\end{pmatrix}}\frac{5n^{2}-x^{2}}{n^{2}-x^{2}}\prod_{m=1}^{n-1}(1-\frac{x^{2}}{m^{2}}) \end{eqnarray}

[1]$\displaystyle C(n;k)\coloneqq\frac{1}{(n;k)^{2}}$の様に置くと
\begin{eqnarray} C(n;k)&=&\frac{(n-k-1)^{2}}{(n+k)^{2}}C(n-1;k)\\ &=&\frac{(n+k-2k-1)^{2}}{(n+k)^{2}}C(n-1;k)\\ &=&\{1-2\frac{2k+1}{n+k}+\frac{(2k+1)^{2}}{(n+k)^{2}}\}C(n-1;k) \end{eqnarray}
ゆえに
\begin{align} &(n+k-1)^{2}C(n-1;k)-(n+k)^{2}C(n;k)\\ &=\{2(2k+1)(n+k)-(2k+1)^{2}-2(n+k)+1\}C(n-1;k)\\ &=4k(n-1)C(n-1;k) \end{align}
の様に出来るので
\begin{eqnarray} \sum_{n=k+2}^{\infty}\frac{n-1}{(n-1;k)^{2}}&=&\frac{1}{4k}(2k+1)^{2}\frac{1}{\{(2k+1)!\}^{2}}\\ &=&\frac{1}{4k\{(2k)!\}^{2}} \end{eqnarray}
[2]
\begin{align} &(n+A-1)(n+B-1)(n+k-1)^{2}C(n-1;k)-(n+A)(n+B)(n+k)^{2}C(n;k)\\ &=\{4k(n+A)(n+B)(n-1)-(2n+A+B-1)(n+k-1)^{2}\}C(n-1;k) \end{align}
\begin{align} &4k(n+A)(n+B)(n-1)-(2n+A+B-1)(n+k-1)^{2}\\ &=4k(n-1+A+1)(n-1+B+1)(n-1)-\{2(n-1)+A+B+1\}(n-1+k)^{2}\\ &=4k(n-1)^{3}+4k(A+B+2)(n-1)^{2}+4k(A+1)(B+1)(n-1)\\ &-\{2(n-1)+A+B+1\}\{(n-1)^{2}+2k(n-1)+k^{2}\}\\ &=4k(n-1)^{3}+4k(A+B+2)(n-1)^{2}+4k(A+1)(B+1)(n-1)\\ &-2(n-1)^{3}-\{(A+B+1)+4k\}(n-1)^{2}-\{2k^{2}+2k(A+B+1)\}(n-1)-(A+B+1)k^{2}\\ &=2(2k-1)(n-1)^{3}+\{4(A+B+1)k-(A+B+1)\}(n-1)^{2}-[2k^{2}+k\{2(A+B+1)-2(A+1)(B+1)\}](n-1)-(A+B+1) \end{align}
そこで$A+B+1=0$が成り立つ様に$A=0,B=-1$とすると次の漸化式を得る。
\begin{align} &(n-2)(n-1)(n+k-1)^{2}C(n-1;k)-(n-1)n(n+k)^{2}C(n;k)\\ &=\{2(2k-1)(n-1)^{3}-2k^{2}(n-1)\}C(n-1;k) \end{align}
よって
\begin{align} &2(2k-1)\sum_{n=k+2}^{\infty}\frac{(n-1)^{3}}{(n-1;k)^{2}}-2k^{2}\sum_{n=k+2}^{\infty}\frac{n-1}{(n-1;k)^{2}}\\ &=2(2k-1)\sum_{n=k+2}^{\infty}\frac{(n-1)^{3}}{(n-1;k)^{2}}-\frac{k}{2\{(2k)!\}^{2}}\\ &=k(k+1)(2k+1)\frac{1}{\{(2k+1)!\}^{2}}\\ &=\frac{k(k+1)}{\{(2k)!\}^{2}} \end{align}
整理すると
\begin{equation} \sum_{n=k+1}^{\infty}\frac{n^{3}}{(n;k)^{2}}=\frac{k(2k+3)}{4(2k-1)\{(2k)!\}^{2}} \end{equation}

\begin{eqnarray} \varphi_{k-1}^{(2)}(x)&=&\frac{k}{(k^{2}-x^{2})(k;k-1)^{2}}+\sum_{n=k+1}^{\infty}\frac{n(n^{2}-k^{2})^{2}}{(n^{2}-x^{2})(n;k)^{2}}\\ &=&\frac{k}{(k^{2}-x^{2})\{(2k-1)!\}^{2}}+\sum_{n=k+1}^{\infty}\frac{n(n^{2}-x^{2})}{(n;k)^{2}}+2(x^{2}-k^{2})\sum_{n=k+1}^{\infty}\frac{n}{(n;k)^{2}}+(x^{2}-k^{2})^{2}\sum_{n=k+1}^{\infty}\frac{n}{(n^{2}-x^{2})(n;k)^{2}}\\ &=&\frac{k}{(k^{2}-x^{2})\{(2k-1)!\}^{2}}+\sum_{n=k+1}^{\infty}\frac{n^{3}}{(n;k)^{2}}+(x^{2}-2k^{2})\sum_{n=k+1}^{\infty}\frac{n}{(n;k)^{2}}+(x^{2}-k^{2})^{2}\varphi_{k}^{(2)}(x)\\ &=&\frac{k}{(k^{2}-x^{2})\{(2k-1)!\}^{2}}+\frac{k(2k+3)}{4(2k-1)\{(2k)!\}^{2}}+(x^{2}-2k^{2})\frac{1}{4k\{(2k)!\}^{2}}+(x^{2}-k^{2})^{2}\varphi_{k}^{(2)}(x)\\ &=&\frac{1}{\{(2k)!\}^{2}}\{\frac{4k^{3}}{k^{2}-x^{2}}+\frac{k(2k+3)}{4(2k-1)}+\frac{x^{2}-2k^{2}}{4k}\}+(x^{2}-k^{2})^{2}\varphi_{k}^{(2)}(x)\\ &=&\frac{30k^{5}-11k^{4}+4x^{2}k^{3}-6x^{2}k^{2}-2x^{4}k+x^{4}}{4k(2k-1)(k^{2}-x^{2})\{(2k)!\}^{2}}+(x^{2}-k^{2})^{2}\varphi_{k}^{(2)}(x) \end{eqnarray}

$k=1$を補題6で代入し帰納的に漸化式を適用するだけ。

[1]
\begin{eqnarray} (n;k)_{a}&=&n(an-1)(an-2)\cdots(an-k)(an+1)(an+2)\cdots(an+k)\\ &=&n\frac{(an)!}{an(an-k-1)!}\frac{(an+k)!}{(an)!}\\ &=&\frac{1}{a}\frac{(an+k)!}{(an-k-1)!} \end{eqnarray}
[2]
\begin{eqnarray} \frac{1}{(n;k-1)_{a}}&=&\frac{a^{2}n^{2}-k^{2}}{(n;k)_{a}} \end{eqnarray}
[3]$\displaystyle C(n;k)\coloneqq\frac{1}{(n;k)_{2}}$とおくと
\begin{eqnarray} C(n;k)&=&\frac{(2n-k-1)(2n-k-2)}{(2n+k)(2n+k-1)}C(n-1;k)\\ &=&\frac{(2n+k-2k-1)(2n+k-1-2k-1)}{(2n+k)(2n+k-1)}C(n-1;k)\\ &=&\{1-\frac{(4n+2k-1)(2k+1)-(2k+1)^{2}}{(2n+k)(2n+k-1)}\}C(n-1;k)\\ &=&\{1-\frac{(4n-2)(2k+1)}{(2n+k)(2n+k-1)}\}C(n-1;k) \end{eqnarray}
\begin{equation} C(n;k)-C(n-1;k)=\frac{(4n-2)(2k+1)}{(2n+k)(2n+k-1)}C(n-1;k) \end{equation}
少し変形して
\begin{align} &(2n+k-2)(2n+k-3)C(n-1;k)-(2n+k)(2n+k-1)C(n;k)\\ &=\{(4n-2)(2k+1)-8n-4k+6)\}C(n-1;k)\\ &=(8nk-4n-8k+4)C(n-1;k)\\ &=4\{(2k-1)n-(2k-1)\}C(n-1;k)\\ &=4(2k-1)(n-1)C(n-1;k) \end{align}
なので
\begin{eqnarray} \sum_{n=k+2}^{\infty}\frac{n-1}{(n-1;k)_{2}}&=&\sum_{n=k+1}^{\infty}\frac{n}{(n;k)_{2}}\\ &=&\frac{1}{4(2k-1)}(3k+2)(3k+1)2\frac{(k+1)!}{(3k+2)!}\\ &=&\frac{1}{2(2k-1)}\frac{(k+1)!}{(3k)!} \end{eqnarray}

\begin{eqnarray} \varphi_{k-1;2}(x)&=&\frac{k}{(4k^{2}-x^{2})(k;k-1)_{2}}+\sum_{n=k+1}^{\infty}\frac{n(4n^{2}-k^{2})}{(4n^{2}-x^{2})(n;k)_{2}}\\ &=&\frac{2kk!}{(3k-1)!(4k^{2}-x^{2})}+\sum_{n=k+1}^{\infty}\frac{n}{(n;k)_{2}}+(x^{2}-k^{2})\sum_{n=k+1}^{\infty}\frac{n}{(4n^{2}-x^{2})(n;k)_{2}}\\ &=&\frac{2kk!}{(3k-1)!(4k^{2}-x^{2})}+\frac{1}{2(2k-1)}\frac{(k+1)!}{(3k)!}+(x^{2}-k^{2})\varphi_{k;2}(x)\\ &=&\frac{k!}{(3k)!}\{\frac{6k^{2}}{4k^{2}-x^{2}}+\frac{k+1}{2(2k-1)}\}+(x^{2}-k^{2})\varphi_{k;2}(x)\\ &=&\frac{(28k^{3}-8k^{2}-x^{2}k-x^{2})k!}{2(2k-1)(4k^{2}-x^{2})(3k)!}-(k^{2}-x^{2})\varphi_{k;2}(x) \end{eqnarray}

定理8で$k=1$を代入し漸化式を帰納的に使うだけ。

\begin{align} &T_{N}(n-1;k|\{a_{n}\},\{b_{n}\},\{c_{n}\})\\ &=\prod_{m=1}^{k}\prod_{s=1}^{N}(a_{s}n+b_{s}m+c_{s}-a_{s})\\ &=\prod_{m=1}^{k}\prod_{s=1}^{N}\frac{a_{s}n+b_{s}m+c_{s}-a_{s}}{a_{s}n+b_{s}m+c_{s}-a_{s}+1}\frac{a_{s}n+b_{s}m+c_{s}-a_{s}+1}{a_{s}n+b_{s}m+c_{s}-a_{s}+2}\cdots\frac{a_{s}n+b_{s}m+c_{s}-1}{a_{s}n+b_{s}m+c_{s}}(a_{s}n+b_{s}m+c_{s})\\ &=T_{N}(n;k|\{a_{n}\},\{b_{n}\},\{c_{n}\})\prod_{m=1}^{k}\prod_{s=1}^{N}\frac{a_{s}n+b_{s}m+c_{s}-a_{s}}{a_{s}n+b_{s}m+c_{s}-a_{s}+1}\frac{a_{s}n+b_{s}m+c_{s}-a_{s}+1}{a_{s}n+b_{s}m+c_{s}-a_{s}+2}\cdots\frac{a_{s}n+b_{s}m+c_{s}-1}{a_{s}n+b_{s}m+c_{s}}\\ &=T_{N}(n;k|\{a_{n}\},\{b_{n}\},\{c_{n}\})\times (n,kに関する有理関数) \end{align}

これは見たまま計算するだけ。

これもみたまま計算するだけ。

[1]
\begin{eqnarray} (n+a)T_{2}(n;k|\{1,1\};\{-1,1\},\{a,b\})&=&(n+a)\prod_{m=1}^{k}(n-m+a)(n+m+b)\\ &=&(n+a)(n+a-1)(n+a-2)\cdots(n+a-k)(n+b+1)(n+b+2)\cdots(n+b+k)\\ &=&\frac{(n+a)!}{(n+a-k-1)!}\frac{(n+b+k)!}{(n+b)!} \end{eqnarray}
[2]$\displaystyle C(n;k)\coloneqq\frac{1}{(n+a)T_{2}(n;k|\{1,1\};\{-1,1\},\{a,b\})}$とおくと
\begin{eqnarray} C(n;k)&=&\frac{(n+a-k-1)(n+b)}{(n+a)(n+b+k)}C(n-1;k)\\ &=&\frac{(n+a+b-a)(n+b+k-2k+a-b-1)}{(n+a)(n+b+k)}C(n-1;k)\\ &=&(1+\frac{b-a}{n+a})(1-\frac{2k+1+b-a}{n+b+k})C(n-1;k)\\ &=&\{1+\frac{b-a}{n+a}-\frac{2k+1+b-a}{n+b+k}-\frac{b-a}{n+a}\frac{2k+1+b-a}{n+b+k}\}C(n-1;k)\\ &=&\{1+\frac{b-a}{n+a}-\frac{(n+b)(2k+1+b-a)}{(n+a)(n+b+k)}\}C(n-1;k)\\ &=&\{1-\frac{(2k+1)n+(a+b)k+b}{(n+a)(n+b+k)}\}C(n-1;k) \end{eqnarray}
ゆえに
\begin{equation} C(n-1;k)-C(n;k)=\frac{(2k+1)n+(a+b)k+b}{(n+a)(n+b+k)}C(n-1;k) \end{equation}
少し整理すると
\begin{align} &(n+a-1)(n+b+k-1)C(n-1;k)-(n+a)(n+b+k)C(n;k)\\ &=\{(2k+1)n+(a+b)k+b-(2n+a+b+k)+1\}C(n-1;k)\\ &=\{(2k-1)n+(a+b-1)k+1-a\}C(n-1;k)\\ &=\{(2k-1)(n-1)+(a+b+1)k-a\}C(n-1;k) \end{align}
特に$a=0,b=-1$として
\begin{equation} (n-1)(n+k-2)C(n-1;k)-n(n+k-1)C(n;k)=(2k-1)(n-1)C(n-1;k) \end{equation}
ゆえに
\begin{eqnarray} \sum_{n=k+2}^{\infty}\frac{1}{T_{2}(n-1;k|\{1,1\};\{-1,1\},\{0,-1\})}&=& \sum_{n=k+1}^{\infty}\frac{1}{T_{2}(n;k|\{1,1\};\{-1,1\},\{0,-1\})}\\ &=&\frac{2(k+1)k}{(2k-1)(k+1)(2k)!}\\ &=&\frac{1}{(2k-1)(2k-1)!} \end{eqnarray}

\begin{align} &\varphi_{k-1}(x|\{1,1\};\{-1,1\},\{0,-1\})\\ &=\frac{1}{(k^{2}-k-x^{2})T_{2}(k;k-1|\{1,1\};\{-1,1\},\{0,-1\})}+\sum_{n=k+1}^{\infty}\frac{n^{2}-n-k^{2}+k}{(n^{2}-n-x^{2})T_{2}(n;k|\{1,1\};\{-1,1\},\{0,-1\})}\\ &=\frac{1}{(k^{2}-k-x^{2})}\frac{1}{(2k-2)!}+\sum_{n=k+1}^{\infty}\frac{1}{T_{2}(n;k|\{1,1\};\{-1,1\},\{0,-1\})}+(x^{2}-k^{2}+k)\varphi_{k}(x|\{1,1\};\{-1,1\},\{a,b\})\\ &=\frac{1}{(k^{2}-k-x^{2})}\frac{1}{(2k-2)!}+\frac{1}{(2k-1)(2k-1)!}+(x^{2}-k^{2}+k)\varphi_{k}(x|\{1,1\};\{-1,1\},\{0,-1\})\\ &=\frac{5k^{2}-5k+1-x^{2}}{(2k-1)(2k-1)!(k^{2}-k-x^{2})}+(x^{2}-k^{2}+k)\varphi_{k}(x|\{1,1\};\{-1,1\},\{0,-1\})\\ \end{align}