なんだろうこんなに簡単な方法で級数作るの素晴らしいのでもっとやってもらっていいですか？

[1]仮定より$\forall m\in\mathbb{N}:\exists R_{m}(n),S_{m}(n)\in\mathbb{K}[n](\gcd{(R_{m},S_{m})}=1)\ s.t.\ Q_{l}(n)=\frac{S_{m}(n)}{R_{m}(n)}$が成り立つ。
[2]さらに$R\coloneqq\gcd{(R_{1},R_{2},...,R_{m})},S\coloneqq\gcd{(S_{1},S_{2},...,S_{m})}$のようにおき、次のように書き直す。
\begin{eqnarray} \left\{ \begin{array}{l} R_{1}(n)=R(n)\tilde{R_{1}}(n)\\ R_{2}(n)=R(n)\tilde{R_{2}}(n)\\ \cdots\\ R_{m}(n)=R(n)\tilde{R_{m}}(n) \end{array} \right. \end{eqnarray}
\begin{eqnarray} \left\{ \begin{array}{l} S_{1}(n)=S(n)\tilde{S_{1}}(n)\\ S_{2}(n)=S(n)\tilde{S_{2}}(n)\\ \cdots\\ S_{m}(n)=S(n)\tilde{S_{m}}(n) \end{array} \right. \end{eqnarray}
[3]次に一次結合$T(n)\coloneqq\sum_{l=1}^{m}p_{l}(n)Q_{l}(n)$として両辺$\frac{R(n)}{S(n)}$をかけて以下の式を得る
\begin{equation} \frac{R(n)}{S(n)}T(n)=\sum_{l=1}^{m}p_{l}(n)\frac{\tilde{S}_{l}(n)}{\tilde{R_{l}}(n)} \end{equation}
[4]ゆえに
\begin{eqnarray} \left\{ \begin{array}{l} \displaystyle p_{l}(n)=\tilde{R_{l}}(n)\prod_{\substack{1\leq k\leq m\\k\neq l}}\tilde{S_{k}}(n)\quad(k=1,2,...,m-1)\\ \displaystyle p_{m}(n)=-(m-1)\tilde{R_{m}}(n)\prod_{1\leq k\leq m-1}\tilde{S_{k}}(n) \end{array} \right. \end{eqnarray}
とおけば$\frac{R(n)}{S(n)}T(n)=0$なので証明完了。

[1]
\begin{equation} p_{2}(n)z^{2}\frac{\prod_{k=1}^{M}(a_{k}+n+1)(a_{k}+n)}{\prod_{l=1}^{N}(b_{l}+n+1)(b_{l}+n)}+ p_{1}(n)z\frac{\prod_{k=1}^{M}(a_{k}+n)}{\prod_{l=1}^{N}(b_{l}+n)}+p_{0}(n)=0 \end{equation}
[2]
\begin{eqnarray} p_{2}(n)z^{2}\prod_{k=1}^{M}(a_{k}+n+1)(a_{k}+n)+p_{1}(n)z\prod_{k=1}^{M}(a_{k}+n)\prod_{l=1}^{N}(b_{l}+n+1)+p_{0}(n)\prod_{l=1}^{N}(b_{l}+n+1)(b_{l}+n)=0 \end{eqnarray}
[3]
\begin{eqnarray} \left\{ \begin{array}{l} p_{0}(n)=-2z^{2}\prod_{k=1}^{M}(a_{k}+n+1)(a_{k}+n)\\ p_{1}(n)=z\prod_{k=1}^{M}(a_{k}+n+1)\prod_{k=1}^{N}(b_{l}+n)\\ p_{2}(n)=\prod_{l=1}^{N}(b_{l}+n+1)(b_{l}+n) \end{array} \right. \end{eqnarray}

[1]
\begin{equation} \frac{A_{n}}{A_{n-1}}=\frac{n^{4}}{4(n+1)^{2}(2n-1)^{2}} \end{equation}
[2]上記計算より
\begin{equation} A_{n}=\frac{1}{16^{n}}\frac{(1)_{n}^{4}}{(2)_{n}^{2}(\frac{1}{2})_{n}^{2}} \end{equation}
\begin{eqnarray} \left\{ \begin{array}{l} p_{0}(n)=-\frac{1}{128}(n+2)^{4}(n+1)^{4}\\ p_{1}(n)=\frac{1}{16}(n+1)^{4}(n+\frac{3}{2})^{2}(n+3)^{2}\\ p_{2}(n)=(n+\frac{1}{2})^{2}(n+\frac{3}{2})^{2}(n+2)^{2}(n+3)^{2} \end{array} \right. \end{eqnarray}
[3]
\begin{eqnarray} p_{2}(n)A_{n+2}+p_{1}(n)A_{n+1}+p_{0}(n)A_{n}=0 \end{eqnarray}
[4]ゆえに整理すると以下の級数を得る。
\begin{equation} \frac{1}{128}\sum_{n=0}^{\infty}\frac{(n+2)^{4}(n+1)^{4}}{(n+1)^{2}\begin{pmatrix}2n\\n\end{pmatrix}^{2}}\frac{1}{16^{n}}=\sum_{n=0}^{N}\frac{(n+\frac{1}{2})^{2}(n+\frac{3}{2})^{2}(n+2)^{2}(n+3)^{2}}{(n+3)^{2}\begin{pmatrix}2n+4\\n+2\end{pmatrix}^{2}}\frac{1}{16^{n+2}}+\frac{1}{16}\sum_{n=0}^{N}\frac{(n+1)^{4}(n+\frac{3}{2})^{2}(n+3)^{2}}{(n+2)^{2}\begin{pmatrix}2n+2\\n+1\end{pmatrix}^{2}}\frac{1}{16^{n+1}} \end{equation}