部分分数展開を用いた級数の計算方法
部分分数展開
任意の多項式$f(x),g(x)\in\mathbb{C[x]}$に対して以下の様な集合を定める。
\begin{equation}
I=\{h_{1}(x)f(x)+h_{2}(x)g(x)|h_{1}(x),h_{2}(x)\in\mathbb{C}[x]\}
\end{equation}
すると$\gcd{(f(x)),g(x))}=d(x)$とすれば以下の式が成り立つ。
\begin{equation}
I=d(x)\mathbb{C}[x]
\end{equation}
[1]$I$に含まれる多項式のうち最小次数を持つものを$h(x)$とする。すると以下の式が成り立つ。
\begin{equation}
I=h(x)\mathbb{C}[x]
\end{equation}
A. $I\supset h(x)\mathbb{C}[x]$
まず仮定より、$\exists h_{1}(x),h_{2}(x)\in\mathbb{C}[x]\ s.t.\ h_{1}(x)f(x)+h_{2}(x)g(x)=h(x)$が成り立つので$\forall q(x)\in\mathbb{C}[x]: h_{1}(x)q(x)f(x)+h_{2}(x)q(x)g(x)=h(x)q(x)$が成り立つ。ゆえに$I\supset h(x)\mathbb{C}[x]$
B. $I\subset h(x)\mathbb{C}[x]$
$\exists p(x)\in I\setminus h(x)\mathbb{C}[x]$とすると$\exists q(x),r(x)\in\mathbb{C}[x](\deg{r(x)})\lt \deg{h(x)}\ s.t.\ p(x)=q(x)h(x)+r(x)$が成り立つ。しかし仮定より$p(x),h(x)\in I$なので$r(x)=p(x)-q(x)h(x)\in I$。しかし$h(x)$の次数の最小性に反する。ゆえに$\forall p(x)\in I\Rightarrow p(x)\in\mathbb{C}[x]$
[2][1]において$h(x)=d(x)$を示す。まず$d(x)|f(x),g(x)$なので$\exists q(x)\in\mathbb{C}\ s.t.\ h(x)=q(x)d(x)$を得る。
この時、$f(x)=r_{1}(x)q(x)h(x)\land g(x)=r_{2}(x)q(x)h(x)$も成立せねばならない。$q(x)|f(x),g(x)$なので結局$q(x)=1$が得られるので終了。
有理関数$f(x)=\frac{q(x)}{p(x)}\in\mathbb{C}(x)$について$p(x)$の根$\alpha$が$m$重根を持つとする。そして$\exists p^{(1)}(x)\ s.t.\ p(x)=(x-\alpha)^{m}p_{1}(x)$が成り立つとすると以下の式が成り立つ。
\begin{equation}
\exists q_{1}(x),q_{2}(x)\in\mathbb{C}[x]\ s.t.\ \frac{q_{1}(x)}{(x-\alpha)^{m}}+\frac{q_{2}(x)}{p_{1}(x)}=\frac{q(x)}{p(x)}
\end{equation}
仮定より$\gcd\{((x-\alpha)^{m},p_{1}(x))\}=1$なので
\begin{equation}
\exists q_{1}(x),q_{2}(x)\in\mathbb{C}[x]\ s.t.\ q_{1}(x)p_{1}(x)+q_{2}(x)(x-\alpha)^{m}=q(x)\cdot 1
\end{equation}
を満たす。ゆえに証明終了。
$p(x)=(x-\alpha_{1})^{m_{1}}(x-\alpha_{2})^{m_{2}}\cdots(x-\alpha_{n})^{m_{n}}$とすると以下の式が成り立つ。
\begin{equation}
\exists q_{1}(x),q_{2}(x),...,q_{n}(x)\in\mathbb{C}[x]\ s.t.\ \frac{q_{1}(x)}{(x-\alpha_{1})^{m_{1}}}+\frac{q_{2}(x)}{(x-\alpha_{2})^{m_{2}}}+\cdots+\frac{q_{n}(x)}{(x-\alpha_{n})^{m_{n}}}=\frac{q(x)}{p(x)}
\end{equation}
定理2で$\frac{q_{2}(x)}{p_{1}(x)}$にも繰り返し同じ議論を繰り返すと証明される。
多項式$p(x)$について以下の式が成り立つ。
\begin{equation}
\exists A_{1},A_{2},...,A_{m}\in\mathbb{C}\ s.t.\ \frac{p(x)}{(x-\alpha)^{m}}=\frac{A_{m}}{(x-\alpha)^{m}}+\frac{A_{m-1}}{(x-\alpha)^{m-1}}+\cdots+\frac{A_{1}}{x-\alpha}+A_{0}
\end{equation}
ただし$\deg{p(x)}\lt m$
\begin{eqnarray} x^{k}&=&(x-\alpha+\alpha)^{k}\\ &=&\sum_{l=0}^{k}\begin{pmatrix}k\\l\end{pmatrix}\alpha^{k-l}(x-\alpha)^{l} \end{eqnarray}
応用
正整数$\alpha,\beta(\alpha\lt\beta)$に対して以下の式が成り立つ。
\begin{equation}
\sum_{n=1}^{\infty}\frac{1}{(n+\alpha)^{3})(n+\beta)^{3}}=\frac{1}{(\beta-\alpha)^{3}}\{\frac{1}{(\alpha+1)^{3}}+\frac{1}{(\alpha+2)^{3}}+\cdots+\frac{1}{\beta^{3}}\}-\frac{3}{(\beta-\alpha)^{4}}[2\zeta(2)-(\frac{1}{1^{2}}+\frac{1}{2^{2}}+\cdots+\frac{1}{\alpha^{2}})-2\{\frac{1}{(\alpha+1)^{2}}+\frac{1}{(\alpha+2)^{2}}+\cdots+\frac{1}{\beta^{2}}\}]+\frac{6}{(\beta-\alpha)^{5}}\{\frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\cdots+\frac{1}{\beta}\}
\end{equation}
[1]
\begin{eqnarray}
\frac{1}{(n+\alpha)^{3}(n+\beta)^{3}}&=&\frac{A}{(n+\alpha)^{3}}+\frac{B}{(n+\alpha)^{2}}+\frac{C}{n+\alpha}+\frac{D}{(n+\beta)^{3}}+\frac{E}{(n+\beta)^{2}}+\frac{F}{n+\beta}
\end{eqnarray}
[2]下記を満たす様に$A,B,C,D,E,F$を定めればいい。
\begin{eqnarray}
\left\{
\begin{array}{l}
A=\frac{1}{(\beta-\alpha)^{3}}\\
B=-\frac{3}{(\beta-\alpha)^{4}}\\
C=\frac{6}{(\beta-\alpha)^{5}}\\
D=-\frac{1}{(\beta-\alpha)^{3}}\\
E=-\frac{3}{(\beta-\alpha)^{4}}\\
F=-\frac{6}{(\beta-\alpha)^{5}}\\
\end{array}
\right.
\end{eqnarray}
[3]
\begin{align}
&\sum_{n=1}^{\infty}\frac{1}{(n+\alpha)^{3}(n+\beta)^{3}}\\
&=\frac{1}{(\beta-\alpha)^{3}}\sum_{n=1}^{\infty}\{\frac{1}{(n+\alpha)^{3}}-\frac{1}{(n+\beta)^{3}}\}-\frac{3}{(\beta-\alpha)^{4}}\sum_{n=1}^{\infty}\{\frac{1}{(n+\alpha)^{2}}+\frac{1}{(n+\beta)^{2}}\}+\frac{6}{(\beta-\alpha)^{5}}\sum_{n=1}^{\infty}\{\frac{1}{n+\alpha}-\frac{1}{n+\beta}\}\\
&=\frac{1}{(\beta-\alpha)^{3}}\{\frac{1}{(\alpha+1)^{3}}+\frac{1}{(\alpha+2)^{3}}+\cdots+\frac{1}{\beta^{3}}\}-\frac{3}{(\beta-\alpha)^{4}}[2\zeta(2)-(\frac{1}{1^{2}}+\frac{1}{2^{2}}+\cdots+\frac{1}{\alpha^{2}})-2\{\frac{1}{(\alpha+1)^{2}}+\frac{1}{(\alpha+2)^{2}}+\cdots+\frac{1}{\beta^{2}}\}]+\frac{6}{(\beta-\alpha)^{5}}\{\frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\cdots+\frac{1}{\beta}\}
\end{align}
正整数$\alpha,\beta(\alpha\lt \beta)$に対して以下の式が成り立つことを証明せよ。
\begin{equation}
\sum_{n=1}^{\infty}\frac{1}{(n+\alpha)(n+\beta)^{3}}=\frac{1}{(\beta-\alpha)^{3}}(\frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\cdots+\frac{1}{\beta})-\frac{1}{\beta-\alpha}\{\zeta(3)-(\frac{1}{1^{3}}+\frac{1}{2^{3}}+\cdots+\frac{1}{\beta^{3}})\}-\frac{1}{(\beta-\alpha)^{2}}\{\zeta(2)-(\frac{1}{1^{2}}+\frac{1}{2^{2}}+\cdots+\frac{1}{\beta^{2}})\}
\end{equation}
[1]
\begin{equation}
\frac{1}{(n+\alpha)(n+\beta)^{3}}=\frac{A}{n+\alpha}+\frac{B}{(n+\beta)^{3}}+\frac{C}{(n+\beta)^{2}}+\frac{D}{n+\beta}
\end{equation}
[2]
\begin{eqnarray}
\left\{
\begin{array}{l}
A=\frac{1}{(\beta-\alpha)^{3}}\\
B=-\frac{1}{\beta-\alpha}\\
C=-\frac{1}{(\beta-\alpha)^{2}}\\
D=-\frac{1}{(\beta-\alpha)^{3}}
\end{array}
\right.
\end{eqnarray}
[3]
\begin{eqnarray}
\sum_{n=1}^{\infty}\frac{1}{(n+\alpha)(n+\beta)^{3}}&=&-\frac{1}{\beta-\alpha}\sum_{n=1}^{\infty}\frac{1}{(n+\beta)^{3}}-\frac{1}{(\beta-\alpha)^{2}}\sum_{n=1}^{\infty}\frac{1}{(n+\beta)^{2}}+\frac{1}{(\beta-\alpha)^{3}}\sum_{n=1}^{\infty}(\frac{1}{n+\alpha}-\frac{1}{n+\beta})\\
&=&\frac{1}{(\beta-\alpha)^{3}}(\frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\cdots+\frac{1}{\beta})-\frac{1}{\beta-\alpha}\{\zeta(3)-(\frac{1}{1^{3}}+\frac{1}{2^{3}}+\cdots+\frac{1}{\beta^{3}})\}-\frac{1}{(\beta-\alpha)^{2}}\{\zeta(2)-(\frac{1}{1^{2}}+\frac{1}{2^{2}}+\cdots+\frac{1}{\beta^{2}})\}
\end{eqnarray}
正整数$\alpha,\beta(\alpha\lt \beta)$に対して以下の式が成り立つことを証明せよ。
\begin{equation}
\sum_{n=1}^{\infty}\frac{1}{(n+\alpha)^{2}(n+\beta)^{3}}=\frac{1}{(\beta-\alpha)^{2}}\{\zeta(3)-(\frac{1}{1^{3}}+\frac{1}{2^{3}}+\cdots+\frac{1}{\beta^{3}})\}+\frac{1}{(\beta-\alpha)^{3}}\{3\zeta(2)-(\frac{1}{1^{2}}+\frac{1}{2^{2}}+\cdots+\frac{1}{\alpha^{2}})-3[\frac{1}{(\alpha+1)^{2}}+\frac{1}{(\alpha+2)^{2}}+\cdots+\frac{1}{\beta^{2}}\}]-\frac{3}{(\beta-\alpha)^{4}}(\frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\cdots+\frac{1}{\beta})
\end{equation}
[1]
\begin{equation}
\frac{1}{(n+\alpha)^{2}(n+\beta)^{3}}=\frac{A}{(n+\alpha)^{2}}+\frac{B}{(n+\alpha)}+\frac{C}{(n+\beta)^{3}}+\frac{D}{(n+\beta)^{2}}+\frac{E}{n+\beta}
\end{equation}
[2]
\begin{eqnarray}
\left\{
\begin{array}{l}
A=\frac{1}{(\beta-\alpha)^{3}}\\
B=-\frac{3}{(\beta-\alpha)^{4}}\\
C=\frac{1}{(\beta-\alpha)^{2}}\\
D=\frac{2}{(\beta-\alpha)^{3}}\\
E=\frac{3}{(\beta-\alpha)^{4}}
\end{array}
\right.
\end{eqnarray}
[3]
\begin{eqnarray}
\sum_{n=1}^{\infty}\frac{1}{(n+\alpha)^{2}(n+\beta)^{3}}&=&\frac{1}{(\beta-\alpha)^{2}}\sum_{n=1}^{\infty}\frac{1}{(n+\beta)^{3}}+\frac{1}{(\beta-\alpha)^{3}}\sum_{n=1}^{\infty}\{\frac{1}{(n+\alpha)^{2}}+\frac{2}{(n+\beta)^{2}}\}-\frac{3}{(\beta-\alpha)^{4}}\sum_{n=1}^{\infty}(\frac{1}{n+\alpha}-\frac{1}{n+\beta})\\
&=&\frac{1}{(\beta-\alpha)^{2}}\{\zeta(3)-(\frac{1}{1^{3}}+\frac{1}{2^{3}}+\cdots+\frac{1}{\beta^{3}})\}+\frac{1}{(\beta-\alpha)^{3}}\{3\zeta(2)-(\frac{1}{1^{2}}+\frac{1}{2^{2}}+\cdots+\frac{1}{\alpha^{2}})-3[\frac{1}{(\alpha+1)^{2}}+\frac{1}{(\alpha+2)^{2}}+\cdots+\frac{1}{\beta^{2}}\}]-\frac{3}{(\beta-\alpha)^{4}}(\frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\cdots+\frac{1}{\beta})
\end{eqnarray}
まとめ
\begin{eqnarray} \left\{ \begin{array}{l} A=\frac{1}{(\beta-\alpha)^{3}}\\ B=-\frac{3}{(\beta-\alpha)^{4}}\\ C=\frac{6}{(\beta-\alpha)^{5}}\\ D=-\frac{1}{(\beta-\alpha)^{3}}\\ E=-\frac{3}{(\beta-\alpha)^{4}}\\ F=-\frac{6}{(\beta-\alpha)^{5}}\\ \end{array} \right. \end{eqnarray}
\begin{eqnarray} \left\{ \begin{array}{l} A=\frac{1}{(\beta-\alpha)^{3}}\\ B=-\frac{3}{(\beta-\alpha)^{4}}\\ C=\frac{1}{(\beta-\alpha)^{2}}\\ D=\frac{2}{(\beta-\alpha)^{3}}\\ E=\frac{3}{(\beta-\alpha)^{4}} \end{array} \right. \end{eqnarray}
\begin{eqnarray} \left\{ \begin{array}{l} A=\frac{1}{(\beta-\alpha)^{3}}\\ B=-\frac{1}{\beta-\alpha}\\ C=-\frac{1}{(\beta-\alpha)^{2}}\\ D=-\frac{1}{(\beta-\alpha)^{3}} \end{array} \right. \end{eqnarray}
統合
下記の式を証明せよ。
\begin{eqnarray}
\left\{
\begin{array}{l}
\sum_{n=1}^{\infty}\frac{1}{(n)_{N}^{3}}=A_{0,N}\zeta(3)+B_{0,N}\zeta(2)+\sum_{k=1}^{N-1}[A_{k,N}\{\zeta(3)-(\frac{1}{1^{3}}+\frac{1}{2^{3}}+\cdots+\frac{1}{k^{3}})\}+B_{k,N}\{\zeta(2)-(\frac{1}{1^{2}}+\frac{1}{2^{2}}+\cdots+\frac{1}{k^{2}})\}]+\sum_{k=0}^{N-1}\sum_{n=1}^{\infty}\frac{C_{k,N}}{n+k}\\
A_{k,N+1}=\frac{A_{k,N}}{(N-k)^{3}}\quad(k=0,1,2,...,N-1)\\
B_{k,N+1}=-\frac{3A_{k,N}}{(N-k)^{4}}+\frac{B_{k,N}}{(N-k)^{3}}\quad(k=0,1,2,...,N-1)\\
C_{k,N+1}=\frac{6A_{k,N}}{(N-k)^{5}}-\frac{3B_{k,N}}{(N-k)^{4}}+\frac{C_{k,N}}{(N-k)^{3}}\quad(k=0,1,2,...,N-1)\\
A_{N,N+1}=\sum_{k=0}^{N-1}\{-\frac{A_{k,N}}{(N-k)^{3}}+\frac{B_{k,N}}{(N-k)^{2}}-\frac{C_{k,N}}{N-k}\}\\
B_{N,N+1}=\sum_{k=0}^{N-1}\{-\frac{3A_{k,N}}{(N-k)^{4}}+\frac{2B_{k,N}}{(N-k)^{3}}-\frac{C_{k,N}}{(N-k)^{2}}\}\\
C_{N,N+1}=\sum_{k=0}^{N-1}\{-\frac{6A_{k,N}}{(N-k)^{5}}+\frac{3B_{k,N}}{(N-k)^{4}}-\frac{C_{k,N}}{(N-k)^{3}}\}\\
A_{0,1}=1\\
B_{0,1}=0\\
C_{0,1}=0
\end{array}
\right.
\end{eqnarray}
下記の様に部分分数展開できたとする。
\begin{equation}
\frac{1}{(n)_{N}^{3}}=\sum_{k=0}^{N-1}\{\frac{A_{k,N}}{(n+k)^{3}}+\frac{B_{k,N}}{(n+k)^{2}}+\frac{C_{k,N}}{n+k}\}
\end{equation}
すると
\begin{align}
&\frac{1}{(n)_{N+1}^{3}}\\
&=\frac{1}{(n+N)^{3}}\sum_{k=0}^{N-1}\{\frac{A_{k,N}}{(n+k)^{3}}+\frac{B_{k,N}}{(n+k)^{2}}+\frac{C_{k,N}}{n+k}\}\\
&=\sum_{k=0}^{N-1}\{\frac{A_{k,N}}{(n+k)^{3}}\frac{1}{(n+N)^{3}}+\frac{B_{k,N}}{(n+k)^{2}}\frac{1}{(n+N)^{3}}+\frac{C_{k,N}}{n+k}\frac{1}{(n+N)^{3}}\}\\
&=\sum_{k=0}^{N-1}A_{k,N}\{\frac{1}{(N-k)^{3}}\frac{1}{(n+k)^{3}}-\frac{3}{(N-k)^{4}}\frac{1}{(n+k)^{2}}+\frac{6}{(N-k)^{5}}\frac{1}{n+k}-\frac{1}{(N-k)^{3}}\frac{1}{(n+N)^{3}}-\frac{3}{(N-k)^{4}}\frac{1}{(n+N)^{2}}-\frac{6}{(N-k)^{5}}\frac{1}{n+N}\}\\
&+\sum_{k=0}^{N-1}B_{k,N}\{\frac{1}{(N-k)^{3}}\frac{1}{(n+k)^{2}}-\frac{3}{(N-k)^{4}}\frac{1}{n+k}+\frac{1}{(N-k)^{2}}\frac{1}{(n+N)^{3}}+\frac{2}{(N-k)^{3}}\frac{1}{(n+N)^{2}}+\frac{3}{(N-k)^{4}}\frac{1}{n+N}\}\\
&+\sum_{k=0}^{N-1}C_{k,N}\{\frac{1}{(N-k)^{3}}\frac{1}{n+k}-\frac{1}{N-k}\frac{1}{(n+N)^{3}}-\frac{1}{(N-k)^{2}}\frac{1}{(n+N)^{2}}-\frac{1}{(N-k)^{3}}\frac{1}{n+N}\}\\
&=\sum_{k=0}^{N-1}\frac{A_{k,N}}{(N-k)^{3}}\frac{1}{(n+k)^{3}}+\sum_{k=0}^{N-1}\{-\frac{3A_{k,N}}{(N-k)^{4}}+\frac{B_{k,N}}{(N-k)^{3}}\}\frac{1}{(n+k)^{2}}+\sum_{k=0}^{N-1}\{\frac{6A_{k,N}}{(N-k)^{5}}-\frac{3B_{k,N}}{(N-k)^{4}}+\frac{C_{k,N}}{(N-k)^{3}}\}\frac{1}{n+k}\\
&+\frac{1}{(n+N)^{3}}\sum_{k=0}^{N-1}\{-\frac{A_{k,N}}{(N-k)^{3}}+\frac{B_{k,N}}{(N-k)^{2}}-\frac{C_{k,N}}{N-k}\}\\
&+\frac{1}{(n+N)^{2}}\sum_{k=0}^{N-1}\{-\frac{3A_{k,N}}{(N-k)^{4}}+\frac{2B_{k,N}}{(N-k)^{3}}-\frac{C_{k,N}}{(N-k)^{2}}\}\\
&+\frac{1}{n+N}\sum_{k=0}^{N-1}\{-\frac{6A_{k,N}}{(N-k)^{5}}+\frac{3B_{k,N}}{(N-k)^{4}}-\frac{C_{k,N}}{(N-k)^{3}}\}\\
\end{align}
以上より下記の漸化式を得る。
\begin{eqnarray}
\left\{
\begin{array}{l}
A_{k,N+1}=\frac{A_{k,N}}{(N-k)^{3}}\quad(k=0,1,2,...,N-1)\\
B_{k,N+1}=-\frac{3A_{k,N}}{(N-k)^{4}}+\frac{B_{k,N}}{(N-k)^{3}}\quad(k=0,1,2,...,N-1)\\
C_{k,N+1}=\frac{6A_{k,N}}{(N-k)^{5}}-\frac{3B_{k,N}}{(N-k)^{4}}+\frac{C_{k,N}}{(N-k)^{3}}\quad(k=0,1,2,...,N-1)\\
A_{N,N+1}=\sum_{k=0}^{N-1}\{-\frac{A_{k,N}}{(N-k)^{3}}+\frac{B_{k,N}}{(N-k)^{2}}-\frac{C_{k,N}}{N-k}\}\\
B_{N,N+1}=\sum_{k=0}^{N-1}\{-\frac{3A_{k,N}}{(N-k)^{4}}+\frac{2B_{k,N}}{(N-k)^{3}}-\frac{C_{k,N}}{(N-k)^{2}}\}\\
C_{N,N+1}=\sum_{k=0}^{N-1}\{-\frac{6A_{k,N}}{(N-k)^{5}}+\frac{3B_{k,N}}{(N-k)^{4}}-\frac{C_{k,N}}{(N-k)^{3}}\}
\end{array}
\right.
\end{eqnarray}
この漸化式を用いると
\begin{align}
&\sum_{n=1}^{\infty}\frac{1}{(n)_{N}^{3}}\\
&=\sum_{n=1}^{\infty}\sum_{k=0}^{N-1}\{\frac{A_{k,N}}{(n+k)^{3}}+\frac{B_{k,N}}{(n+k)^{2}}+\frac{C_{k,N}}{n+k}\}\\
&=\sum_{k=0}^{N-1}\sum_{n=1}^{\infty}\{\frac{A_{k,N}}{(n+k)^{3}}+\frac{B_{k,N}}{(n+k)^{2}}+\frac{C_{k,N}}{n+k}\}\\
&=A_{0,N}\zeta(3)+B_{0,N}\zeta(2)+\sum_{k=1}^{N-1}[A_{k,N}\{\zeta(3)-(\frac{1}{1^{3}}+\frac{1}{2^{3}}+\cdots+\frac{1}{k^{3}})\}+B_{k,N}\{\zeta(2)-(\frac{1}{1^{2}}+\frac{1}{2^{2}}+\cdots+\frac{1}{k^{2}})\}]+\sum_{k=0}^{N-1}\sum_{n=1}^{\infty}\frac{C_{k,N}}{n+k}
\end{align}
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