ゼータさんリベンジ
あいさつ
んちゃろはー
今回は前回だめだった$\zeta(3)$についてリベンジします。
再チャレンジ
\begin{align} &\lambda_{0,0}a_{0}\\ &+(\lambda_{1,0}+\lambda_{0,1})a_{1}\\ &+(\lambda_{2,0}+\lambda_{1,1}+\lambda_{0,2})a_{2}\\ &+\cdots\\ &+(\lambda_{N-1,0}+\lambda_{N-2,1}+\cdots+\lambda_{0,N-1})a_{N-1}\\ &+\sum_{n=N}^{\infty}(\lambda_{0,n}+\lambda_{1,n-1}+\cdots+\lambda_{N,n-N})a_{n}\\ &+\sum_{n=0}^{\infty}(1-\sum_{m=0}^{N}\lambda_{m,n}\frac{a_{n+m}}{a_{n}})a_{n} \end{align}
\begin{eqnarray} \sum_{n=0}^{\infty}a_{n}&=&\sum_{n=0}^{\infty}(1-x_{n}+x_{n})a_{n}\\ &=&\sum_{n=0}^{\infty}x_{n}a_{n}+\sum_{n=0}^{\infty}(1-x_{n})a_{n}\\ &=&\sum_{n=0}^{\infty}(\sum_{m=0}^{N}\lambda_{m,n}\frac{a_{n+m}}{a_{n}})a_{n}+\sum_{n=0}^{\infty}(1-\sum_{m=0}^{N}\lambda_{m,n}\frac{a_{n+m}}{a_{n}})a_{n}\\ &=&\lambda_{0,0}a_{0}+(\lambda_{0,1}+\lambda_{1,0})a_{1}+(\lambda_{0,2}+\lambda_{1,1}+\lambda_{2,0})a_{2}+\cdots+(\lambda_{0,N-1}+\lambda_{1,N-2}+\cdots+\lambda_{N-1,0})a_{N-1}+\sum_{n=N}^{\infty}(\lambda_{0,n}+\lambda_{1,n-1}+\cdots+\lambda_{N,n-N})a_{n}+\sum_{n=0}^{\infty}(1-\sum_{m=0}^{N}\lambda_{m,n}\frac{a_{n+m}}{a_{n}})a_{n} \end{eqnarray}
前回とほとんど同じだけど再掲
定理1の$a_{n}$が超幾何数列であるとする。この時、以下の式が成り立つ。
\begin{eqnarray}
\sum_{n=0}^{\infty}a_{n}&=&\lambda_{0,0}a_{0}+\sum_{n=1}^{\infty}\mu_{n}a_{n}+\sum_{n=0}^{\infty}\{1-\lambda_{0,n}-(\mu_{n+1}-\lambda_{0,n+1})\frac{a_{n+1}}{a_{n}}\}a_{n}
\end{eqnarray}
$\lambda_{0,n}+\lambda_{1,n-1}=\mu_{n},\lambda_{0,n}\coloneqq\nu_{n}$とおくと
\begin{eqnarray}
\sum_{n=0}^{\infty}a_{n}&=&\nu_{0}a_{0}+\sum_{n=1}^{\infty}\mu_{n}a_{n}+\sum_{n=0}^{\infty}\{1-\nu_{n}-(\mu_{n+1}-\nu_{n+1})\frac{a_{n+1}}{a_{n}}\}a_{n}
\end{eqnarray}
以下の式を証明せよ。
\begin{eqnarray}
\left\{
\begin{array}{l}
\zeta(3)=2\sum_{n=1}^{\infty}\frac{n!}{3^{n-1}(3n-1)(\frac{2}{3})_{n-1}n^{3}}+2\sum_{m=1}^{\infty}\frac{(m!)^{4}}{3^{m}(\frac{2}{3})_{m}m}T_{m+1}\\
S_{N}=\frac{2N}{(3N-1)(N!)^{3}}+\frac{2N^{3}}{3N-1}T_{N}+\frac{N^{4}}{3N-1}S_{N+1}\\
T_{N}=\frac{2N-1}{(3N-2)(N!)^{3}}+\frac{2N^{3}}{(3N-2)}(T_{N}+U_{N})+\frac{N^{3}}{3N-2}T_{N+1}\\
U_{N}=\frac{6N^{2}-6N+1}{3(N-1)(3N-2)(N!)^{3}}+\frac{6N^{3}-10N^{2}+5N-1}{3(3N-2)}(V_{N}+2U_{N}+T_{N})+\frac{N-1}{3(3N-2)}U_{N+1}\\
V_{N}=S_{N-1}-\frac{1}{\{(N-1)!\}^{3}}
S_{1}=\zeta(3)\\
T_{1}=\zeta(2)\\
U_{2}=1-\zeta(2)-\zeta(3)+2\\
V_{2}=\zeta(3)-1
\end{array}
\right.
\end{eqnarray}
[1]
\begin{eqnarray}
w_{n}&=&1-\nu_{n}-(\mu_{n+1}-\nu_{n+1})\frac{a_{n+1}}{a_{n}}\\
&=&1-\nu_{n}+\nu_{n+1}\frac{n^{3}}{(n+1)^{3}}-\mu_{n+1}\frac{n^{3}}{(n+1)^{3}}
\end{eqnarray}
[2]$\nu_{n}=An^{2}+Bn+C$とおく。
\begin{align}
&\frac{(1-\nu_{n})(n+1)^{3}+\nu_{n+1}n^{3}}{(n+1)^{3}}\\
&=\frac{\{-An^{2}-Bn+(1-C)\}(n^{3}+3n^{2}+3n+1)+\{An^{2}+(2A+B)n+A+B+C\}n^{3}}{(n+1)^{3}}\\
&=\frac{-An^{4}+(1-2A-2B)n^{3}-(A+3B+3C-3)n^{2}-(B+3C-3)n+1-C}{(n+1)^{3}}
\end{align}
より$A=0,B=\frac{1}{2},C=\frac{1}{2}$として
\begin{equation}
\frac{(1-\nu_{n})(n+1)^{3}+\nu_{n+1}n^{3}}{(n+1)^{3}}=\frac{n+\frac{1}{2}}{(n+1)^{3}}
\end{equation}
$\mu_{n}=0$として
\begin{eqnarray}
w_{n}&=&\frac{n+\frac{1}{2}}{(n+1)^{3}}-\mu_{n+1}\frac{n^{3}}{(n+1)^{3}}\\
&=&\frac{n+\frac{1}{2}-n^{3}\mu_{n+1}}{(n+1)^{3}}
\end{eqnarray}
そこで$\mu_{n+1}=\frac{1}{n^{2}}$として
\begin{eqnarray}
\zeta(3)&=&1+\sum_{n=2}^{\infty}\frac{1}{n^{3}(n-1)^{2}}+\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n^{3}(n+1)^{3}}\\
&=&1+\sum_{n=1}^{\infty}\frac{1}{n^{2}(n+1)^{3}}+\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n^{3}(n+1)^{3}}
\end{eqnarray}
[3]同様に以下の様に置く。
\begin{eqnarray}
1-\nu_{n}+\nu_{n+1}\frac{n^{3}}{(n+N)^{3}}&=&\frac{(1-\nu_{n})(n^{3}+3Nn^{2}+3N^{2}n+N^{3})+\nu_{n+1}n^{3}}{(n+N)^{3}}\\
&=&\frac{(-An^{2}-Bn+1-C)(n^{3}+3Nn^{2}+3N^{2}n+N^{3})+\{An^{2}+(2A+B)n+A+B+C\}n^{3}}{(n+N)^{3}}\\
&=&\frac{-(3N-2)An^{4}-\{(3N^{2}-1)A+(3N-1)B-1\}n^{3}+\{-N^{3}A-3N^{2}B+3N(1-C)\}n^{2}-\{N^{3}B-3N^{2}(1-C)\}n+N^{3}(1-C)}{(n+N)^{3}}
\end{eqnarray}
$A=0,B=\frac{1}{3N-1},C=\frac{2N-1}{3N-1}$
\begin{eqnarray}
1-\nu_{n}+\nu_{n+1}\frac{n^{3}}{(n+N)^{3}}&=&\frac{\frac{2N^{3}}{3N-1}n+\frac{N^{4}}{3N-1}}{(n+N)^{3}}\\
&=&\frac{N^{3}(2n+N)}{(3N-1)(n+N)^{3}}
\end{eqnarray}
を得る。よって以下の様に書ける事がわかる。
\begin{eqnarray}
w_{n}&=&\frac{N^{3}(2n+N)}{(3N-1)(n+N)^{3}}-\mu_{n+1}\frac{n^{3}}{(n+N)^{3}}
\end{eqnarray}
であるから$\mu_{n+1}=\frac{2N^{3}}{(3N-1)n^{2}}$とすると
\begin{eqnarray}
\sum_{n=1}^{\infty}\frac{1}{(n)_{N}^{3}}&=&\frac{2N}{(3N-1)(N!)^{3}}+\frac{2N^{3}}{3N-1}\sum_{n=2}^{\infty}\frac{1}{(n-1)^{2}(n)_{N}^{3}}+\frac{N^{4}}{3N-1}\sum_{n=1}^{\infty}\frac{1}{(n)_{N+1}^{3}}\\
&=&\frac{2N}{(3N-1)(N!)^{3}}+\frac{2N^{3}}{3N-1}\sum_{n=1}^{\infty}\frac{n}{(n)_{N+1}^{3}}+\frac{N^{4}}{3N-1}\sum_{n=1}^{\infty}\frac{1}{(n)_{N+1}^{3}}
\end{eqnarray}
[4]
\begin{eqnarray}
\zeta(3)&=&\sum_{n=1}^{\infty}\frac{1}{n^{3}}\\
&=&\sum_{n=1}^{\infty}\frac{1}{(n)_{1}^{3}}\\
&=&1+\sum_{n=1}^{\infty}\frac{n}{(n)_{2}^{3}}+\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{(n)_{2}^{3}}\\
&=&1+\sum_{n=1}^{\infty}\frac{n}{(n)_{2}^{3}}+\frac{1}{2}\{\frac{1}{10}+\frac{16}{5}\sum_{n=1}^{\infty}\frac{n}{(n)_{3}^{3}}+\frac{16}{5}\sum_{n=1}^{\infty}\frac{1}{(n)_{3}^{3}}\}\\
&=&1+\frac{1}{20}+\sum_{n=1}^{\infty}\frac{n}{(n)_{2}^{3}}+\frac{8}{5}\sum_{n=1}^{\infty}\frac{n}{(n)_{3}^{3}}+\frac{8}{5}\sum_{n=1}^{\infty}\frac{1}{(n)_{3}^{3}}\\
&=&\alpha_{N}+\sum_{m=1}^{N}\beta_{m}\sum_{n=1}^{\infty}\frac{n}{(n)_{m+1}}+\gamma_{N}\sum_{n=1}^{\infty}\frac{1}{(n)_{N+1}}
\end{eqnarray}
とおく。すると
\begin{eqnarray}
\zeta(3)&=&\alpha_{N}+\sum_{m=1}^{N}\beta_{m}\sum_{n=1}^{\infty}\frac{n}{(n)_{m+1}^{3}}+\gamma_{N}\sum_{n=1}^{\infty}\frac{1}{(n)_{N+1}^{3}}\\
&=&\alpha_{N}+\frac{2(N+1)}{(3N+2)\{(N+1)!\}^{3}}\gamma_{N}+\sum_{m=1}^{N}\beta_{m}\sum_{n=1}^{\infty}\frac{n}{(n)_{m+1}^{3}}+\gamma_{N}\frac{2(N+1)^{3}}{3N+2}\sum_{n=1}^{\infty}\frac{n}{(n)_{N+2}^{3}}+\frac{(N+1)^{4}}{3N+2}\gamma_{N}\sum_{n=1}^{\infty}\frac{1}{(n)_{N+2}^{3}}
\end{eqnarray}
ゆえに
\begin{eqnarray}
\left\{
\begin{array}{l}
\alpha_{N+1}=\alpha_{N}+\frac{2(N+1)}{(3N+2)\{(N+1)!\}^{3}}\gamma_{N}\\
\beta_{N+1}=\frac{2(N+1)^{3}}{3N+2}\gamma_{N}\\
\gamma_{N+1}=\frac{(N+1)^{4}}{3N+2}\gamma_{N}
\end{array}
\right.
\end{eqnarray}
これを解くと
\begin{eqnarray}
\left\{
\begin{array}{l}
\alpha_{N}=2\sum_{n=1}^{N}\frac{n!}{3^{n-1}(3n-1)(\frac{2}{3})_{n-1}n^{3}}\\
\beta_{N}=\frac{2}{N}\frac{(N!)^{4}}{3^{N}(\frac{2}{3})_{N}}\\
\gamma_{N}=\frac{(N!)^{4}}{3^{N}(\frac{2}{3})_{N}}
\end{array}
\right.
\end{eqnarray}
ゆえに
\begin{eqnarray}
\zeta(3)&=&2\sum_{n=1}^{N}\frac{n!}{3^{n-1}(3n-1)(\frac{2}{3})_{n-1}n^{3}}+2\sum_{m=1}^{N}\frac{(m!)^{4}}{3^{m}(\frac{2}{3})_{m}m}\sum_{n=1}^{\infty}\frac{n}{(n)_{m+1}^{3}}+\frac{(N!)^{4}}{3^{N}(\frac{2}{3})_{N}}\sum_{n=1}^{\infty}\frac{1}{(n)_{N+1}^{3}}\\
&=&2\sum_{n=1}^{\infty}\frac{n!}{3^{n-1}(3n-1)(\frac{2}{3})_{n-1}n^{3}}+2\sum_{m=1}^{\infty}\frac{(m!)^{4}}{3^{m}(\frac{2}{3})_{m}m}\sum_{n=1}^{\infty}\frac{n}{(n)_{m+1}^{3}}
\end{eqnarray}
[5]これで終わらせるのは勿体無い。そこで$\sum_{n=1}^{\infty}\frac{n}{(n)_{N}^{3}}$に関する漸化式を求める。
\begin{align}
&1-\nu_{n}+\nu_{n+1}\frac{n+1}{n}\frac{n^{3}}{(n+N)^{3}}\\
&=(-An+1-B)+(An+A+B)\frac{n^{2}(n+1)}{(n+N)^{3}}\\
&=\frac{(-An+1-B)(n+N)^{3}+(An+A+B)(n^{3}+n^{2})}{(n+N)^{3}}\\
&=\frac{(-An+1-B)(n^{3}+3Nn^{2}+3N^{2}n+N^{3})+(An+A+B)(n^{3}+n^{2})}{(n+N)^{3}}\\
&=\frac{\{(-3N+2)A+1\}n^{3}+3N\{-NA+(1-B)\}n^{2}+N^{2}\{-NA+3(1-B)\}n+N^{3}(1-B)}{(n+N)^{3}}
\end{align}
より以下の様に記号を定める。
\begin{eqnarray}
\left\{
\begin{array}{l}
A=\frac{1}{3N-2}\\
B=\frac{2(N-1)}{3N-2}
\end{array}
\right.
\end{eqnarray}
すると以下の式が得られる。
\begin{align}
&1-\nu_{n}+\nu_{n+1}\frac{n+1}{n}\frac{n^{3}}{(n+N)^{3}}\\
&=\frac{\frac{2N^{3}}{3N-2}n+\frac{N^{3}}{3N-2}}{(n+N)^{3}}\\
&=\frac{N^{3}(2n+1)}{(3N-2)(n+N)^{3}}
\end{align}
ゆえに$\mu_{n+1}=\frac{2N^{3}}{(3N-2)n^{2}}$
\begin{eqnarray}
\sum_{n=1}^{\infty}\frac{n}{(n)_{N}^{3}}&=&\frac{2N-1}{(3N-2)(N!)^{3}}+\frac{2N^{3}}{(3N-2)}\sum_{n=2}^{\infty}\frac{n}{(n)_{N}^{3}(n-1)^{2}}+\frac{N^{3}}{3N-2}\sum_{n=1}^{\infty}\frac{1}{(n)_{N+1}^{3}}\\
&=&\frac{2N-1}{(3N-2)(N!)^{3}}+\frac{2N^{3}}{(3N-2)}\sum_{n=1}^{\infty}\frac{n(n+1)}{(n)_{N+1}^{3}}+\frac{N^{3}}{3N-2}\sum_{n=1}^{\infty}\frac{1}{(n)_{N+1}^{3}}
\end{eqnarray}
以下次の様な記号を定める。
\begin{eqnarray}
\left\{
\begin{array}{l}
S_{N}\coloneqq\sum_{n=1}^{\infty}\frac{1}{(n)_{N}^{3}}\\
T_{N}\coloneqq\sum_{n=1}^{\infty}\frac{n}{(n)_{N}^{3}}\\
U_{N}\coloneqq\sum_{n=1}^{\infty}\frac{n^{2}}{(n)_{N}^{3}}\\
V_{N}\coloneqq\sum_{n=1}^{\infty}\frac{n^{3}}{(n)_{N}^{3}}=\sum_{n=1}^{\infty}\frac{1}{(n+1)_{N-1}^{3}}=\sum_{n=2}^{\infty}\frac{1}{(n)_{N-1}^{3}}=-\frac{1}{\{(N-1)!\}^{3}}+S_{N-1}\\
\end{array}
\right.
\end{eqnarray}
また以下の計算より
\begin{align}
&1-\nu_{n}+\nu_{n+1}\frac{(n+1)^{2}}{n^{2}}\frac{n^{3}}{(n+N)^{3}}\\
&=(-An+1-B)+(An+A+B)\frac{n(n+1)^{2}}{(n+N)^{3}}\\
&=\frac{(-An+1-B)(n^{3}+3Nn^{2}+3N^{2}n+N^{3})+(An+A+B)(n^{3}+2n^{2}+n)}{(n+N)^{3}}\\
&=\frac{\{-3(N-1)A+1\}n^{3}+\{-3(N^{2}-1)A-(3N-2)B+3N\}n^{2}+\{-(N^{3}-1)A-(3N^{2}-1)B+3N^{2}\}n+N^{3}(1-B)}{(n+N)^{3}}
\end{align}
\begin{eqnarray}
\left\{
\begin{array}{l}
A=\frac{1}{3(N-1)}\\
B=\frac{2N-1}{3N-2}
\end{array}
\right.
\end{eqnarray}
\begin{align}
&1-\nu_{n}+\nu_{n+1}\frac{(n+1)^{2}}{n^{2}}\frac{n^{3}}{(n+N)^{3}}\\
&=\frac{(6N^{3}-10N^{2}+5N-1)n+N-1}{3(3N-2)(n+N)^{3}}
\end{align}
ゆえに$\mu_{n+1}=\frac{6N^{3}-10N^{2}+5N-1}{3(3N-2)n^{2}}$
[6]
\begin{eqnarray}
\left\{
\begin{array}{l}
S_{N}=\frac{2N}{(3N-1)(N!)^{3}}+\frac{2N^{3}}{3N-1}T_{N}+\frac{N^{4}}{3N-1}S_{N+1}\\
T_{N}=\frac{2N-1}{(3N-2)(N!)^{3}}+\frac{2N^{3}}{(3N-2)}(T_{N}+U_{N})+\frac{N^{3}}{3N-2}T_{N+1}\\
U_{N}=\frac{6N^{2}-6N+1}{3(N-1)(3N-2)(N!)^{3}}+\frac{6N^{3}-10N^{2}+5N-1}{3(3N-2)}(V_{N}+2U_{N}+T_{N})+\frac{N-1}{3(3N-2)}U_{N+1}\\
V_{N}=S_{N-1}-\frac{1}{\{(N-1)!\}^{3}}
\end{array}
\right.
\end{eqnarray}
[7]
\begin{eqnarray}
\left\{
\begin{array}{l}
\zeta(3)=2\sum_{n=1}^{\infty}\frac{n!}{3^{n-1}(3n-1)(\frac{2}{3})_{n-1}n^{3}}+2\sum_{m=1}^{\infty}\frac{(m!)^{4}}{3^{m}(\frac{2}{3})_{m}m}T_{m+1}\\
S_{N}=\frac{2N}{(3N-1)(N!)^{3}}+\frac{2N^{3}}{3N-1}T_{N}+\frac{N^{4}}{3N-1}S_{N+1}\\
T_{N}=\frac{2N-1}{(3N-2)(N!)^{3}}+\frac{2N^{3}}{(3N-2)}(T_{N}+U_{N})+\frac{N^{3}}{3N-2}T_{N+1}\\
U_{N}=\frac{6N^{2}-6N+1}{3(N-1)(3N-2)(N!)^{3}}+\frac{6N^{3}-10N^{2}+5N-1}{3(3N-2)}(V_{N}+2U_{N}+T_{N})+\frac{N-1}{3(3N-2)}U_{N+1}\\
V_{N}=S_{N-1}-\frac{1}{\{(N-1)!\}^{3}}
S_{1}=\zeta(3)\\
T_{1}=\zeta(2)\\
U_{2}=1-\zeta(2)-\zeta(3)+2\\
V_{2}=\zeta(3)-1
\end{array}
\right.
\end{eqnarray}
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