ζ(3)の加速級数を積分を使って求めてみた

[1]順次積分加速法を使って級数を変形する。
\begin{eqnarray} \zeta(3)&=&\frac{1}{2}+\sum_{n=1}^{\infty}\{\frac{1}{n^{3}}-\frac{1}{2}(\frac{1}{n^{2}}-\frac{1}{(n+1)^{2}})\}\\ &=&\frac{1}{2}+\sum_{n=1}^{\infty}\{\frac{1}{n^{3}}-\frac{2n+1}{2n^{2}(n+1)^{2}}\}\\ &=&\frac{1}{2}+\frac{1}{2}\sum_{n=1}^{\infty}\frac{3n+2}{n^{3}(n+1)^{2}} \end{eqnarray}
[2]
\begin{align} &\frac{1}{2}\sum_{n=1}^{\infty}\frac{3n+2}{n^{3}(n+1)^{2}}\\ &=\frac{1}{2}-\frac{1}{2}+\frac{1}{4}+\sum_{n=1}^{\infty}[\frac{3n+2}{2n^{3}(n+1)^{2}}-\int_{n}^{n+1}\{\frac{1}{x^{3}}-\frac{1}{2x^{2}}+\frac{1}{2(x+1)^{2}}\}dx]\\ &=\frac{1}{4}+\sum_{n=1}^{\infty}[\frac{3n+2}{2n^{3}(n+1)^{2}}-\frac{1}{2n^{2}}+\frac{1}{2(n+1)^{2}}+\frac{1}{2n}-\frac{1}{2(n+1)}-\frac{1}{2(n+1)}+\frac{1}{2(n+2)}]\\ &=\frac{1}{4}+\sum_{n=1}^{\infty}[\frac{3n+2}{2n^{3}(n+1)^{2}}-\frac{2n+1}{2n^{2}(n+1)^{2}}+\frac{1}{2n(n+1)}-\frac{1}{2(n+1)(n+2)}\\ &=\frac{1}{4}+\sum_{n=1}^{\infty}[\frac{3n+2}{2n^{3}(n+1)^{2}}-\frac{2n+1}{2n^{2}(n+1)^{2}}+\frac{2}{2n(n+1)(n+2)} \end{align}
\begin{align} &(3n+2)(n+2)-(2n+1)n(n+2)+2n^{2}(n+1)\\ &=3n^{2}+8n+4\\ &-2n^{3}-5n^{2}-2n\\ &+2n^{3}+2n^{2}\\ &=6n+4 \end{align}
より
\begin{equation} \frac{1}{2}\sum_{n=1}^{\infty}\frac{3n+2}{n^{3}(n+1)^{2}}=\frac{1}{4}+\sum_{n=1}^{\infty}\frac{3n+2}{n^{3}(n+1)^{2}(n+2)} \end{equation}
なので
\begin{equation} \zeta(3)=\frac{1}{2}+\frac{1}{4}+\sum_{n=1}^{\infty}\frac{3n+2}{n^{3}(n+1)^{2}(n+2)} \end{equation}
[3]ここで止めずにもう一度積分加速法を適用する。
\begin{align} &\frac{3x+2}{x^{3}(x+1)^{2}(x+2)}\\ &=\frac{A}{x^{3}}+\frac{B}{x^{2}}+\frac{C}{x}+\frac{D}{(x+1)^{2}}+\frac{E}{x+1}+\frac{F}{x+2} \end{align}
とすると
\begin{eqnarray} \left\{ \begin{array}{l} \displaystyle A=\lim_{x\rightarrow 0}\frac{3x+2}{(x+1)^{2}(x+2)}=1\\ \displaystyle B=\lim_{x\rightarrow 0}\frac{d}{dx}\frac{3x+2}{(x+1)^{2}(x+2)}=\frac{3}{2}-2\cdot\frac{5}{4}=-1\\ \displaystyle C=\frac{1}{2}\lim_{x\rightarrow 0}\frac{d^{2}}{dx^{2}}\frac{3x+2}{(x+1)^{2}(x+2)}=\frac{1}{2}\\ \displaystyle D=\lim_{x\rightarrow-1}\frac{3x+2}{x^{3}(x+2)}=1\\ \displaystyle E=\lim_{x\rightarrow-1}\frac{d}{dx}\frac{3x+2}{x^{3}(x+2)}=-3+(-1)\cdot(-2)=-1\\ \displaystyle F=\lim_{x\rightarrow-2}\frac{3x+2}{x^{3}(x+1)^{2}}=\frac{1}{2} \end{array} \right. \end{eqnarray}
なので
\begin{align} &\int_{n}^{n+1}\frac{3x+2}{x^{3}(x+1)^{2}(x+2)}dx\\ &=\frac{1}{2n^{2}}-\frac{1}{2(n+1)^{2}}-\frac{1}{n}+\frac{1}{n+1}+\frac{1}{2}\log{\frac{n+1}{n}}+\frac{1}{n+1}-\frac{1}{n+2}-\log{\frac{n+2}{n+1}}+\frac{1}{2}\log{\frac{n+3}{n+2}}\\ &=\frac{2n+1}{2n^{2}(n+1)^{2}}-\frac{1}{n(n+1)}+\frac{1}{(n+1)(n+2)}+\frac{1}{2}\log{\frac{(n+1)^{3}(n+3)}{n(n+2)^{3}}}\\ &=\frac{2n+1}{2n^{2}(n+1)^{2}}-\frac{2}{n(n+1)(n+2)}+\frac{1}{2}\log{\frac{(n+1)^{3}(n+3)}{n(n+2)^{3}}}\\ &=\frac{(2n+1)(n+2)-4n(n+1)}{2n^{2}(n+1)^{2}(n+2)}+\frac{1}{2}\log{\frac{(n+1)^{3}(n+3)}{n(n+2)^{3}}}\\ &=\frac{-2n^{2}+n+2}{2n^{2}(n+1)^{2}(n+2)}+\frac{1}{2}\log{\frac{(n+1)^{3}(n+3)}{n(n+2)^{3}}} \end{align}
を代入すると
\begin{align} &\sum_{n=1}^{\infty}\frac{3n+2}{n^{3}(n+1)^{2}(n+2)}\\ &=\log{2}-\frac{1}{2}\log{3}+\sum_{n=1}^{\infty}[\frac{3n+2}{n^{3}(n+1)^{2}(n+2)}-\{\frac{-2n^{2}+n+2}{2n^{2}(n+1)^{2}(n+2)}+\frac{1}{2}\log{\frac{(n+1)^{3}(n+3)}{n(n+2)^{3}}}\}]\\ &=\log{2}-\frac{1}{2}\log{3}+\sum_{n=1}^{\infty}\{\frac{2n^{3}-n^{2}+4n+4}{2n^{3}(n+1)^{2}(n+2)}-\frac{1}{2}\log{\frac{(n+1)^{3}(n+3)}{n(n+2)^{3}}}\} \end{align}
つまり
\begin{equation} \zeta(3)=\frac{1}{2}+\frac{1}{4}+\log{2}-\frac{1}{2}\log{3}+\sum_{n=1}^{\infty}\{\frac{2n^{3}-n^{2}+4n+4}{2n^{3}(n+1)^{2}(n+2)}-\frac{1}{2}\log{\frac{(n+1)^{3}(n+3)}{n(n+2)^{3}}}\} \end{equation}