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高校数学解説
級数botⅡの級数計算まとめ
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級数botⅡ が投稿する数式の証明をまとめる。
No. 20
$\BA\D\\
\zeta(3)=\frac{2\pi}{7}\sum_{0\le m\le n}\frac{\binom{2m}{m}^2}{2^{4m}(2n+1)^2}
\EA$
証明
$\BA\D\\
\frac{\pi}{2}\sum_{0\le m\le n}\frac{\binom{2m}{m}^2}{2^{4m}(2n+1)^2}
&=\sum_{0\le n}\frac{1}{(2n+1)^2}\sum_{m=0}^n \frac{\binom{2m}{m}}{2^{2m}}\int_0^1 \frac{x^{2m}}{\sqrt{1-x^2}}\,dx\\
&=\sum_{0\le n}\frac{1}{(2n+1)^2}\int_0^1 \frac{1}{1-x^2}\L(1-(2n+1)\frac{\binom{2n}{n}}{2^{2n}}\int_0^x \frac{t^{2n+1}}{\sqrt{1-t^2}}\,dt\R)dx\\
&=\int_0^1 \frac{1}{1-x^2}\L(\frac{\pi^2}{8}-\int_0^x \frac{\sin^{-1}t}{\sqrt{1-t^2}}\,dt\R)dx\\
&=\frac{1}{2}\int_0^1 \frac{1}{1-x^2}\L(\frac{\pi^2}{4}-(\sin^{-1}x)^2\R)dx\\
&=\frac{1}{2}\int_0^\frac{\pi}{2} \frac{\frac{\pi^2}{4}-x^2}{\cos x}\,dx\\
&=\frac{7}{4}\zeta(3)
\EA$
No. 22
$\BA\D\\
\zeta(3)=9\sum_{0< m< n}\frac{\binom{2m}{m}}{mn^2\binom{2n}{n}}
\EA$
証明
補題
$\BA\D\\
&1.\qquad \sum_{0< k}\frac{2}{k\binom{k+n}{k}^2}=\frac{1}{n}-n\binom{2n}{n}\sum_{n< m}\frac{3}{m^2\binom{2m}{m}}\\
&2.\qquad \sum_{0< k}\frac{1}{k^2\binom{k+n}{k}^2}=\binom{2n}{n}\sum_{n< m}\frac{3}{m^2\binom{2m}{m}}\\
\EA$
$\BA\D\\
S:&=\sum_{0< m< n}\frac{\binom{2m}{m}}{mn^2\binom{2n}{n}}\\
&=\frac{1}{3}\sum_{0< k}\frac{1}{k^2}\sum_{0< m}\frac{1}{m\binom{k+m}{k}^2}\\
&=\frac{1}{6}\sum_{0< k}\frac{1}{k^2}\L(\frac{1}{k}-k\binom{2k}{k}\sum_{k< m}\frac{3}{m^2\binom{2m}{m}}\R)\\
&=\frac{\zeta(3)}{6}-\frac{S}{2}
\EA$
よって
$\BA\D\\
\zeta(3)=9S
\EA$
No. 34
$\BA\D\\
\zeta(4)=4\sum_{0< m< n}\frac{\binom{2m}{m}}{m^2n^2\binom{2n}{n}}+12\sum_{0< k< m< n}\frac{\binom{2m}{m}}{kmn^2\binom{2n}{n}}
\EA$
証明
補題
$\BA\D\\
&1.\qquad \sum_{0< k}\frac{2}{k\binom{k+n}{k}^2}=\frac{1}{n}-n\binom{2n}{n}\sum_{n< m}\frac{3}{m^2\binom{2m}{m}}\\
&2.\qquad \sum_{0< k}\frac{2}{k^3\binom{k+n}{k}^2}=\sum_{n< k}\frac{1}{n^3}+9\sum_{n< k< m}\frac{\binom{2k}{k}}{km^2\binom{2m}{m}}\\
\EA$
$\BA\D\\
\sum_{0< m,n}\frac{2}{mn^3\binom{m+n}{m}^2}
&=\sum_{0< n}\frac{1}{n^3}\L(\frac{1}{n}-n\binom{2n}{n}\sum_{n< m}\frac{3}{m^2\binom{2m}{m}}\R)\\
&=\zeta(4)-3\sum_{0< m< n}\frac{\binom{2m}{m}}{m^2n^2\binom{2n}{n}}\\
\sum_{0< m,n}\frac{2}{mn^3\binom{m+n}{m}^2}
&=\sum_{0< m}\frac{1}{m}\L(\sum_{m< n}\frac{1}{n^3}+9\sum_{m< k< n}\frac{\binom{2k}{k}}{kn^2\binom{2n}{n}}\R)\\
&=\zeta(1,3)+9\sum_{0< k< m< n}\frac{\binom{2m}{m}}{kmn^2\binom{2n}{n}}\\
&=\frac{\zeta(4)}{4}+9\sum_{0< k< m< n}\frac{\binom{2m}{m}}{kmn^2\binom{2n}{n}}
\EA$
よって
$\BA\D\\
\zeta(4)=4\sum_{0< m< n}\frac{\binom{2m}{m}}{m^2n^2\binom{2n}{n}}+12\sum_{0< k< m< n}\frac{\binom{2m}{m}}{kmn^2\binom{2n}{n}}
\EA$
No. 41
$\BA\D\\
\beta(2)=\frac{1}{2}\sum_{0\le m\le n}\frac{2^n}{(2m+1)(2n+1)\binom{2n}{n}}
\EA$
証明
$\BA\D\\
\sum_{0\le m\le n}\frac{2^n}{(2m+1)(2n+1)\binom{2n}{n}}
&=\sum_{0\le m}\frac{1}{2m+1}\frac{2^{2m+1}}{\binom{2m}{m}}\int_0^\frac{1}{\sqrt{2}}\frac{x^{2m}}{\sqrt{1-x^2}}\,dx\\
&=2\int_0^\frac{1}{\sqrt{2}}\frac{1}{\sqrt{1-x^2}}\frac{\sin^{-1}x}{x\sqrt{1-x^2}}\,dx\\
&=2\int_0^\frac{\pi}{4} \frac{x}{\sin x\cos x}\,dx\\
&=\int_0^\frac{\pi}{2}\frac{x}{\sin x}\,dx\\
&=2\beta(2)
\EA$
No. 42
$\BA\D\\
\beta(2)=\frac{1}{2}\sum_{0< m< n}\frac{2^{m+n}}{mn\binom{2n}{n}}
\EA$
証明
$\BA\D\\
\sum_{0< m< n}\frac{2^{m+n}}{mn\binom{2n}{n}}
&=\sum_{0< m}\frac{2^{2m}}{m\binom{2m}{m}}\int_0^\frac{\pi}{2} \frac{\cos x\sin^{2m}x}{\sqrt{1+\cos^2 x}}\,dx\\
&=2\int_0^\frac{\pi}{2} \frac{x\sin x}{\sqrt{1+\cos^2 x}}\,dx\\
&=-2\Big[x\sinh^{-1}\cos x\Big]_0^\frac{\pi}{2}+2\int_0^\frac{\pi}{2}\sinh^{-1}\cos x\,dx\\
&=2\sum_{0\le n}\frac{(-1)^n\binom{2n}{n}}{2^{2n}(2n+1)}\int_0^\frac{\pi}{2} \cos^{2n+1}x\,dx\\
&=2\sum_{0\le n}\frac{(-1)^n}{(2n+1)^2}\\
&=2\beta(2)
\EA$
No. 50
$\BA\D\\
\sum_{0\le n}\frac{\binom{2n}{n}}{2^{4n}(2n+1)^3}=\frac{7\pi^3}{216}
\EA$
証明
$\BA\D\\
\sum_{0\le n}\frac{\binom{2n}{n}}{2^{4n}(2n+1)^3}
&=\sum_{0\le n}\frac{\binom{2n}{n}}{2^{2n}}\int_0^\frac{1}{2} x^{2n}\ln^2\frac{1}{2x}\,dx\\
&=\int_0^\frac{1}{2} \frac{\ln^2\frac{1}{2x}}{\sqrt{1-x^2}}\,dx\\
&=\int_0^\frac{\pi}{6} \ln^2\frac{1}{2\sin x}\,dx\\
&=\int_0^\frac{\pi}{6} \ln^2\frac{i}{e^{ix}-e^{-ix}}\,dx\\
&=\int_1^{e^{\frac{\pi i}{6}}} \ln^2\frac{ix}{x^2-1}\,\frac{dx}{ix}\\
&=2i\int_{e^{\frac{\pi i}{3}}}^1 \frac{1}{x}\ln^2\frac{-x}{(1-x)^2}\,dx
\EA$
$\ln^2$を展開し，各項は多重対数関数で書くことができる．$\D {\rm Li}_3(e^{\frac{\pi i}{3}})=\frac{\zeta(3)}{3}+\frac{5\pi^3i}{162}$などの各特殊値を代入して$\D\frac{7\pi^3}{216}$となる．
投稿日：2022年3月21日
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