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高校数学解説

𝐿𝑖𝑠𝑡 𝑜𝑓 𝒓ₖ,ₙ 𝑎𝑛𝑑 𝒔ₖ,ₙ 𝑓𝑜𝑟 1/𝜋, 𝜋, 𝜁(2), 𝜁(3), 𝛽(2), ･･･

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この記事で示した$r_{k,n},~s_{k,n}$と，それにより得られる級数を示す。

級数	$r_{k,n}$	$s_{k,n}$
$\D\frac{1}{\pi}=\sum_{n=0}^\infty \L(\frac{32n^3}{(2n-1)^2}+\frac{10n+1}{4}\R)\frac{\binom{2n}{n}^3}{2^{12n}}$	$\D\frac{(2k+2n-1)^2}{4(2k+n+1)^2}$	$\D1+\frac{(10k+4n+7)(2k+2n-1)^2}{32(k+1)(2k+n+1)^2}$
$\D\frac{1}{\pi}=\sum_{n=0}^\infty \frac{(6n+1)\binom{2n}{n}^3}{2^{8n+2}}$	$\D\frac{(2n-1)^2}{4(k+n+1)^2}$	$\D\frac{6k+4n+3}{8(k+1)}$
$\D\pi=\sum_{n=0}^\infty \frac{2^{2n+2}(3n+2)}{(4n+1)(4n+3)\binom{4n}{2n}}$	$\D-\frac{2n-1}{4k+2n+1}$	$\D\frac{3k+n+1}{4k+2n+1}$
$\D \pi=\sum_{n=0}^\infty \frac{2(5n+3)}{(3n+1)(3n+2)2^n\binom{3n}{n}}$	$\D-\frac{(k+2n-1)(k+2n)}{(3k+2n)(3k+2n+1)}$	$\D\frac{5k+2n+1}{2(3k+2n)}$
$\D \pi=\sum_{n=0}^\infty \frac{(-1)^n2^{2n+3}(7n+6)}{3(6n+1)(6n+5)}\frac{\binom{2n}{n}}{\binom{6n}{3n}\binom{3n}{n}}$	$\D\frac{(2n+1)(2k+2n+1)}{8(k+n+1)(6k+2n+3)}$	$\D1+\frac{(2k+2n+1)(2k+2n+3)}{3(6k+2n+3)(6k+2n+5)}$
$\D\sum_{n=1}^\infty\frac{1}{(n-a)(n-b)}=\sum_{n=1}^\infty \frac{3n-a-b}{n\binom{2n}{n}}\frac{(1+a-b,1-a+b)_{n-1}}{(1-a,1-b)_n}$	$\D\frac{n(n+x)}{(k+n+1)(k+n+1+x)}$	$\D\frac{3k+2n+1+x}{2(2k+1)}$
$\D \zeta(2)=2\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2\binom{2n}{n}\binom{3n+1}{n}}-2\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^2\binom{2n}{n}\binom{3n}{n}}$	$\D-\frac{(k+n)(2k+n)}{(2k+n+1)(3k+n+1)}$	$\D1-\frac{(k+n)(2k+n)(4k+n+3)}{2(2k+n+1)(3k+n+1)(3k+n+2)}$
$\D\zeta(2)=\sum_{n=1}^\infty \frac{21n-8}{n^3\binom{2n}{n}^3}$	$\D\frac{(k+n)^2}{(2k+n)^2}$	$\D1+\frac{(k+n)^2(5k+2n)}{2(2k-1)(2k+n)^2}$
$\D \zeta(2)=\sum_{n=0}^\infty \frac{1}{(3n+1)^2\binom{2n}{n}\binom{3n}{n}^2}+\sum_{n=1}^\infty \frac{32n-9}{2n^3\binom{2n}{n}\binom{3n}{n}^2}$	$\BA\D r_{2k,n}&=\frac{(2k+n)^2}{(3k+n+1)^2} \\ r_{2k+1,n}&=\frac{(2k+n+1)^2}{(3k+n+2)(3k+n+3)} \EA$	$\BA\D s_{2k,n}&=1+\frac{(2k+n)^2}{(2k+1)(3k+n+1)} \\ s_{2k+1,n}&=1+\frac{(2k+n+1)^2}{2(k+1)(3k+n+2)} \EA$
$\D\zeta(2)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2\binom{2n}{n}}+\frac{3}{2}\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^2\binom{2n}{n}}$	$\D\frac{(k+n)(2k+n)}{(2k+n+1)^2}$	$\D1+\frac{(k+n)(2k+n)(3k+n+2)}{(k+1)(2k+n+1)^2}$
$\D \beta(2)=\sum_{n=1}^\infty \frac{(-1)^{n-1}2^{6n}(4n-1)}{16n^3\binom{2n}{n}^3}$	$\D-\frac{k+2n+1}{k+2n-1}\frac{(2n-1)^3}{(2k+2n+1)^3}$	$\D\frac{4k+2n+1}{2(k+2n-1)}$
$\D \beta(2)=\sum_{n=1}^\infty \frac{(-1)^{n-1}2^{3n}(3n-1)}{2n^3\binom{2n}{n}^3}$	$\D-\frac{3k+2n+1}{3k+2n-1}\frac{(2k+2n-1)^3}{(4k+2n+1)^3}$	$\D1-\frac{6k+2n+3}{2(3k+2n-1)}\frac{(2k+2n-1)^3}{(4k+2n+1)^3}$
$\D \beta(2)=\sum_{n=1}^\infty \frac{(-1)^{n-1}2^{8n-5}(6n-1)}{n^3\binom{2n}{n}\binom{4n}{2n}^2}+\sum_{n=0}^\infty \frac{(-1)^n2^{8n+1}}{(2n+1)^2\binom{2n}{n}\binom{4n+2}{2n+1}^2}$	$\D-\frac{n^2}{(2k+n+1)^2}$	$\D\frac{1}{2}\L(1+\frac{(2k+1)(3k+n+2)}{(2k+n+1)^2}\R)$
$\D\zeta(3)=\frac{5}{2}\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^3\binom{2n}{n}}$	$\D\frac{n(k+n)^2}{(k+n+1)^2(2k+n+1)}$	$\D1+\frac{(k+n)^2}{2(k+1)(2k+n+1)}$
$\D\zeta(3)=\frac{11}{4}\sum_{n=1}^\infty \frac{1}{n^3\binom{2n}{n}^2}+\frac{1}{2}\sum_{n=0}^\infty \frac{1}{(2n+1)^3\binom{2n}{n}^2}$	$\BA\D r_{2k,n}&=\frac{n^2(k+n)}{(k+n+1)(2k+n+1)^2}\\r_{2k+1,n}&=\frac{n^2(k+n)(2k+2n+3)}{(k+n+2)(2k+n+2)^2(2k+2n+1)} \EA$	$\BA\D s_{2k,n}&=\frac{3k+n+1}{2(2k+1)}\\ s_{2k+1,n}&=\frac{(k+n+1)(3k+n+2)}{2(k+1)(2k+2n+1)}\EA$
	$\D \frac{n^2(k+n)}{(k+n+1)(2k+n+1)^2}$	$\D \frac{3k+n+1}{2(2k+1)}+\frac{(k+1)(3k+n+2)}{4(2k+n+1)^2}$

投稿日：2022年10月15日

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𝐿𝑖𝑠𝑡 𝑜𝑓 𝒓ₖ,ₙ 𝑎𝑛𝑑 𝒔ₖ,ₙ 𝑓𝑜𝑟 1/𝜋, 𝜋, 𝜁(2), 𝜁(3), 𝛽(2), ･･･