＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿

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

$$$$

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	$r_{k,n}$	$s_{k,n}$
${\displaystyle \frac{1}{\pi }=\sum _{n=0}^{\mathrm{\infty }}(\frac{32n^3}{(2n-1{)}^2}+\frac{10n+1}{4})\frac{{(\genfrac{}{}{0ex}{}{2n}{n})}^3}{2^{12n}}}$	${\displaystyle \frac{(2k+2n-1{)}^2}{4(2k+n+1{)}^2}}$	${\displaystyle 1+\frac{(10k+4n+7)(2k+2n-1{)}^2}{32(k+1)(2k+n+1{)}^2}}$
${\displaystyle \frac{1}{\pi }=\sum _{n=0}^{\mathrm{\infty }}\frac{(6n+1){(\genfrac{}{}{0ex}{}{2n}{n})}^3}{2^{8n+2}}}$	${\displaystyle \frac{(2n-1{)}^2}{4(k+n+1{)}^2}}$	${\displaystyle \frac{6k+4n+3}{8(k+1)}}$
${\displaystyle \pi =\sum _{n=0}^{\mathrm{\infty }}\frac{2^{2n+2}(3n+2)}{(4n+1)(4n+3)(\genfrac{}{}{0ex}{}{4n}{2n})}}$	${\displaystyle -\frac{2n-1}{4k+2n+1}}$	${\displaystyle \frac{3k+n+1}{4k+2n+1}}$
${\displaystyle \pi =\sum _{n=0}^{\mathrm{\infty }}\frac{2(5n+3)}{(3n+1)(3n+2)2^n(\genfrac{}{}{0ex}{}{3n}{n})}}$	${\displaystyle -\frac{(k+2n-1)(k+2n)}{(3k+2n)(3k+2n+1)}}$	${\displaystyle \frac{5k+2n+1}{2(3k+2n)}}$
${\displaystyle \pi =\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n{2}^{2n+3}(7n+6)}{3(6n+1)(6n+5)}\frac{(\genfrac{}{}{0ex}{}{2n}{n})}{(\genfrac{}{}{0ex}{}{6n}{3n})(\genfrac{}{}{0ex}{}{3n}{n})}}$	${\displaystyle \frac{(2n+1)(2k+2n+1)}{8(k+n+1)(6k+2n+3)}}$	${\displaystyle 1+\frac{(2k+2n+1)(2k+2n+3)}{3(6k+2n+3)(6k+2n+5)}}$
${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{1}{(n-a)(n-b)}=\sum _{n=1}^{\mathrm{\infty }}\frac{3n-a-b}{n(\genfrac{}{}{0ex}{}{2n}{n})}\frac{(1+a-b,1-a+b{)}_{n-1}}{(1-a,1-b{)}_n}}$	${\displaystyle \frac{n(n+x)}{(k+n+1)(k+n+1+x)}}$	${\displaystyle \frac{3k+2n+1+x}{2(2k+1)}}$
${\displaystyle \zeta (2)=2\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{(2n+1{)}^2(\genfrac{}{}{0ex}{}{2n}{n})(\genfrac{}{}{0ex}{}{3n+1}{n})}-2\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n-1}}{n^2(\genfrac{}{}{0ex}{}{2n}{n})(\genfrac{}{}{0ex}{}{3n}{n})}}$	${\displaystyle -\frac{(k+n)(2k+n)}{(2k+n+1)(3k+n+1)}}$	${\displaystyle 1-\frac{(k+n)(2k+n)(4k+n+3)}{2(2k+n+1)(3k+n+1)(3k+n+2)}}$
${\displaystyle \zeta (2)=\sum _{n=1}^{\mathrm{\infty }}\frac{21n-8}{n^3{(\genfrac{}{}{0ex}{}{2n}{n})}^3}}$	${\displaystyle \frac{(k+n{)}^2}{(2k+n{)}^2}}$	${\displaystyle 1+\frac{(k+n{)}^2(5k+2n)}{2(2k-1)(2k+n{)}^2}}$
${\displaystyle \zeta (2)=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(3n+1{)}^2(\genfrac{}{}{0ex}{}{2n}{n}){(\genfrac{}{}{0ex}{}{3n}{n})}^2}+\sum _{n=1}^{\mathrm{\infty }}\frac{32n-9}{2n^3(\genfrac{}{}{0ex}{}{2n}{n}){(\genfrac{}{}{0ex}{}{3n}{n})}^2}}$	$\begin{array}{rl}{\displaystyle r_{2k,n}}& =\frac{(2k+n{)}^2}{(3k+n+1{)}^2}\\ r_{2k+1,n}& =\frac{(2k+n+1{)}^2}{(3k+n+2)(3k+n+3)}\end{array}$	$\begin{array}{rl}{\displaystyle s_{2k,n}}& =1+\frac{(2k+n{)}^2}{(2k+1)(3k+n+1)}\\ s_{2k+1,n}& =1+\frac{(2k+n+1{)}^2}{2(k+1)(3k+n+2)}\end{array}$
${\displaystyle \zeta (2)=\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{(2n+1{)}^2(\genfrac{}{}{0ex}{}{2n}{n})}+\frac{3}{2}\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n-1}}{n^2(\genfrac{}{}{0ex}{}{2n}{n})}}$	${\displaystyle \frac{(k+n)(2k+n)}{(2k+n+1{)}^2}}$	${\displaystyle 1+\frac{(k+n)(2k+n)(3k+n+2)}{(k+1)(2k+n+1{)}^2}}$
${\displaystyle \beta (2)=\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n-1}{2}^{6n}(4n-1)}{16n^3{(\genfrac{}{}{0ex}{}{2n}{n})}^3}}$	${\displaystyle -\frac{k+2n+1}{k+2n-1}\frac{(2n-1{)}^3}{(2k+2n+1{)}^3}}$	${\displaystyle \frac{4k+2n+1}{2(k+2n-1)}}$
${\displaystyle \beta (2)=\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n-1}{2}^{3n}(3n-1)}{2n^3{(\genfrac{}{}{0ex}{}{2n}{n})}^3}}$	${\displaystyle -\frac{3k+2n+1}{3k+2n-1}\frac{(2k+2n-1{)}^3}{(4k+2n+1{)}^3}}$	${\displaystyle 1-\frac{6k+2n+3}{2(3k+2n-1)}\frac{(2k+2n-1{)}^3}{(4k+2n+1{)}^3}}$
${\displaystyle \beta (2)=\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n-1}{2}^{8n-5}(6n-1)}{n^3(\genfrac{}{}{0ex}{}{2n}{n}){(\genfrac{}{}{0ex}{}{4n}{2n})}^2}+\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n{2}^{8n+1}}{(2n+1{)}^2(\genfrac{}{}{0ex}{}{2n}{n}){(\genfrac{}{}{0ex}{}{4n+2}{2n+1})}^2}}$	${\displaystyle -\frac{n^2}{(2k+n+1{)}^2}}$	${\displaystyle \frac{1}{2}(1+\frac{(2k+1)(3k+n+2)}{(2k+n+1{)}^2})}$
${\displaystyle \zeta (3)=\frac{5}{2}\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n-1}}{n^3(\genfrac{}{}{0ex}{}{2n}{n})}}$	${\displaystyle \frac{n(k+n{)}^2}{(k+n+1{)}^2(2k+n+1)}}$	${\displaystyle 1+\frac{(k+n{)}^2}{2(k+1)(2k+n+1)}}$
${\displaystyle \zeta (3)=\frac{11}{4}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^3{(\genfrac{}{}{0ex}{}{2n}{n})}^2}+\frac{1}{2}\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(2n+1{)}^3{(\genfrac{}{}{0ex}{}{2n}{n})}^2}}$	$\begin{array}{rl}{\displaystyle r_{2k,n}}& =\frac{n^2(k+n)}{(k+n+1)(2k+n+1{)}^2}\\ r_{2k+1,n}& =\frac{n^2(k+n)(2k+2n+3)}{(k+n+2)(2k+n+2{)}^2(2k+2n+1)}\end{array}$	$\begin{array}{rl}{\displaystyle s_{2k,n}}& =\frac{3k+n+1}{2(2k+1)}\\ s_{2k+1,n}& =\frac{(k+n+1)(3k+n+2)}{2(k+1)(2k+2n+1)}\end{array}$
	${\displaystyle \frac{n^2(k+n)}{(k+n+1)(2k+n+1{)}^2}}$	${\displaystyle \frac{3k+n+1}{2(2k+1)}+\frac{(k+1)(3k+n+2)}{4(2k+n+1{)}^2}}$

投稿日：2022年10月15日

投稿者

＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿

282

17698

42Followers0Follow