美しすぎる二項係数付き級数一覧

1355

Notation

$\begin{array}{r} β_{x} : = \frac{Γ (x + \frac{1}{2})}{Γ (\frac{1}{2}) Γ (x + 1)}, β_{n} = \frac{(\binom{2 n}{n})}{2^{2 n}}, β_{n + \frac{1}{2}} = \frac{2^{2 n}}{π (n + \frac{1}{2}) (\binom{2 n}{n})} \end{array}$

$\begin{array}{r} β_{x}^{r, k} : = γ_{r, k} (x) β_{x}^{r} \end{array}$

\begin{aligned} γ_{2 r, 1} (x) = π (2 x + \frac{1}{2}), & γ_{2 r + 1, - 1} (x) = (- 1)^{⌊ x ⌋} π (2 x + \frac{1}{2}) \\ γ_{5, - 3} (x) = (- 1)^{⌊ x ⌋} π^{3} \frac{(4 x + 1)^{3} + (4 x + 1)}{16} \\ γ_{3, 1} (x) = π^{2} (x + \frac{1}{4}) \end{aligned}

\begin{array}{r} G = \sum_{0 \leq n} \frac{(- 1)^{n}}{(2 n + 1)^{2}}, A = \sum_{0 \leq n} β_{n}^{3} = \frac{π}{Γ {(\frac{3}{4})}^{4}} \end{array}

$\begin{array}{r} \sum_{n \leq k} β_{k + \frac{1}{2}}^{5, - 3} : = \sum_{0 \leq k} (β_{k + \frac{1}{2}}^{5, - 3} - 2 β_{k}^{1, - 1}) - \sum_{k = 0}^{n - 1} β_{k + \frac{1}{2}}^{5, - 3} \end{array}$

Series

$\begin{aligned} \sum_{0 \leq n} β_{n}^{3} (\frac{8 G}{π^{2}} + \sum_{k = 0}^{n - 1} β_{k + \frac{1}{2}}^{2}) & = \sum_{0 \leq n} β_{n}^{3} \\ \sum_{0 \leq n} (- 1)^{n} β_{n}^{3} (\frac{8 G}{π^{2}} + \sum_{k = 0}^{n - 1} β_{k + \frac{1}{2}}^{2}) & = \frac{1}{\sqrt{2}} \sum_{0 \leq n} (- 1)^{n} β_{n}^{3} \end{aligned}$
$\begin{aligned} \sum_{0 \leq n} β_{n + \frac{1}{2}}^{3} \sum_{k = 0}^{n} β_{k}^{4, 1} & = \sum_{0 \leq n} β_{n}^{3} \\ \sum_{0 \leq n} (- 1)^{n} β_{n + \frac{1}{2}}^{3} \sum_{k = 0}^{n} β_{k}^{4, 1} & = \frac{1}{3} \sum_{0 \leq n} (- 1)^{n} β_{n}^{3} \end{aligned}$

$\begin{array}{r} \sum_{0 \leq n} β_{n + \frac{1}{2}}^{3, - 1} = \frac{8 G}{π^{2}} \end{array}$

$\begin{array}{r} \sum_{0 \leq n} (β_{n + \frac{1}{2}}^{4, 1} - \frac{2}{π} \frac{1}{n + 1}) = \frac{6 \ln 2}{π} - \frac{28 ζ (3)}{π^{3}} \end{array}$
$\begin{array}{r} \sum_{0 \leq n} (β_{n + \frac{1}{2}}^{4, 1} {(\sum_{k = 0}^{n} β_{k}^{3, - 1})}^{2} - \frac{2}{π} \frac{1}{n + 1}) = \frac{6 \ln 2}{π} \end{array}$

\begin{aligned} \sum_{0 \leq n} \frac{(- 1)^{n}}{(n + \frac{1}{2})^{4} β_{n}^{3}} \sum_{k = 0}^{n} β_{k}^{3, - 1} + \sum_{0 < n} \frac{(- 1)^{n - 1}}{n^{4} β_{n}^{3}} \sum_{k = 0}^{n - 1} β_{k}^{3, - 1} = 4 π \sum_{0 < n} \frac{1}{n^{2}} \sum_{m = 0}^{n - 1} \frac{(- 1)^{m}}{m + \frac{1}{2}} \end{aligned}

$\begin{array}{r} \sum_{0 \leq n} \frac{1}{(n + \frac{1}{2})^{3} β_{n}^{2}} {(\sum_{k = 0}^{n} β_{k}^{3, - 1})}^{2} + \sum_{0 < n} \frac{1}{n^{3} β_{n}^{2}} {(\sum_{k = 0}^{n - 1} β_{k}^{3, - 1})}^{2} = 28 ζ (3) \end{array}$
$\begin{aligned} \sum_{0 \leq n} \frac{(- 1)^{n}}{(n + \frac{1}{2})^{4} β_{n}^{3}} {(\sum_{k = 0}^{n} β_{k}^{3, - 1})}^{2} + \sum_{0 < n} \frac{(- 1)^{n - 1}}{n^{4} β_{n}^{3}} {(\sum_{k = 0}^{n - 1} β_{k}^{3, - 1})}^{2} & = π^{4} - 4 π \sum_{0 < n} \frac{1}{n^{2}} \sum_{m = 0}^{n - 1} \frac{(- 1)^{m}}{m + \frac{1}{2}} \end{aligned}$
$\begin{array}{r} \sum_{0 \leq n} \frac{1}{(n + \frac{1}{2})^{5} β_{n}^{4}} {(\sum_{k = 0}^{n} β_{k}^{3, - 1})}^{2} + \sum_{0 < n} \frac{1}{n^{5} β_{n}^{4}} {(\sum_{k = 0}^{n - 1} β_{k}^{3, - 1})}^{2} = 16 π^{3} G - 28 π^{2} ζ (3) \end{array}$

\begin{array}{r} \sum_{0 \leq n} \frac{1}{(2 n + \frac{3}{2}) (n + \frac{1}{2})^{4} β_{n}^{4}} (\sum_{k = 0}^{n} β_{k}^{3, 1}) (\sum_{k = 0}^{n} β_{k}^{2}) = \frac{π^{5}}{12} A \end{array}

\begin{array}{r} \sum_{0 \leq n} β_{n + \frac{1}{2}}^{5, - 1} \sum_{k = 0}^{n} β_{k}^{5, - 1} = \frac{A^{2}}{3} \end{array}

\begin{array}{r} \sum_{0 \leq n} β_{n + \frac{1}{2}}^{5, - 1} \sum_{k = 0}^{n} β_{k}^{5, - 1} \sum_{m = 0}^{k - 1} β_{m + \frac{1}{2}}^{3, - 1} \sum_{l = 0}^{m} β_{l}^{3, - 1} = \frac{A^{2}}{30} \end{array}

\begin{array}{r} \sum_{0 \leq n} β_{n + \frac{1}{2}}^{5, - 1} (\sum_{k = 0}^{n} β_{k}^{5, - 1}) (\sum_{k = 0}^{n} β_{k}^{4, 1}) = \frac{A^{2}}{2} \end{array}

\begin{aligned} \sum_{0 \leq n} β_{n + \frac{1}{2}}^{5, - 1} (\sum_{k = 0}^{n} β_{k}^{5, - 1}) (\sum_{k = 0}^{n} β_{k}^{3, - 1}) & = \frac{2}{3} A^{2} \end{aligned}

\begin{array}{r} \sum_{0 \leq n} β_{n + \frac{1}{2}}^{5, - 1} (\sum_{k = 0}^{n} β_{k}^{3, - 1}) (\sum_{k = 0}^{n} β_{k}^{5, - 1} \cdot \frac{1}{π} (2 \ln 2 + \sum_{l = 1}^{2 k} \frac{1}{l})) = \frac{3}{20} A^{2} \end{array}

\begin{array}{r} \sum_{0 \leq n} β_{n + \frac{1}{2}}^{5, - 1} {(\sum_{k = 0}^{n} β_{k}^{3, - 1})}^{3} = \sum_{0 \leq n} (2 β_{n}^{3} - β_{n + \frac{1}{2}}^{3}) \end{array}

\begin{array}{r} \sum_{0 \leq n} β_{n}^{5, - 1} {(\sum_{n \leq k} β_{k + \frac{1}{2}}^{3, - 1})}^{2} = \sum_{0 \leq n} β_{n + \frac{1}{2}}^{5, - 1} {(\sum_{k = 0}^{n} β_{k}^{4, 1})}^{2} = \sum_{0 \leq n} (β_{n}^{3} - β_{n + \frac{1}{2}}^{3}) \end{array}

\begin{array}{r} \sum_{0 \leq n} β_{n + \frac{1}{2}}^{4} (\sum_{k = 0}^{n} β_{k}^{4, 1}) (\sum_{k = 0}^{n} β_{k}^{2}) = \frac{1}{2} \sum_{0 \leq n} β_{n}^{4} \end{array}

\begin{aligned} \frac{1}{A} \sum_{0 \leq n} β_{n}^{6, 1} \sum_{n \leq k} β_{k + \frac{1}{2}}^{5, - 1} & = 2 \sum_{0 \leq n} β_{n}^{4} - \sum_{0 \leq n} β_{n}^{6, 1} \\ A \sum_{0 \leq n} β_{n}^{6, 1} \sum_{n \leq k} β_{k + \frac{1}{2}}^{5, - 3} & = 2 \sum_{0 \leq n} β_{n}^{4} + \sum_{0 \leq n} β_{n}^{6, 1} \end{aligned}

$\begin{array}{r} \sum_{0 \leq n} β_{n + \frac{1}{2}}^{6, 1} \sum_{k = 0}^{n} β_{k}^{6, 1} + 2 \sum_{0 \leq n} β_{n + \frac{1}{2}}^{4} \sum_{k = 0}^{n} β_{k}^{4} = {(\sum_{0 \leq n} β_{n}^{4})}^{2} \end{array}$
$\begin{aligned} \sum_{0 \leq n} β_{n + \frac{1}{2}}^{6, 1} (\sum_{k = 0}^{n} β_{k}^{6, 1}) (\sum_{k = 0}^{n} β_{k}^{3, - 1}) = {(\sum_{0 \leq n} β_{n}^{4})}^{2} - \sum_{0 \leq n} β_{n + \frac{1}{2}}^{4} \sum_{m = 0}^{n} β_{m}^{4} \\ = \sum_{0 \leq n} β_{n + \frac{1}{2}}^{6, 1} (\sum_{k = 0}^{n} β_{k}^{6, 1} \sum_{k \leq m} β_{m + \frac{1}{2}}^{3, - 1}) (\sum_{k = 0}^{n} β_{k}^{4, 1}) \end{aligned}$

$\begin{array}{r} \sum_{0 \leq n} β_{n}^{4} (\frac{8 G}{π^{2}} + \sum_{k = 0}^{n - 1} β_{k + \frac{1}{2}}^{2}) (\frac{28 ζ (3)}{π^{3}} + \sum_{k = 0}^{n - 1} β_{k + \frac{1}{2}}^{4, 1}) = \sum_{0 \leq n} β_{n}^{6, 1} {(\sum_{n \leq k} β_{k + \frac{1}{2}}^{3, - 1})}^{3} + \frac{1}{2} \sum_{0 \leq n} β_{n}^{4} \end{array}$
$\begin{array}{r} \sum_{0 \leq n} β_{n + \frac{1}{2}}^{6, 1} {(\sum_{k = 0}^{n} β_{k}^{3, - 1})}^{3} = \frac{3}{4} \sum_{0 \leq n} β_{n}^{6, 1} \end{array}$
$\begin{array}{r} \sum_{0 \leq n} β_{n + \frac{1}{2}}^{4} {(\sum_{k = 0}^{n} β_{k}^{2})}^{3} = \frac{1}{4} \sum_{0 \leq n} β_{n}^{6, 1} \end{array}$
$\begin{array}{r} \sum_{0 \leq n} β_{n}^{6, 1} {(\frac{8 G}{π^{2}} + \sum_{k = 0}^{n - 1} β_{k + \frac{1}{2}}^{2})}^{2} = \sum_{0 \leq n} β_{n + \frac{1}{2}}^{6, 1} {(\sum_{k = 0}^{n} β_{k}^{2})}^{4} \end{array}$

\begin{array}{r} \sum_{0 \leq n} \frac{(- 1)^{n} β_{n}^{5}}{n + \frac{1}{2}} + \sum_{0 < n} \frac{(- 1)^{n} β_{n}^{5}}{n} = 10 \ln 2 - \frac{π}{2} (A \sum_{0 \leq n} β_{n + \frac{1}{2}}^{5, - 3} + \frac{1}{A} \sum_{0 \leq n} β_{n + \frac{1}{2}}^{5, - 1}) \end{array}

\begin{array}{r} \sum_{0 \leq n} \frac{(- 1)^{n}}{(n + \frac{1}{2})^{6} β_{n}^{5}} + \sum_{0 < n} \frac{(- 1)^{n - 1}}{n^{6} β_{n}^{5}} = \frac{π^{6}}{2} (1 - (\sum_{0 \leq n} β_{n + \frac{1}{2}}^{5, - 3}) (\sum_{0 \leq n} β_{n + \frac{1}{2}}^{5, - 1})) \end{array}

\begin{array}{r} \sum_{0 \leq n} β_{n}^{7, - 1} (\frac{28 ζ (3)}{π^{3}} + \sum_{k = 0}^{n - 1} β_{k + \frac{1}{2}}^{4, 1}) - \sum_{0 \leq n} β_{n}^{7, - 1} \sum_{n \leq k} β_{k + \frac{1}{2}}^{3, - 1} = \sum_{0 \leq n} β_{n + \frac{1}{2}}^{7, - 1} {(\sum_{k = 0}^{n} β_{k}^{4, 1})}^{2} \end{array}

$\begin{aligned} 4 \sum_{0 \leq n} β_{n}^{8, 1} (\frac{28 ζ (3)}{π^{3}} + \sum_{k = 0}^{n - 1} β_{k + \frac{1}{2}}^{4, 1}) - 2 \sum_{0 \leq n} β_{n}^{8, 1} \sum_{n \leq k} β_{k + \frac{1}{2}}^{3, - 1} & = \sum_{0 \leq n} β_{n}^{8, 1} (A \sum_{n \leq k} β_{k + \frac{1}{2}}^{5, - 3} + \frac{1}{A} \sum_{n \leq k} β_{k + \frac{1}{2}}^{5, - 1}) \\ 3 \sum_{0 \leq n} β_{n}^{8, 1} - \sum_{0 \leq n} β_{n}^{8, 1} {(\sum_{n \leq k} β_{k + \frac{1}{2}}^{3, - 1})}^{2} & = \sum_{0 \leq n} β_{n}^{8, 1} (A \sum_{n \leq k} β_{k + \frac{1}{2}}^{5, - 3} - \frac{1}{A} \sum_{n \leq k} β_{k + \frac{1}{2}}^{5, - 1}) \end{aligned}$

$\begin{array}{r} A^{2} \sum_{0 \leq n} β_{n + \frac{1}{2}}^{8, 1} {(\sum_{k = 0}^{n} β_{k}^{5, - 3})}^{2} - \frac{1}{A^{2}} \sum_{0 \leq n} β_{n + \frac{1}{2}}^{8, 1} {(\sum_{k = 0}^{n} β_{k}^{5, - 1})}^{2} = 4 \sum_{0 \leq n} β_{n}^{8, 1} \sum_{n \leq k} β_{k + \frac{1}{2}}^{3, - 1} \end{array}$

$\begin{array}{r} 3 \sum_{0 \leq n} β_{n}^{8, 1} {(\sum_{n \leq k} β_{k + \frac{1}{2}}^{3, - 1})}^{2} - 2 \sum_{0 \leq n} β_{n + \frac{1}{2}}^{8, 1} {(\sum_{k = 0}^{n} β_{k}^{3, - 1})}^{3} = \sum_{0 \leq n} β_{n}^{8, 1} \end{array}$
$\begin{aligned} \sum_{0 \leq n} β_{n + \frac{1}{2}}^{8, 1} {(\sum_{k = 0}^{n} β_{k}^{3, - 1})}^{2} + \sum_{0 \leq n} β_{n}^{8, 1} (\sum_{n \leq k} β_{k + \frac{1}{2}}^{5, - 1}) (\sum_{n \leq k} β_{k + \frac{1}{2}}^{5, - 3}) & = \sum_{0 \leq n} β_{n}^{8, 1} \\ \sum_{0 \leq n} β_{n}^{8, 1} {(\sum_{n \leq k} β_{k + \frac{1}{2}}^{3, - 1})}^{2} + \sum_{0 \leq n} β_{n + \frac{1}{2}}^{8, 1} (\sum_{k = 0}^{n} β_{k}^{5, - 3}) (\sum_{k = 0}^{n} β_{k}^{5, - 1}) & = \sum_{0 \leq n} β_{n}^{8, 1} \end{aligned}$

$\begin{aligned} \sum_{0 \leq n} β_{n}^{9, - 1} {(\sum_{n \leq k} β_{k + \frac{1}{2}}^{3, - 1})}^{3} - \sum_{0 \leq n} β_{n + \frac{1}{2}}^{9, - 1} {(\sum_{k = 0}^{n} β_{k}^{3, - 1})}^{3} \\ = \frac{1}{4} \sum_{0 \leq n} β_{n + \frac{1}{2}}^{9, - 1} {(A \sum_{k = 0}^{n} β_{k}^{5, - 3} - \frac{1}{A} \sum_{k = 0}^{n} β_{k}^{5, - 1})}^{2} \end{aligned}$

$\begin{array}{r} \sum_{0 \leq n} β_{n + \frac{1}{2}}^{10, 1} {(\sum_{k = 0}^{n} β_{k}^{4, 1})}^{3} + \sum_{0 \leq n} β_{n}^{10, 1} (A \sum_{n \leq k} β_{k + \frac{1}{2}}^{5, - 3} + \frac{1}{A} \sum_{n \leq k} β_{k + \frac{1}{2}}^{5, - 1}) = 3 \sum_{0 \leq n} β_{n}^{10, 1} (\frac{28 ζ (3)}{π^{3}} + \sum_{k = 0}^{n - 1} β_{k + \frac{1}{2}}^{4, 1}) \end{array}$

投稿日：2024年7月11日

更新日：2024年7月17日

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美しすぎる二項係数付き級数一覧

Notation
Series