二項係数の3乗が入った二重級数まとめ

453

$\begin{aligned} β_{n} := \frac{(\binom{2 n}{n})}{2^{2 n}}, A := \sum_{n = 0}^{\infty} β_{n}^{3}, B & := \sum_{n = 1}^{\infty} \frac{1}{n^{3} β_{n}^{3}}, C := \sum_{n = 0}^{\infty} \frac{(- 1)^{n}}{{(n + \frac{1}{2})}^{2}} \end{aligned}$
とする.

分母が整数なもの

$\begin{aligned} \sum_{n = 0}^{\infty} β_{n}^{3} & = A \\ \sum_{n = 0}^{\infty} β_{n}^{3} \sum_{k = 1}^{n} \frac{1}{k} & = 2 A (\frac{π}{3} - \ln 2) \\ \sum_{n = 0}^{\infty} β_{n}^{3} \sum_{k = 1}^{n} \frac{1}{k^{2}} & = A (C - \frac{π^{2}}{3}) \\ \sum_{n = 0}^{\infty} β_{n}^{3} \sum_{k = 1}^{n} \frac{1}{k^{3}} & = ? \\ \sum_{n = 0}^{\infty} β_{n}^{3} \sum_{k = 1}^{n} \frac{1}{k^{2} β_{k}} & = \frac{A π^{2}}{2} - \frac{B}{π} \\ \sum_{n = 0}^{\infty} β_{n}^{3} \sum_{k = 1}^{n} \frac{1}{k^{3} β_{k}} & = ? \\ \sum_{n = 0}^{\infty} β_{n}^{3} \sum_{k = 1}^{n} \frac{1}{k^{2} β_{k}^{2}} & = (π^{2} - 2 C) A \\ \sum_{n = 0}^{\infty} β_{n}^{3} \sum_{k = 1}^{n} \frac{1}{k^{3} β_{k}^{2}} & = (π^{3} - 14 ζ (3)) A - B \\ \sum_{n = 0}^{\infty} β_{n}^{3} \sum_{k = 1}^{n} \frac{1}{k^{3} β_{k}^{3}} & = A (B - \frac{A π^{3}}{3}) \end{aligned}$

分母が半整数なもの

$\begin{aligned} \sum_{n = 0}^{\infty} β_{n}^{3} \sum_{k = 0}^{n - 1} \frac{1}{k + \frac{1}{2}} & = \frac{A π}{3} \\ \sum_{n = 0}^{\infty} β_{n}^{3} \sum_{k = 0}^{n - 1} \frac{1}{{(k + \frac{1}{2})}^{2}} & = (\frac{π^{2}}{2} - C) A \\ \sum_{n = 0}^{\infty} β_{n}^{3} \sum_{k = 0}^{n - 1} \frac{1}{{(k + \frac{1}{2})}^{3}} & = ? \\ \sum_{n = 0}^{\infty} β_{n}^{3} \sum_{k = 0}^{n - 1} \frac{1}{{(k + \frac{1}{2})}^{2} β_{k}} & = (2 C - π^{2}) A + \frac{B}{π} \\ \sum_{n = 0}^{\infty} β_{n}^{3} \sum_{k = 0}^{n - 1} \frac{1}{{(k + \frac{1}{2})}^{3} β_{k}} & = ? \\ \sum_{n = 0}^{\infty} β_{n}^{3} \sum_{k = 0}^{n - 1} \frac{1}{{(k + \frac{1}{2})}^{2} β_{k}^{2}} & = (π^{2} - 2 C) A \\ \sum_{n = 0}^{\infty} β_{n}^{3} \sum_{k = 0}^{n - 1} \frac{1}{{(k + \frac{1}{2})}^{3} β_{k}^{2}} & = (28 ζ (3) - π^{3}) A \\ \sum_{n = 0}^{\infty} β_{n}^{3} \sum_{k = 0}^{n - 1} \frac{1}{{(k + \frac{1}{2})}^{3} β_{k}^{3}} & = A (B - \frac{A π^{3}}{3}) \end{aligned}$

$(k + \frac{1}{4}) β_{k}^{3}$ が入っているもの

$\begin{aligned} \sum_{n = 1}^{\infty} \frac{1}{n^{2}} \sum_{k = 0}^{n - 1} (k + \frac{1}{4}) β_{k}^{3} & = A - \frac{1}{A} (\frac{1}{2} + \frac{C}{π^{2}}) \\ \sum_{n = 1}^{\infty} \frac{1}{n^{3}} \sum_{k = 0}^{n - 1} (k + \frac{1}{4}) β_{k}^{3} & = ? \\ \sum_{n = 1}^{\infty} (\frac{1}{n^{2} β_{n}} \sum_{k = 0}^{n - 1} (k + \frac{1}{4}) β_{k}^{3} - \frac{2}{π n}) & = ? \\ \sum_{n = 1}^{\infty} \frac{1}{n^{3} β_{n}} \sum_{k = 0}^{n - 1} (k + \frac{1}{4}) β_{k}^{3} & = ? \\ \sum_{n = 1}^{\infty} \frac{1}{n^{3} β_{n}^{2}} \sum_{k = 0}^{n - 1} (k + \frac{1}{4}) β_{k}^{3} & = ? \\ \sum_{n = 1}^{\infty} (\frac{1}{n^{3} β_{n}^{3}} \sum_{k = 0}^{n - 1} (k + \frac{1}{4}) β_{k}^{3} - \frac{2}{n}) & = 4 \ln 2 - π \\ \sum_{n = 0}^{\infty} \frac{1}{{(n + \frac{1}{2})}^{2}} \sum_{k = 0}^{n} (k + \frac{1}{4}) β_{k}^{3} & = \frac{C}{π^{2} A} + A \\ \sum_{n = 0}^{\infty} \frac{1}{{(n + \frac{1}{2})}^{3}} \sum_{k = 0}^{n} (k + \frac{1}{4}) β_{k}^{3} & = ? \\ \sum_{n = 0}^{\infty} (\frac{1}{{(n + \frac{1}{2})}^{2} β_{n}} \sum_{k = 0}^{n} (k + \frac{1}{4}) β_{k}^{3} - \frac{2}{π (n + 1)}) & = ? \\ \sum_{n = 0}^{\infty} \frac{1}{{(n + \frac{1}{2})}^{3} β_{n}} \sum_{k = 0}^{n} (k + \frac{1}{4}) β_{k}^{3} & = ? \\ \sum_{n = 0}^{\infty} \frac{1}{{(n + \frac{1}{2})}^{3} β_{n}^{2}} \sum_{k = 0}^{n} (k + \frac{1}{4}) β_{k}^{3} & = \frac{1}{A} (\frac{2 C}{π} + π) \\ \sum_{n = 0}^{\infty} (\frac{1}{{(n + \frac{1}{2})}^{3} β_{n}^{3}} \sum_{k = 0}^{n} (k + \frac{1}{4}) β_{k}^{3} - \frac{2}{n + 1}) & = 4 \ln 2 - π \end{aligned}$