二项式系数级数连带奇数倒数平方和 The Binomial Coefficient Series Is Associated with a Sum of Odd Reciprocal Squares

doi:10.12677/PM.2019.97108

Pure Mathematics
Vol. 09 No. 07 ( 2019 ), Article ID: 32262 , 18 pages
10.12677/PM.2019.97108

The Binomial Coefficient Series Is Associated with a Sum of Odd Reciprocal Squares

Zuoyuan Xue¹, Xue Xue², Wanhui Ji¹

●How to Cite this Article

¹Department of Electricity, Yinchuan Energy College, Yinchuan Ningxia

²Binhe Institute of Science and Technology, Yinchuan Ningxia

Received: Aug. 26^th, 2019; accepted: Sep. 16^th, 2019; published: Sep. 23^rd, 2019

ABSTRACT

Using one known series, we can structure several new binomial coefficient series which is associated with a sum of odd reciprocal squares. Their denominator has parity indefinite linear factors 1, 2, 3, 4, 5. Using relation of inverse Trigonometric and Hyperbolic function, we get that alternating the binomial coefficient series is associated with a sum of odd reciprocal squares. The numerical identities of binomial coefficient series with odd reciprocal square are given.

Keywords:Binomial Coefficients, Sum of Odd Reciprocal Squares, Differential, Split Terms, Series

二项式系数级数连带奇数倒数平方和

薛坐远¹，薛雪²，及万会¹

¹银川能源学院电力学院，宁夏银川

²滨河科技学院，宁夏银川

收稿日期：2019年8月26日；录用日期：2019年9月16日；发布日期：2019年9月23日

摘要

根据一个已知级数，使用裂项方法得到分母含奇偶性不定因子 $(m + i) (i = 1, 2, 3, 4, 5)$ 1个，2个，3个，4个，5个线性因子的二项式系数级数连带奇数倒数平方和。利用反正弦与反双曲正弦关系给出交错二项式系数级数连带奇数倒数平方和。所给出级数的和式是封闭形的。并给出二项式系数级数连带奇数倒数平方和数值恒等式。

关键词 :二项式系数，奇数倒数平方和，微分，裂项，级数

This work is licensed under the Creative Commons Attribution International License (CC BY).

http://creativecommons.org/licenses/by/4.0/

1. 引言及引理

二项式系数在数论，图论，统计和概率等数学分支扮演重要角色。二项式系数变换研究有大量文献

[1] - [10] 。文献 [2] [3] [4] [5] 中都提到被称为Lehmer级数恒等式 $\sum_{n = 1}^{\infty} \frac{{(2 x)}^{2 n}}{n (\begin{matrix} 2 n \\ n \end{matrix})} = \frac{2 x \arcsin x}{\sqrt{1 - x^{2}}}$ ， $| x | < 1$ ，一些作者

使用微分，积分，发生函数，白塔–伽马函数，递推等数学工具得到二项式系数倒数级数的重要结果。文 [3] 给出二项式系数倒数数值级数用积分表示，并给出递推公式。文 [4] 给出二项式系数倒数数值级数用反三角函数表示的公式；文 [5] 给出二项式系数倒数数值级数用积分表示。文 [5] [6] [7] [8] 利已知级数裂项构造出一批新二项式系数的倒数级数。文 [9] 利用白塔函数建立非中心型二项式倒数级数。文 [10] 利用“变换核”函数导出无穷级数恒等式。我们利用一个已知级数，用微分裂项法，将分式的化成部分分式经过一定程序转化分母含奇偶性不定的1个，2个，3个，4个，5个线性因子的二项式系数级数连带奇数倒数平方和。利用反正弦与反双曲正弦关系给出交错二项式系数级数连带奇数倒数平方和。所给出级数是封闭形的。并给出二项式系数级数连带奇数倒数平方和数值恒等式。

引理1 [11] $\sum_{k = 1}^{\infty} \frac{(\begin{matrix} 2 k \\ k \end{matrix})}{2^{2 k}} (\sum_{j = 1}^{k} \frac{1}{{(2 j - 1)}^{2}}) \frac{x^{2 k + 1}}{2 k + 1} = \frac{1}{6} (\arcsin^{3} x)$

2. 主要结果和证明

2.1. 定理

设 $B = \frac{\arcsin^{2} x}{2 \sqrt{1 - x^{2}}}$ ，则二项式系数级数连带续奇数倒数平方和

1) 分母含有1个线性因子的二项式系数级数连带奇数倒数平方和

$\sum_{m = 1}^{\infty} \frac{(2 m)!}{2^{2 m} {(m!)}^{2} (m + 1)} (\sum_{j = 1}^{m + 1} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} = (- \frac{2}{x^{2}} + 2) B + 1$ (1)

$\sum_{m = 1}^{\infty} \frac{(2 m)!}{2^{2 m} {(m!)}^{2} (m + 2)} (\sum_{j = 1}^{m + 2} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} = (- \frac{4}{3 x^{4}} + \frac{2}{3 x^{2}} + \frac{2}{3}) B + \frac{2}{3 x^{2}} + \frac{2}{9}$ (2)

$\sum_{m = 1}^{\infty} \frac{(2 m)!}{2^{2 m} {(m!)}^{2} (m + 3)} (\sum_{j = 1}^{m + 3} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} = (- \frac{16}{15 x^{6}} + \frac{8}{15 x^{4}} + \frac{2}{15 x^{2}} + \frac{2}{5}) B + \frac{8}{15 x^{4}} + \frac{8}{45 x^{2}} + \frac{64}{675}$ (3)

$\begin{array}{l} \sum_{m = 1}^{\infty} \frac{(2 m)!}{2^{2 m} {(m!)}^{2} (m + 4)} (\sum_{j = 1}^{m + 4} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} \\ = (\frac{- 32}{35 x^{8}} + \frac{16}{35 x^{6}} + \frac{4}{35 x^{4}} + \frac{2}{35 x^{2}} + \frac{2}{7}) B + \frac{16}{35 x^{6}} + \frac{16}{105 x^{4}} + \frac{128}{1575 x^{2}} + \frac{64}{1225} \end{array}$ (4)

$\begin{array}{l} \sum_{m = 1}^{\infty} \frac{(2 m)!}{2^{2 m} {(m!)}^{2} (m + 5)} (\sum_{j = 1}^{m + 5} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} \\ = (- \frac{256}{315 x^{10}} + \frac{128}{315 x^{8}} + \frac{32}{315 x^{6}} + \frac{16}{315 x^{4}} + \frac{2}{63 x^{2}} + \frac{2}{9}) B \\ + \frac{128}{315 x^{10}} + \frac{128}{945 x^{6}} + \frac{1024}{14175 x^{4}} + \frac{512}{11025 x^{2}} + \frac{16384}{496125} \end{array}$ (5)

2) 分母含有2个线性因子的二项式系数级数连带奇数倒数平方和

$\sum_{m = 1}^{\infty} \frac{(2 m)!}{2^{2 m} {(m!)}^{2} (m + 1) (m + 2)} (\sum_{j = 1}^{m + 2} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} = (\frac{4}{3 x^{4}} - \frac{2}{3 x^{2}} - \frac{2}{3}) B - \frac{2}{3 x^{2}} + \frac{7}{9}$ (6)

$\begin{array}{l} \sum_{m = 1}^{\infty} \frac{(2 m)!}{2^{2 m} {(m!)}^{2} (m + 1) (m + 3)} (\sum_{j = 1}^{m + 2} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} \\ = (\frac{8}{15 x^{6}} - \frac{4}{15 x^{4}} - \frac{16}{15 x^{2}} + \frac{4}{5}) B - \frac{4}{15 x^{4}} - \frac{4}{45 x^{2}} + \frac{611}{1350} \end{array}$ (7)

$\begin{array}{l} \sum_{m = 1}^{\infty} \frac{(2 m)!}{2^{2 m} {(m!)}^{2} (m + 2) (m + 3)} (\sum_{j = 1}^{m + 2} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} \\ = (\frac{16}{15 x^{6}} - \frac{28}{15 x^{4}} + \frac{8}{15 x^{2}} + \frac{4}{15}) B - \frac{8}{15 x^{4}} + \frac{22}{45 x^{2}} + \frac{86}{675} \end{array}$ (8)

$\begin{array}{l} \sum_{m = 1}^{\infty} \frac{(2 m)!}{2^{2 m} {(m!)}^{2} (m + 1) (m + 4)} (\sum_{j = 1}^{m + 4} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} \\ = (\frac{32}{105 x^{8}} - \frac{16}{105 x^{6}} - \frac{4}{105 x^{4}} - \frac{24}{35 x^{2}} + \frac{4}{7}) B - \frac{16}{105 x^{6}} - \frac{16}{315 x^{4}} - \frac{128}{4725 x^{2}} + \frac{387}{1225} \end{array}$ (9)

$\begin{array}{l} \sum_{m = 1}^{\infty} \frac{(2 m)!}{2^{2 m} {(m!)}^{2} (m + 2) (m + 4)} (\sum_{j = 1}^{m + 4} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} \\ = (\frac{16}{35 x^{8}} - \frac{8}{35 x^{6}} - \frac{76}{105 x^{4}} + \frac{32}{105 x^{2}} + \frac{4}{21}) B - \frac{8}{35 x^{6}} - \frac{8}{105 x^{4}} + \frac{461}{1575 x^{2}} + \frac{937}{11025} \end{array}$ (10)

$\begin{array}{l} \sum_{m = 1}^{\infty} \frac{(2 m)!}{2^{2 m} {(m!)}^{2} (m + 3) (m + 4)} (\sum_{j = 1}^{m + 4} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} \\ = (\frac{32}{35 x^{8}} - \frac{32}{21 x^{6}} + \frac{44}{105 x^{4}} + \frac{8}{105 x^{2}} + \frac{4}{35}) B - \frac{16}{35 x^{6}} + \frac{8}{21 x^{4}} + \frac{152}{1575 x^{2}} + \frac{1408}{33075} \end{array}$ (11)

3) 分母含有3个线性因子的二项式系数级数连带奇数倒数平方和

$\begin{array}{l} \sum_{m = 1}^{\infty} \frac{(2 m)!}{2^{2 m} {(m!)}^{2} (m + 1) (m + 2) (m + 3)} (\sum_{j = 1}^{m + 3} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} \\ = (- \frac{8}{15 x^{6}} + \frac{8}{5 x^{4}} - \frac{8}{5 x^{2}} + \frac{8}{15}) B + \frac{4}{15 x^{4}} - \frac{26}{45 x^{2}} + \frac{439}{1350} \end{array}$ (12)

$\begin{array}{l} \sum_{m = 1}^{\infty} \frac{(2 m)!}{2^{2 m} {(m!)}^{2} (m + 1) (m + 2) (m + 4)} (\sum_{j = 1}^{m + 4} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} \\ = (- \frac{16}{105 x^{8}} + \frac{8}{105 x^{6}} + \frac{24}{35 x^{4}} - \frac{104}{105 x^{2}} + \frac{8}{21}) B + \frac{8}{105 x^{6}} + \frac{568}{105 x^{4}} - \frac{1511}{4725 x^{2}} + \frac{2546}{11025} \end{array}$ (13)

$\begin{array}{l} \sum_{m = 1}^{\infty} \frac{(2 m)!}{2^{2 m} {(m!)}^{2} (m + 1) (m + 3) (m + 4)} (\sum_{j = 1}^{m + 4} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} \\ = (- \frac{32}{105 x^{8}} + \frac{24}{35 x^{6}} - \frac{8}{35 x^{4}} - \frac{8}{21 x^{2}} + \frac{8}{35}) B + \frac{16}{105 x^{6}} - \frac{68}{315 x^{4}} - \frac{292}{4725 x^{2}} + \frac{9041}{66150} \end{array}$ (14)

$\begin{array}{l} \sum_{m = 1}^{\infty} \frac{(2 m)!}{2^{2 m} {(m!)}^{2} (m + 2) (m + 3) (m + 4)} (\sum_{j = 1}^{m + 4} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} \\ = (- \frac{16}{35 x^{8}} + \frac{136}{105 x^{6}} - \frac{8}{7 x^{4}} + \frac{8}{35 x^{2}} + \frac{8}{105}) B + \frac{8}{35 x^{6}} - \frac{16}{35 x^{4}} + \frac{103}{525 x^{2}} + \frac{1403}{33075} \end{array}$ (15)

4) 分母含有4个线性因子的二项式系数级数连带奇数倒数平方和

$\begin{array}{l} \sum_{m = 1}^{\infty} \frac{(2 m)!}{2^{2 m} {(m!)}^{2} (m + 1) (m + 2) (m + 3) (m + 4)} (\sum_{j = 1}^{m + 4} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} \\ = (\frac{16}{105 x^{8}} - \frac{64}{105 x^{6}} + \frac{32}{35 x^{4}} - \frac{64}{105 x^{2}} + \frac{16}{105}) B \\ - \frac{8}{105 x^{6}} + \frac{76}{315 x^{4}} - \frac{1219}{4725 x^{2}} + \frac{1247}{13230} \end{array}$ (16)

$\begin{array}{l} \sum_{m = 1}^{\infty} \frac{(2 m)!}{2^{2 m} {(m!)}^{2} (m + 1) (m + 2) (m + 4) (m + 5)} (\sum_{j = 1}^{m + 5} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} \\ = (\frac{64}{945 x^{10}} - \frac{176}{945 x^{8}} + \frac{64}{945 x^{6}} + \frac{32}{135 x^{4}} - \frac{256}{945 x^{2}} + \frac{16}{189}) B \\ - \frac{32}{945 x^{8}} + \frac{184}{2835 x^{6}} + \frac{828}{42525 x^{4}} - \frac{671}{6615 x^{2}} + \frac{311081}{5953500} \end{array}$ (17)

$\begin{array}{l} \sum_{m = 1}^{\infty} \frac{(2 m)!}{2^{2 m} {(m!)}^{2} (m + 1) (m + 3) (m + 4) (m + 5)} (\sum_{j = 1}^{m + 5} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} \\ = (\frac{32}{315 x^{10}} - \frac{16}{45 x^{8}} + \frac{128}{315 x^{6}} - \frac{32}{315 x^{4}} - \frac{32}{315 x^{2}} + \frac{16}{315}) B \\ - \frac{16}{315 x^{8}} + \frac{128}{945 x^{6}} - \frac{1298}{14175 x^{4}} - \frac{766}{33075 x^{2}} + \frac{124031}{3969000} \end{array}$ (18)

$\begin{array}{l} \sum_{m = 1}^{\infty} \frac{(2 m)!}{2^{2 m} {(m!)}^{2} (m + 2) (m + 3) (m + 4) (m + 5)} (\sum_{j = 1}^{m + 5} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} \\ = (\frac{128}{945 x^{10}} - \frac{496}{945 x^{8}} + \frac{704}{945 x^{6}} - \frac{416}{945 x^{4}} + \frac{64}{945 x^{2}} + \frac{16}{945}) B \\ - \frac{64}{945 x^{8}} + \frac{584}{2835 x^{6}} - \frac{8612}{42525 x^{4}} + \frac{1823}{33075 x^{2}} + \frac{2179}{212625} \end{array}$ (19)

5) 分分母含有5个线性因子的二项式系数级数连带奇数倒数平方和

$\begin{array}{l} \sum_{m = 1}^{\infty} \frac{(2 m)!}{2^{2 m} {(m!)}^{2} (m + 1) (m + 2) (m + 3) (m + 4) (m + 5)} (\sum_{j = 1}^{m + 5} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} \\ = (- \frac{32}{945 x^{10}} + \frac{32}{189 x^{8}} - \frac{64}{189 x^{6}} + \frac{64}{189 x^{4}} - \frac{32}{189 x^{2}} + \frac{32}{945}) B \\ + \frac{16}{945 x^{8}} - \frac{40}{567 x^{6}} + \frac{674}{6075 x^{4}} - \frac{863}{11025 x^{2}} + \frac{250069}{11907000} \end{array}$ (20)

推论1 设 $b = \frac{- \ln^{2} (x + \sqrt{1 + x^{2}})}{2 \sqrt{1 + x^{2}}}$ ，则交错二项式系数连带续奇数倒数平方和级数

1) 分母含有1个因子交错的二项式系数级数连带奇数倒数平方和

$\sum_{m = 1}^{\infty} \frac{{(- 1)}^{m} (2 m)!}{2^{2 m} {(m!)}^{2} (m + 1)} (\sum_{j = 1}^{m + 1} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} = (\frac{2}{x^{2}} + 2) b + 1$ (21)

$\sum_{m = 1}^{\infty} \frac{{(- 1)}^{m} (2 m)!}{2^{2 m} {(m!)}^{2} (m + 2)} (\sum_{j = 1}^{m + 2} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} = (- \frac{4}{3 x^{4}} - \frac{2}{3 x^{2}} + \frac{2}{3}) b - \frac{2}{3 x^{2}} + \frac{2}{9}$ (22)

$\sum_{m = 1}^{\infty} \frac{{(- 1)}^{m} (2 m)!}{2^{2 m} {(m!)}^{2} (m + 3)} (\sum_{j = 1}^{m + 3} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} = (\frac{16}{15 x^{6}} + \frac{8}{15 x^{4}} - \frac{2}{15 x^{2}} + \frac{2}{5}) b + \frac{8}{15 x^{4}} - \frac{8}{45 x^{2}} + \frac{64}{675}$ (23)

$\begin{array}{l} \sum_{m = 1}^{\infty} \frac{{(- 1)}^{m} (2 m)!}{2^{2 m} {(m!)}^{2} (m + 4)} (\sum_{j = 1}^{m + 4} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} \\ = (\frac{- 32}{35 x^{8}} - \frac{16}{35 x^{6}} + \frac{4}{35 x^{4}} - \frac{2}{35 x^{2}} + \frac{2}{7}) b - \frac{16}{35 x^{6}} + \frac{16}{105 x^{4}} - \frac{128}{1575 x^{2}} + \frac{64}{1225} \end{array}$ (24)

$\begin{array}{l} \sum_{m = 1}^{\infty} \frac{{(- 1)}^{m} (2 m)!}{2^{2 m} {(m!)}^{2} (m + 5)} (\sum_{j = 1}^{m + 5} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} \\ = (\frac{256}{315 x^{10}} + \frac{128}{315 x^{8}} - \frac{32}{315 x^{6}} + \frac{16}{315 x^{4}} - \frac{2}{63 x^{2}} + \frac{2}{9}) b \\ + \frac{128}{315 x^{8}} - \frac{128}{945 x^{6}} + \frac{1024}{14175 x^{4}} - \frac{512}{11025 x^{2}} + \frac{16384}{496125} \end{array}$ (25)

2) 母含有2个因子交错的二项式系数级数连带奇数倒数平方和

$\sum_{m = 1}^{\infty} \frac{{(- 1)}^{m} (2 m)!}{2^{2 m} {(m!)}^{2} (m + 1) (m + 2)} (\sum_{j = 1}^{m + 2} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} = (\frac{4}{3 x^{4}} + \frac{2}{3 x^{2}} - \frac{2}{3}) b + \frac{2}{3 x^{2}} + \frac{7}{9}$ (26)

$\begin{array}{l} \sum_{m = 1}^{\infty} \frac{{(- 1)}^{m} (2 m)!}{2^{2 m} {(m!)}^{2} (m + 1) (m + 3)} (\sum_{j = 1}^{m + 2} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} \\ = (- \frac{8}{15 x^{6}} - \frac{4}{15 x^{4}} + \frac{16}{15 x^{2}} + \frac{4}{5}) b - \frac{4}{15 x^{4}} + \frac{4}{45 x^{2}} + \frac{611}{1350} \end{array}$ (27)

$\begin{array}{l} \sum_{m = 1}^{\infty} \frac{{(- 1)}^{m} (2 m)!}{2^{2 m} {(m!)}^{2} (m + 2) (m + 3)} (\sum_{j = 1}^{m + 2} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} \\ = (- \frac{16}{15 x^{6}} - \frac{28}{15 x^{4}} - \frac{8}{15 x^{2}} + \frac{4}{15}) b - \frac{8}{15 x^{4}} - \frac{22}{45 x^{2}} + \frac{86}{675} \end{array}$ (28)

$\begin{array}{l} \sum_{m = 1}^{\infty} \frac{{(- 1)}^{m} (2 m)!}{2^{2 m} {(m!)}^{2} (m + 1) (m + 4)} (\sum_{j = 1}^{m + 4} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} \\ = (\frac{32}{105 x^{8}} + \frac{16}{105 x^{6}} - \frac{4}{105 x^{4}} + \frac{24}{35 x^{2}} + \frac{4}{7}) b + \frac{16}{105 x^{6}} - \frac{16}{315 x^{4}} + \frac{128}{4725 x^{2}} + \frac{387}{1225} \end{array}$ (29)

$\begin{array}{l} \sum_{m = 1}^{\infty} \frac{{(- 1)}^{m} (2 m)!}{2^{2 m} {(m!)}^{2} (m + 2) (m + 4)} (\sum_{j = 1}^{m + 4} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} \\ = (\frac{16}{35 x^{8}} + \frac{8}{35 x^{6}} - \frac{76}{105 x^{4}} - \frac{32}{105 x^{2}} + \frac{4}{21}) b + \frac{8}{35 x^{6}} - \frac{8}{105 x^{4}} - \frac{461}{1575 x^{2}} + \frac{937}{11025} \end{array}$ (30)

$\begin{array}{l} \sum_{m = 1}^{\infty} \frac{{(- 1)}^{m} (2 m)!}{2^{2 m} {(m!)}^{2} (m + 3) (m + 4)} (\sum_{j = 1}^{m + 4} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} \\ = (\frac{32}{35 x^{8}} + \frac{32}{21 x^{6}} + \frac{44}{105 x^{4}} - \frac{8}{105 x^{2}} + \frac{4}{35}) b + \frac{16}{35 x^{6}} + \frac{8}{21 x^{4}} - \frac{152}{1575 x^{2}} + \frac{1408}{33075} \end{array}$ (31)

3) 母含有3个因子交错的二项式系数级数连带奇数倒数平方和

$\begin{array}{l} \sum_{m = 1}^{\infty} \frac{{(- 1)}^{m} (2 m)!}{2^{2 m} {(m!)}^{2} (m + 1) (m + 2) (m + 3)} (\sum_{j = 1}^{m + 3} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} \\ = (\frac{8}{15 x^{6}} + \frac{8}{5 x^{4}} + \frac{8}{5 x^{2}} + \frac{8}{15}) b + \frac{4}{15 x^{4}} + \frac{26}{45 x^{2}} + \frac{439}{1350} \end{array}$ (32)

$\begin{array}{l} \sum_{m = 1}^{\infty} \frac{{(- 1)}^{m} (2 m)!}{2^{2 m} {(m!)}^{2} (m + 1) (m + 2) (m + 4)} (\sum_{j = 1}^{m + 4} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} \\ = (- \frac{16}{105 x^{8}} - \frac{8}{105 x^{6}} + \frac{24}{35 x^{4}} + \frac{104}{105 x^{2}} + \frac{8}{21}) b - \frac{8}{105 x^{6}} + \frac{568}{105 x^{4}} + \frac{1511}{4725 x^{2}} + \frac{2546}{11025} \end{array}$ (33)

$\begin{array}{l} \sum_{m = 1}^{\infty} \frac{{(- 1)}^{m} (2 m)!}{2^{2 m} {(m!)}^{2} (m + 1) (m + 3) (m + 4)} (\sum_{j = 1}^{m + 4} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} \\ = (- \frac{32}{105 x^{8}} - \frac{24}{35 x^{6}} - \frac{8}{35 x^{4}} + \frac{8}{21 x^{2}} + \frac{8}{35}) b - \frac{16}{105 x^{6}} - \frac{68}{315 x^{4}} + \frac{292}{4725 x^{2}} + \frac{9041}{66150} \end{array}$ (34)

$\begin{array}{l} \sum_{m = 1}^{\infty} \frac{{(- 1)}^{m} (2 m)!}{2^{2 m} {(m!)}^{2} (m + 2) (m + 3) (m + 4)} (\sum_{j = 1}^{m + 4} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} \\ = (- \frac{16}{35 x^{8}} - \frac{136}{105 x^{6}} - \frac{8}{7 x^{4}} - \frac{8}{35 x^{2}} + \frac{8}{105}) b - \frac{8}{35 x^{6}} - \frac{16}{35 x^{4}} - \frac{103}{525 x^{2}} + \frac{1403}{33075} \end{array}$ (35)

4) 分母含有4个因子交错的二项式系数级数连带奇数倒数平方和

$\begin{array}{l} \sum_{m = 1}^{\infty} \frac{{(- 1)}^{m} (2 m)!}{2^{2 m} {(m!)}^{2} (m + 1) (m + 2) (m + 3) (m + 4)} (\sum_{j = 1}^{m + 4} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} \\ = (\frac{16}{105 x^{8}} + \frac{64}{105 x^{6}} + \frac{32}{35 x^{4}} + \frac{64}{105 x^{2}} + \frac{16}{105}) b + \frac{8}{105 x^{6}} + \frac{76}{315 x^{4}} + \frac{1219}{4725 x^{2}} + \frac{1247}{13230} \end{array}$ (36)

$\begin{array}{l} \sum_{m = 1}^{\infty} \frac{{(- 1)}^{m} (2 m)!}{2^{2 m} {(m!)}^{2} (m + 1) (m + 2) (m + 4) (m + 5)} (\sum_{j = 1}^{m + 5} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} \\ = (- \frac{64}{945 x^{10}} + \frac{176}{945 x^{8}} - \frac{64}{945 x^{6}} + \frac{32}{135 x^{4}} + \frac{256}{945 x^{2}} + \frac{16}{189}) b \\ - \frac{32}{945 x^{8}} - \frac{184}{2835 x^{6}} + \frac{828}{42525 x^{4}} + \frac{671}{6615 x^{2}} + \frac{311081}{5953500} \end{array}$ (37)

$\begin{array}{l} \sum_{m = 1}^{\infty} \frac{{(- 1)}^{m} (2 m)!}{2^{2 m} {(m!)}^{2} (m + 1) (m + 3) (m + 4) (m + 5)} (\sum_{j = 1}^{m + 5} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} \\ = (- \frac{32}{315 x^{10}} - \frac{16}{45 x^{8}} - \frac{128}{315 x^{6}} - \frac{32}{315 x^{4}} + \frac{32}{315 x^{2}} + \frac{16}{315}) b \\ - \frac{16}{315 x^{8}} - \frac{128}{945 x^{6}} - \frac{1298}{14175 x^{4}} + \frac{756}{33075 x^{2}} + \frac{124031}{3969000} \end{array}$ (38)

$\begin{array}{l} \sum_{m = 1}^{\infty} \frac{{(- 1)}^{m} (2 m)!}{2^{2 m} {(m!)}^{2} (m + 2) (m + 3) (m + 4) (m + 5)} (\sum_{j = 1}^{m + 5} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} \\ = (- \frac{128}{945 x^{10}} - \frac{496}{945 x^{8}} - \frac{704}{945 x^{6}} - \frac{416}{945 x^{4}} - \frac{64}{945 x^{2}} + \frac{16}{945}) b \\ - \frac{64}{945 x^{8}} - \frac{584}{2835 x^{6}} - \frac{8612}{42525 x^{4}} - \frac{1823}{33075 x^{2}} + \frac{2179}{212625} \end{array}$ (39)

5) 分母含有5个因子交错的二项式系数级数连带奇数倒数平方和

$\begin{array}{l} \sum_{m = 1}^{\infty} \frac{{(- 1)}^{m} (2 m)!}{2^{2 m} {(m!)}^{2} (m + 1) (m + 2) (m + 3) (m + 4) (m + 5)} (\sum_{j = 1}^{m + 5} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} \\ = (\frac{32}{945 x^{10}} + \frac{32}{189 x^{8}} + \frac{64}{189 x^{6}} + \frac{64}{189 x^{4}} + \frac{32}{189 x^{2}} + \frac{32}{945}) b \\ + \frac{16}{945 x^{8}} + \frac{40}{567 x^{6}} + \frac{674}{6075 x^{4}} + \frac{863}{11025 x^{2}} + \frac{250069}{11907000} \end{array}$ (40)

2.2. 定理的证明

由文 [11] 公式 $\sum_{k = 1}^{\infty} \frac{(\begin{matrix} 2 k \\ k \end{matrix})}{2^{2 k}} (\sum_{j = 1}^{k} \frac{1}{{(2 j - 1)}^{2}}) \frac{x^{2 k + 1}}{2 k + 1} = \frac{1}{6} (\arcsin^{3} x)$ 两端关于x微分

$\sum_{k = 1}^{\infty} \frac{(\begin{matrix} 2 k \\ k \end{matrix}) x^{2 k}}{2^{2 k}} (\sum_{j = 1}^{k} \frac{1}{{(2 j - 1)}^{2}}) = \frac{1}{2} \frac{\arcsin^{2} x}{\sqrt{1 - x^{2}}} = B$ (41)

1) 对(41)式左端裂项

$\frac{x^{2}}{2} + \sum_{k = 1}^{\infty} \frac{(2 k - 2)! (2 k - 1) 2 k}{2^{2 k} {((k - 1)!)}^{2} {(k)}^{2}} (\sum_{j = 1}^{k} \frac{1}{{(2 j - 1)}^{2}}) x^{2 k} = B$ 令 $k - 1 = m$ ，化成

$\frac{x^{2}}{2} + \sum_{m = 1}^{\infty} \frac{(2 m)! (2 m + 1) (2 m + 2)}{2^{2 m + 2} {(m!)}^{2} {(m + 1)}^{2}} (\sum_{j = 1}^{m + 1} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m + 2} = B$ ，两端同乘以 $\frac{2}{x^{2}}$ ，得出

$1 + \sum_{m = 1}^{\infty} \frac{(2 m)! (2 m + 1)}{2^{2 m} {(m!)}^{2} (m + 1)} (\sum_{j = 1}^{m + 1} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} = \frac{2}{x^{2}} B$

$1 + \sum_{m = 1}^{\infty} \frac{(2 m)!}{2^{2 m} {(m!)}^{2}} (2 - \frac{1}{m + 1}) (\sum_{j = 1}^{m + 1} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} = \frac{2}{x^{2}} B$ ，整理得到如下(1)式，令其为 $B_{1}$

$\sum_{m = 1}^{\infty} \frac{(2 m)!}{2^{2 m} {(m!)}^{2} (m + 1)} (\sum_{j = 1}^{m + 1} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} = (- \frac{2}{x^{2}} + 2) B + 1$

2) 对(41)式左端裂项

$\frac{x^{2}}{2} + \frac{5 x^{4}}{12} + \sum_{k = 1}^{\infty} \frac{(2 k - 4)! (2 k - 3) (2 k - 2) (2 k - 1) (2 k)}{2^{2 k} {((k - 2)!)}^{2} {(k - 1)}^{2} {(k)}^{2}} (\sum_{j = 1}^{k} \frac{1}{{(2 j - 1)}^{2}}) x^{2 k} = B$ ，令 $k - 2 = m$

$\frac{x^{2}}{2} + \frac{5 x^{4}}{12} + \sum_{m = 1}^{\infty} \frac{(2 m)! (2 m + 1) (2 m + 2) (2 m + 3) (2 m + 4)}{2^{2 m + 4} {(m!)}^{2} {(m + 1)}^{2} {(m + 2)}^{2}} (\sum_{j = 1}^{m + 2} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m + 4} = B$

两端同乘以 $\frac{4}{x^{4}}$ 得

$\frac{2}{x^{2}} + \frac{5}{3} + \sum_{m = 1}^{\infty} \frac{(2 m)! (2 m + 1) (2 m + 3)}{2^{2 m} {(m!)}^{2} (m + 1) (m + 2)} (\sum_{j = 1}^{m + 2} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} = \frac{4}{x^{4}} B$

$\frac{2}{x^{2}} + \frac{5}{3} + \sum_{m = 1}^{\infty} \frac{(2 m)!}{2^{2 m} {(m!)}^{2}} (2 - \frac{1}{m + 1}) (2 - \frac{1}{m + 2}) (\sum_{j = 1}^{m + 2} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} = \frac{4}{x^{4}} B$

$\frac{2}{x^{2}} + \frac{5}{3} + \sum_{m = 1}^{\infty} \frac{(2 m)!}{2^{2 m} {(m!)}^{2}} (4 - \frac{2}{m + 1} - \frac{2}{m + 2} + \frac{1}{(m + 1) (m + 2)}) (\sum_{j = 1}^{m + 2} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} = \frac{4}{x^{4}} B$ (42)

a) (42)式化成部分分式，得出

$\frac{2}{x^{2}} + \frac{5}{3} + \sum_{m = 1}^{\infty} \frac{(2 m)!}{2^{2 m} {(m!)}^{2}} (4 - \frac{1}{m + 1} - \frac{3}{m + 2}) (\sum_{j = 1}^{m + 2} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} = \frac{4}{x^{4}} B$

由于 $B, B_{1}$ 已知，计算得出下面(2)式，并令其为 $B_{2}$

b) 在(42)式， $B, B_{1}, B_{2}$ 已知，易得2个因子乘积的二项式系数级数连带奇数倒数平方和公式(6)。

3) 对(41)式左端裂项

$\frac{x^{2}}{2} + \frac{5 x^{4}}{12} + \frac{259 x^{6}}{720} + \sum_{k = 1}^{\infty} \frac{(2 k - 6)! (2 k - 5) \dots (2 k - 1) (2 k)}{2^{2 k} {((k - 3)!)}^{2} {(k - 2)}^{2} {(k - 1)}^{2} {(k)}^{2}} (\sum_{j = 1}^{k} \frac{1}{{(2 j - 1)}^{2}}) x^{2 k} = B$

令 $k - 2 = m$ ，

$\frac{x^{2}}{2} + \frac{5 x^{4}}{12} + \frac{259 x^{6}}{720} + \sum_{k = 1}^{\infty} \frac{(2 m)! (2 m + 1) (2 m + 2) (2 m + 3) (2 m + 4) (2 m + 5) (2 m + 6)}{2^{2 k + 6} {(m!)}^{2} {(m + 1)}^{2} {(m + 2)}^{2} {(m + 3)}^{2}} (\sum_{j = 1}^{m + 3} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m + 6} = B$

两端同乘以 $\frac{8}{x^{6}}$ 得

$\frac{4}{x^{4}} + \frac{10}{3 x^{2}} + \frac{259}{90} + \sum_{k = 1}^{\infty} \frac{(2 m)! (2 m + 1) (2 m + 3) (2 m + 5)}{2^{2 k} {(m!)}^{2} (m + 1) (m + 2) (m + 3)} (\sum_{j = 1}^{m + 3} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} = \frac{8}{x^{6}} B$

$\frac{4}{x^{4}} + \frac{10}{3 x^{2}} + \frac{259}{90} + \sum_{m = 1}^{\infty} \frac{(2 m)!}{2^{2 m} {(m!)}^{2}} (2 - \frac{1}{m + 1}) (2 - \frac{1}{m + 2}) (2 - \frac{1}{m + 3}) (\sum_{j = 1}^{m + 3} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} = \frac{8}{x^{6}} B$

$\begin{array}{l} \frac{4}{x^{4}} + \frac{10}{3 x^{2}} + \frac{259}{90} + \sum_{m = 1}^{\infty} \frac{(2 m)!}{2^{2 m} {(m!)}^{2}} (8 - \frac{4}{m + 1} - \frac{4}{m + 2} - \frac{4}{m + 3} + \frac{2}{(m + 1) (m + 2)} \\ + \frac{2}{(m + 1) (m + 3)} + \frac{2}{(m + 2) (m + 3)} - \frac{1}{(m + 1) (m + 2) (m + 3)}) (\sum_{j = 1}^{m + 3} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} = \frac{8}{x^{6}} B \end{array}$ (43)

a) 将上式所有分式化成部分分式，得出

$\frac{4}{x^{4}} + \frac{10}{3 x^{2}} + \frac{259}{90} + \sum_{m = 1}^{\infty} \frac{(2 m)!}{2^{2 m} {(m!)}^{2}} (8 - \frac{3 / 2}{m + 1} - \frac{3}{m + 2} - \frac{15 / 2}{m + 3}) (\sum_{j = 1}^{m + 3} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} = \frac{8}{x^{6}} B$

由于 $B, B_{1}, B_{2}$ 已知，计算得出下面(3)式，并令其为 $B_{3}$

为行文简便，今后将 $\sum_{m = 1}^{\infty} \frac{2 m! x^{2 m}}{2^{2 m} {(m!)}^{2} (m + 1) (m + 2)} (\sum_{j = 1}^{m + 2} \frac{1}{{(2 j - 1)}^{2}})$ ，

$\sum_{m = 1}^{\infty} \frac{2 m! x^{2 m}}{2^{2 m} {(m!)}^{2} (m + 1) (m + 2) (m + 3)} (\sum_{j = 1}^{m + 2} \frac{1}{{(2 j - 1)}^{2}})$ 等用符号 $B_{12}$ ， $B_{123}$ 表示。

以 $B_{3}$ 证明级数连带奇数倒数平方和上标增加1项或几项不改变其和式收敛性。

$\begin{array}{l} \sum_{m = 1}^{\infty} \frac{(2 m)!}{2^{2 m} {(m!)}^{2} (m + 3)} (\sum_{j = 1}^{m + 3} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} \\ = (- \frac{16}{15 x^{6}} + \frac{8}{15 x^{4}} + \frac{2}{15 x^{2}} + \frac{2}{5}) B + \frac{8}{15 x^{4}} + \frac{8}{45 x^{2}} + \frac{64}{675} \end{array}$

$\begin{array}{l} \sum_{m = 1}^{\infty} \frac{(2 m)!}{2^{2 m} {(m!)}^{2} (m + 3)} (\sum_{j = 1}^{m + 3} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} \\ = \sum_{m = 1}^{\infty} \frac{(2 m)! x^{2 m}}{2^{2 m} {(m!)}^{2} (m + 3)} (\sum_{j = 1}^{m} \frac{1}{{(2 j - 1)}^{2}} + \frac{1}{{[2 (m + 1) - 1]}^{2}} + \frac{1}{{[2 (m + 2) - 1]}^{2}} + \frac{1}{{[2 (m + 3) - 1]}^{2}}) \\ = \sum_{m = 1}^{\infty} \frac{(2 m)! x^{2 m}}{2^{2 m} {(m!)}^{2} (m + 3)} (\sum_{j = 1}^{m} \frac{1}{{(2 j - 1)}^{2}}) \\ + \frac{2 m! x^{2 m}}{2^{2 m} {(m!)}^{2} (m + 3)} (\frac{1}{{[2 (m + 1) - 1]}^{2}} + \frac{1}{{[2 (m + 2) - 1]}^{2}} + \frac{1}{{[2 (m + 3) - 1]}^{2}}) \end{array}$

前和式为二项式系数级数连带奇数倒数平方和表达式，后式极限趋于0。

根据阶乘斯特林渐进公式 $n! \sim \sqrt{2 n π} n^{n} e^{- n}$ ，注意到变量 $| x | < 1$ ，

$\frac{2 m!}{{(m!)}^{2}} \sim \frac{\sqrt{4 π m} {(2 m)}^{2 m} e^{- 2 m}}{2 m π m^{m} e^{- m} m^{m} e^{- m}} \frac{{(2 m)}^{2 m}}{\sqrt{m π} {(m)}^{2 m}} = \frac{2^{2 m}}{\sqrt{m π}}$

$\begin{array}{l} \lim_{m \to \infty} \frac{2 m! x^{2 m}}{2^{2 m} (m!) (m + 3)} (\frac{1}{{[2 (m + 1) - 1]}^{2}} + \frac{1}{{[2 (m + 2) - 1]}^{2}} + \frac{1}{{[2 (m + 3) - 1]}^{2}}) \\ = \lim_{m \to \infty} \frac{x^{2 m}}{\sqrt{m π} (m + 3)} (\frac{1}{{[2 (m + 1) - 1]}^{2}} + \frac{1}{{[2 (m + 2) - 1]}^{2}} + \frac{1}{{[2 (m + 3) - 1]}^{2}}) \\ = 0 \end{array}$

因此 $\begin{array}{l} \sum_{m = 1}^{\infty} \frac{(2 m)!}{2^{2 m} {(m!)}^{2} (m + 3)} (\sum_{j = 1}^{m + 3} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} \\ = \sum_{m = 1}^{\infty} \frac{(2 m)!}{2^{2 m} {(m!)}^{2} (m + 3)} (\sum_{j = 1}^{m} \frac{1}{{(2 j - 1)}^{2}} + 0) x^{2 m} \\ = \sum_{m = 1}^{\infty} \frac{(2 m)!}{2^{2 m} {(m!)}^{2} (m + 3)} (\sum_{j = 1}^{m} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} \end{array}$

所以，级数连带奇数倒数平方和上标增加1项或几项不改变其和式收敛性。

b) 在(43)保留2个因子分式，(在(41)式中出现的2个因子分式不再计算)，对这些2个因子的分式，每次保留1个，其余化成部分分式，得到：

$\frac{4}{x^{4}} + \frac{10}{3 x^{2}} + \frac{259}{90} + B_{13} + 8 B - 2 B_{1} - 3 B_{2} - 7 B_{3} = \frac{8}{x^{6}} B$

$\frac{4}{x^{4}} + \frac{10}{3 x^{2}} + \frac{259}{90} + B_{23} + 8 B - 4 B_{1} - 3 B_{2} - \frac{13}{2} B_{3} = \frac{8}{x^{6}} B$

由于 $B, B_{1}, B_{2}, B_{3}$ 已知，计算得出公式(7)~(8)

c) 在(43)保留3个因子分式，其他分式化成部分分式得到：

$\frac{4}{x^{4}} + \frac{10}{3 x^{2}} + \frac{259}{90} + B_{123} + 8 B - 2 B_{1} - 2 B_{2} - 8 B_{3} = \frac{8}{x^{6}} B$

由于 $B, B_{1}, B_{2}, B_{3}$ 已知，计算得出公式(12)式。

4) 对(41)式左端裂项

$\frac{x^{2}}{2} + \frac{5 x^{4}}{12} + \frac{259 x^{6}}{720} + \frac{3229 x^{8}}{10080} + \sum_{k = 1}^{\infty} \frac{(2 k - 8)! (2 k - 7) \dots (2 k - 1) (2 k)}{2^{2 k} {((k - 4)!)}^{2} {(k - 3)}^{2} {(k - 2)}^{2} {(k - 1)}^{2} {(k)}^{2}} (\sum_{j = 1}^{k} \frac{1}{{(2 j - 1)}^{2}}) x^{2 k} = B$ ， $k - 4 = m$

$\begin{array}{l} \frac{x^{2}}{2} + \frac{5 x^{4}}{12} + \frac{259 x^{6}}{720} + \frac{3229 x^{8}}{10080} \\ + \sum_{m = 1}^{\infty} \frac{(2 m)! (2 m + 1) (2 m + 2) (2 m + 3) \dots (2 m + 6) (2 m + 7) (2 m + 8)}{2^{2 m + 8} {(m!)}^{2} {(m + 1)}^{2} {(m + 2)}^{2} {(m + 3)}^{2} {(m + 4)}^{2}} (\sum_{j = 1}^{m + 4} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m + 8} = B \end{array}$

$\frac{x^{2}}{2} + \frac{5 x^{4}}{12} + \frac{259 x^{6}}{720} + \frac{3229 x^{8}}{10080} + \sum_{m = 1}^{\infty} \frac{8 (2 m)! (2 m + 1) (2 m + 3) (2 m + 5) (2 m + 7)}{2^{2 m + 8} {(m!)}^{2} (m + 1) (m + 2) (m + 3) (m + 4)} (\sum_{j = 1}^{m + 4} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m + 8} = B$ ，两端同乘以 $\frac{16}{x^{8}}$ 得

$\begin{array}{l} \frac{8}{x^{6}} + \frac{20}{3 x^{4}} + \frac{259}{45 x^{2}} + \frac{3229}{630} \\ + \sum_{m = 1}^{\infty} \frac{(2 m)!}{2^{2 m} {(m!)}^{2}} (2 - \frac{1}{m + 1}) (2 - \frac{1}{m + 2}) (2 - \frac{1}{m + 3}) (2 - \frac{1}{m + 4}) (\sum_{j = 1}^{m + 4} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} = \frac{16}{x^{8}} B \end{array}$

$\begin{array}{l} \frac{8}{x^{6}} + \frac{20}{3 x^{4}} + \frac{259}{45 x^{2}} + \frac{3229}{630} + \sum_{m = 1}^{\infty} \frac{(2 m)!}{2^{2 m} {(m!)}^{2}} (16 - \frac{8}{m + 1} - \frac{8}{m + 2} - \frac{8}{m + 3} - \frac{8}{m + 4} \\ + \frac{4}{(m + 1) (m + 2)} + \frac{4}{(m + 1) (m + 3)} + \frac{4}{(m + 2) (m + 3)} + \frac{4}{(m + 1) (m + 4)} + \frac{4}{(m + 2) (m + 4)} \\ + \frac{4}{(m + 3) (m + 4)} + \frac{- 2}{(m + 1) (m + 2) (m + 3)} + \frac{- 2}{(m + 1) (m + 2) (m + 4)} + \frac{- 2}{(m + 1) (m + 3) (m + 4)} \\ + \frac{- 2}{(m + 2) (m + 3) (m + 4)} + \frac{1}{(m + 1) (m + 2) (m + 3) (m + 4)}) (\sum_{j = 1}^{m + 4} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} = \frac{16}{x^{8}} B \end{array}$ (44)

a) 将上式所有分式化成部分分式，得出

$\frac{8}{x^{6}} + \frac{20}{3 x^{4}} + \frac{259}{45 x^{2}} + \frac{3229}{630} + \sum_{m = 1}^{\infty} \frac{(2 m)! x^{2 m}}{2^{2 m} {(m!)}^{2}} (16 - \frac{5 / 2}{m + 1} - \frac{9 / 2}{m + 2} - \frac{15 / 2}{m + 3} - \frac{35 / 2}{m + 4}) (\sum_{j = 1}^{m + 4} \frac{1}{{(2 j - 1)}^{2}}) = \frac{16}{x^{8}} B$

由于 $B, B_{1}, B_{2}, B_{3}$ 已知，计算得出下面(4)式，并令其为 $B_{4}$

$\begin{array}{l} \sum_{m = 1}^{\infty} \frac{(2 m)! x^{2 m}}{2^{2 m} {(m!)}^{2} (m + 4)} (\sum_{j = 1}^{m + 4} \frac{1}{{(2 j - 1)}^{2}}) \\ = (\frac{- 32}{35 x^{8}} + \frac{16}{35 x^{6}} + \frac{4}{35 x^{4}} + \frac{2}{35 x^{2}} + \frac{2}{7}) B + \frac{16}{35 x^{6}} + \frac{16}{105 x^{4}} + \frac{128}{1575 x^{2}} + \frac{64}{1225} \end{array}$

b) 在(44)式，其余保留2个因子分式，(在(43)式中出现的2个因子分式不再计算)然后对这些2个因子的分式，每次保留1个，其余化成部分分式，得到：

$\frac{8}{x^{6}} + \frac{20}{3 x^{4}} + \frac{259}{45 x^{2}} + \frac{3229}{630} + 16 B + B_{14} - \frac{17}{5} B_{1} - \frac{9}{2} B_{2} - \frac{15}{2} B_{3} - \frac{103}{6} B_{4} = \frac{16}{x^{8}} B$

$\frac{8}{x^{6}} + \frac{20}{3 x^{4}} + \frac{259}{45 x^{2}} + \frac{3229}{630} + 16 B + B_{24} - \frac{5}{2} B_{1} - 5 B_{2} - \frac{15}{2} B_{3} - 17 B_{4} = \frac{16}{x^{8}} B$

$\begin{array}{l} \frac{8}{x^{6}} + \frac{20}{3 x^{4}} + \frac{259}{45 x^{2}} + \frac{3229}{630} + 16 B + B_{34} \\ - \frac{5}{2} B_{1} - \frac{9}{2} B_{2} - \frac{17}{2} B_{3} - \frac{33}{2} B_{4} = \frac{16}{x^{8}} B \end{array}$

由于 $B, B_{1}, B_{2}, B_{3}, B_{4}$ 已知，计算得出公式(9)~(11)式。

c) 在(44)保留3个因子分式(在(43)式出现的3个因子分式不再计算)，然后对这些3个因子的分式，每次保留1个，其余化成部分分式，得到：

$\begin{array}{l} \frac{8}{x^{6}} + \frac{20}{3 x^{4}} + \frac{259}{45 x^{2}} + \frac{3229}{630} + 16 B + B_{124} \\ - \frac{17}{6} B_{1} - 4 B_{2} - \frac{15}{2} B_{3} - \frac{53}{3} B_{4} = \frac{16}{x^{8}} B \end{array}$

$\begin{array}{l} \frac{8}{x^{6}} + \frac{20}{3 x^{4}} + \frac{259}{45 x^{2}} + \frac{3229}{630} + 16 B + B_{134} \\ - \frac{8}{3} B_{1} - \frac{9}{2} B_{2} - 7 B_{3} - \frac{107}{6} B_{4} = \frac{16}{x^{8}} B \end{array}$

$\frac{8}{x^{6}} + \frac{20}{3 x^{4}} + \frac{259}{45 x^{2}} + \frac{3229}{630} + 16 B + B_{234} - \frac{5}{2} B_{1} - 5 B_{2} - \frac{13}{2} B_{3} - 18 B_{4} = \frac{16}{x^{8}} B$

由于 $B, B_{1}, B_{2}, B_{3}, B_{4}, B_{5}$ 已知，计算得出(13)~(15)式。

d) 在(44)保留4个因子分式，然后对这些2个因子的分式，每次保留1个，其余化成部分分式，得到：

$\frac{8}{x^{6}} + \frac{20}{3 x^{4}} + \frac{259}{45 x^{2}} + \frac{3229}{630} + 16 B + B_{1234} - \frac{8}{3} B_{1} - 4 B_{2} - 8 B_{3} - \frac{52}{3} B_{4} = \frac{16}{x^{8}} B$

由于 $B, B_{1}, B_{2}, B_{3}, B_{4}$ ，已知，计算得出公式(16)式。

5) 对(41)式左端裂项

$\begin{array}{l} \frac{x^{2}}{2} + \frac{5 x^{4}}{12} + \frac{259 x^{6}}{720} + \frac{3229 x^{8}}{10080} + \frac{117469}{403200} x^{10} \\ + \sum_{k = 1}^{\infty} \frac{(2 k - 10)! (2 k - 9) \dots (2 k - 1) (2 k)}{2^{2 k} {((k - 5)!)}^{2} {(k - 4)}^{2} {(k - 3)}^{2} {(k - 2)}^{2} {(k - 1)}^{2} {(k)}^{2}} (\sum_{j = 1}^{k} \frac{1}{{(2 j - 1)}^{2}}) x^{2 k} = B \end{array}$ ， $k - 5 = m$

$\begin{array}{l} \frac{x^{2}}{2} + \frac{5 x^{4}}{12} + \frac{259 x^{6}}{720} + \frac{3229 x^{8}}{10080} + \frac{117469}{403200} x^{10} \\ + \sum_{m = 1}^{\infty} \frac{(2 m)! (2 m + 1) (2 m + 2) (2 m + 3) \dots (2 m + 8) (2 m + 9) (2 m + 10)}{2^{2 m + 10} {(m!)}^{2} {(m + 1)}^{2} {(m + 2)}^{2} {(m + 3)}^{2} {(m + 4)}^{2} {(m + 5)}^{2}} (\sum_{j = 1}^{m + 5} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m + 10} = B \end{array}$

两端同乘以 $\frac{32}{x^{10}}$ 得

$\begin{array}{l} \frac{16}{x^{8}} + \frac{40}{3 x^{6}} + \frac{518}{45 x^{4}} + \frac{3229}{315 x^{2}} + \frac{117469}{12600} \\ + \sum_{m = 1}^{\infty} \frac{(2 m)!}{2^{2 m} {(m!)}^{2}} (2 - \frac{1}{m + 1}) (2 - \frac{1}{m + 2}) (2 - \frac{1}{m + 3}) (2 - \frac{1}{m + 4}) (2 - \frac{1}{m + 5}) (\sum_{j = 1}^{m + 5} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} = \frac{32}{x^{10}} B \end{array}$

展开乘积表达式

$\begin{array}{l} \frac{16}{x^{8}} + \frac{40}{3 x^{6}} + \frac{518}{45 x^{4}} + \frac{3229}{315 x^{2}} + \frac{117469}{12600} + \sum_{m = 1}^{\infty} \frac{(2 m)! x^{2 m}}{2^{2 m} {(m!)}^{2}} [32 - \frac{16}{m + 1} - \frac{16}{m + 2} - \frac{16}{m + 3} \\ - \frac{16}{m + 4} - \frac{16}{m + 5} + \frac{8}{(m + 1) (m + 2)} + \frac{8}{(m + 1) (m + 3)} + \frac{8}{(m + 2) (m + 3)} + \frac{8}{(m + 1) (m + 4)} \\ + \frac{8}{(m + 2) (m + 4)} + \frac{8}{(m + 3) (m + 4)} + \frac{8}{(m + 1) (m + 5)} + \frac{8}{(m + 2) (m + 5)} + \frac{8}{(m + 3) (m + 5)} \\ + \frac{8}{(m + 4) (m + 5)} + \frac{- 4}{(m + 1) (m + 2) (m + 3)} + \frac{- 4}{(m + 1) (m + 2) (m + 4)} + \frac{- 4}{(m + 1) (m + 3) (m + 4)} \\ + \frac{- 4}{(m + 2) (m + 3) (m + 4)} + \frac{- 4}{(m + 1) (m + 2) (m + 5)} + \frac{- 4}{(m + 1) (m + 3) (m + 5)} \end{array}$

(45)

在(45)式有1个因子分式，2个因子分式4个，3个因子分式6个，4个因子分式4个，5个因子分式1个。为使论文简短写，我们仅选1个因子分式，4个因子分式，5个因子分式，通过如下操作程序：将分式化成分母为1个因子。4个因子(在(44)式出现的4个因子分式不再计算)，5个因子乘积的二项式系数连带连续奇数倒数平方和级数

a) 将(45)式所有分式化成部分分式，得出：

$\begin{array}{l} \frac{16}{x^{8}} + \frac{40}{3 x^{6}} + \frac{518}{45 x^{4}} + \frac{3229}{315 x^{2}} + \frac{117469}{12600} \\ + \sum_{m = 1}^{\infty} \frac{(2 m)!}{2^{2 m} {(m!)}^{2}} (32 - \frac{35 / 8}{m + 1} - \frac{15 / 2}{m + 2} - \frac{45 / 4}{m + 3} - \frac{35 / 2}{m + 4} - \frac{315 / 8}{m + 5}) (\sum_{j = 1}^{m + 5} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} = \frac{32}{x^{10}} B \end{array}$

$\begin{array}{l} \frac{128}{315 x^{8}} + \frac{320}{945 x^{6}} + \frac{4144}{14175 x^{4}} + \frac{25832}{99225 x^{2}} + \frac{117469}{496125} \\ + \sum_{m = 1}^{\infty} \frac{(2 m)!}{2^{2 m} {(m!)}^{2}} (\frac{256}{315} - \frac{1 / 9}{m + 1} - \frac{4 / 21}{m + 2} - \frac{2 / 7}{m + 3} - \frac{4 / 9}{m + 4} - \frac{1}{m + 5}) (\sum_{j = 1}^{m + 5} \frac{1}{{(2 j - 1)}^{2}}) x^{2 m} = \frac{256}{315 x^{10}} B \end{array}$

由于 $B, B_{1}, B_{2}, B_{3}, B_{4}$ 已知，计算得出下面(5)式，并令其为 $B_{5}$

b) 在(45)式保留4个因子分式，然后对这些4个因子的分式，每次保留1个，其余化成部分分式，得到：

$\frac{16}{x^{8}} + \frac{40}{3 x^{6}} + \frac{518}{45 x^{4}} + \frac{3229}{315 x^{2}} + \frac{117469}{12600} + 32 B + B_{1245} - \frac{107}{24} B_{1} - \frac{22}{3} B_{2} - \frac{45}{4} B_{3} - \frac{53}{3} B_{4} - \frac{943}{24} B_{5} = \frac{32}{x^{10}} B$

$\frac{16}{x^{8}} + \frac{40}{3 x^{6}} + \frac{518}{45 x^{4}} + \frac{3229}{315 x^{2}} + \frac{117469}{12600} + 32 B + B_{1345} - \frac{53}{12} B_{1} - \frac{15}{2} B_{2} - 11 B_{3} - \frac{107}{6} B_{4} - \frac{157}{4} B_{5} = \frac{32}{x^{10}} B$

$\frac{16}{x^{8}} + \frac{40}{3 x^{6}} + \frac{518}{45 x^{4}} + \frac{3229}{315 x^{2}} + \frac{117469}{12600} + 32 B + B_{2345} - \frac{35}{8} B_{1} - \frac{23}{3} B_{2} - \frac{43}{4} B_{3} - 18 B_{4} - \frac{941}{24} B_{5} = \frac{32}{x^{10}} B$

由于 $B, B_{1}, B_{2}, B_{3}, B_{4}, B_{5}$ 已知，计算得出(17)~(19)式。

c) 在(45)式保留5个因子分式，其余化成部分分式，得到：

$\frac{16}{x^{8}} + \frac{40}{3 x^{6}} + \frac{518}{45 x^{4}} + \frac{3229}{315 x^{2}} + \frac{117469}{12600} + 32 B + B_{12345} - \frac{53}{12} B_{1} - \frac{22}{3} B_{2} - \frac{23}{2} B_{3} - \frac{52}{3} B_{4} - \frac{473}{12} B_{5} = \frac{32}{x^{10}} B$

由于 $B, B_{1}, B_{2}, B_{3}, B_{4}, B_{5}$ 已知，计算得出(20)式。定理证毕。

2.3. 推论的证明

在定理的公式用ix代替x，注意到 $\arcsin (i x) = i \arcsin h x$ ，

$\arcsin h x = \ln (x + \sqrt{1 + x^{2}})$ ， $\frac{\arcsin^{2} (i x)}{2 \sqrt{1 - {(i x)}^{2}}} = \frac{- \ln^{2} (x + \sqrt{1 + x^{2}})}{2 \sqrt{1 + x^{2}}} = b$ ，推论成立。

3. 一些数值级数

3.1. 数值级数

在定理公式(1)~(20)，令 $x = 1 / \sqrt{2} = \sqrt{2} / 2$ ， $\arcsin x = \frac{π}{4}$ ， $B = \frac{\sqrt{2}}{32} π^{2}$

则分母为奇偶性不定线性因子乘积的二项式系数级数连带奇数倒数平方和数值恒等式

1) $\sum_{m = 1}^{\infty} \frac{2 m!}{2^{3 m} {(m!)}^{2} (m + 1)} (\sum_{j = 1}^{m} \frac{1}{{(2 j - 1)}^{2}}) = - \frac{\sqrt{2}}{16} π^{2} + 1$ ；

2) $\sum_{m = 1}^{\infty} \frac{2 m!}{2^{3 m} {(m!)}^{2} (m + 2)} (\sum_{j = 1}^{m + 1} \frac{1}{{(2 j - 1)}^{2}}) = - \frac{\sqrt{2}}{16} π^{2} + \frac{14}{9}$ ；

3) $\sum_{m = 1}^{\infty} \frac{2 m!}{2^{3 m} {(m!)}^{2} (m + 3)} (\sum_{j = 1}^{m + 2} \frac{1}{{(2 j - 1)}^{2}}) = - \frac{43 \sqrt{2}}{240} π^{2} + \frac{1729}{675}$ ；

4) $\sum_{m = 1}^{\infty} \frac{2 m!}{2^{3 m} {(m!)}^{2} (m + 4)} (\sum_{j = 1}^{m + 3} \frac{1}{{(2 j - 1)}^{2}}) = - \frac{177 \sqrt{2}}{560} π^{2} + \frac{49408}{11025}$ ；

5) $\sum_{m = 1}^{\infty} \frac{2 m!}{2^{3 m} {(m!)}^{2} (m + 5)} (\sum_{j = 1}^{m + 4} \frac{1}{{(2 j - 1)}^{2}}) = - \frac{2867 \sqrt{2}}{5040} π^{2} + \frac{1323008}{165375}$ ；

6) $\sum_{m = 1}^{\infty} \frac{2 m!}{2^{3 m} {(m!)}^{2} (m + 1) (m + 2)} (\sum_{j = 1}^{m + 1} \frac{1}{{(2 j - 1)}^{2}}) = \frac{5 \sqrt{2}}{48} π^{2} - \frac{5}{9}$ ；

7) $\sum_{m = 1}^{\infty} \frac{2 m!}{2^{3 m} {(m!)}^{2} (m + 1) (m + 3)} (\sum_{j = 1}^{m + 2} \frac{1}{{(2 j - 1)}^{2}}) = \frac{7 \sqrt{2}}{120} π^{2} - \frac{1069}{1350}$ ；

8) $\sum_{m = 1}^{\infty} \frac{2 m!}{2^{3 m} {(m!)}^{2} (m + 2) (m + 3)} (\sum_{j = 1}^{m + 2} \frac{1}{{(2 j - 1)}^{2}}) = \frac{3 \sqrt{2}}{40} π^{2} - \frac{694}{675}$ ；

9) ；

10) $\sum_{m = 1}^{\infty} \frac{2 m!}{2^{3 m} {(m!)}^{2} (m + 2) (m + 4)} (\sum_{j = 1}^{m + 3} \frac{1}{{(2 j - 1)}^{2}}) = \frac{73 \sqrt{2}}{840} π^{2} - \frac{16229}{11025}$ ；

11) $\sum_{m = 1}^{\infty} \frac{2 m!}{2^{3 m} {(m!)}^{2} (m + 3) (m + 4)} (\sum_{j = 1}^{m + 3} \frac{1}{{(2 j - 1)}^{2}}) = \frac{23 \sqrt{2}}{168} π^{2} - \frac{62768}{33075}$ ；

12) $\sum_{m = 1}^{\infty} \frac{2 m!}{2^{3 m} {(m!)}^{2} (m + 1) (m + 2) (m + 3)} (\sum_{j = 1}^{m + 2} \frac{1}{{(2 j - 1)}^{2}}) = - \frac{\sqrt{2}}{60} π^{2} + \frac{319}{1350}$ ；

13) $\sum_{m = 1}^{\infty} \frac{2 m!}{2^{3 m} {(m!)}^{2} (m + 1) (m + 2) (m + 4)} (\sum_{j = 1}^{m + 3} \frac{1}{{(2 j - 1)}^{2}}) = - \frac{3 \sqrt{2}}{140} π^{2} + \frac{722324}{33075}$ ；

14) $\sum_{m = 1}^{\infty} \frac{2 m!}{2^{3 m} {(m!)}^{2} (m + 1) (m + 3) (m + 4)} (\sum_{j = 1}^{m + 3} \frac{1}{{(2 j - 1)}^{2}}) = - \frac{11 \sqrt{2}}{420} π^{2} + \frac{4877}{13230}$ ；

15) $\sum_{m = 1}^{\infty} \frac{2 m!}{2^{3 m} {(m!)}^{2} (m + 2) (m + 3) (m + 4)} (\sum_{j = 1}^{m + 3} \frac{1}{{(2 j - 1)}^{2}}) = - \frac{13 \sqrt{2}}{240} π^{2} + \frac{44621}{33075}$ ；

16) $\sum_{m = 1}^{\infty} \frac{2 m!}{2^{3 m} {(m!)}^{2} (m + 1) (m + 2) (m + 3) (m + 4)} (\sum_{j = 1}^{m + 3} \frac{1}{{(2 j - 1)}^{2}}) = \frac{\sqrt{2}}{210} π^{2} - \frac{1459}{22050}$ ；

17) $\sum_{m = 1}^{\infty} \frac{(2 m)!}{2^{3 m} {(m!)}^{2} (m + 1) (m + 2) (m + 4) (m + 5)} (\sum_{j = 1}^{m + 4} \frac{1}{{(2 j - 1)}^{2}}) = \frac{13 \sqrt{2}}{1890} π^{2} - \frac{587439}{5953500}$ ；

18) $\sum_{m = 1}^{\infty} \frac{(2 m)!}{2^{3 m} {(m!)}^{2} (m + 1) (m + 3) (m + 4) (m + 5)} (\sum_{j = 1}^{m + 4} \frac{1}{{(2 j - 1)}^{2}}) = \frac{\sqrt{2}}{126} π^{2} - \frac{146123}{1323000}$ ；

19) $\sum_{m = 1}^{\infty} \frac{(2 m)!}{2^{3 m} {(m!)}^{2} (m + 2) (m + 3) (m + 4) (m + 5)} (\sum_{j = 1}^{m + 4} \frac{1}{{(2 j - 1)}^{2}}) = \frac{17 \sqrt{2}}{1890} π^{2} - \frac{62119}{496125}$ ；

20) $\sum_{m = 1}^{\infty} \frac{(2 m)!}{2^{3 m} {(m!)}^{2} (m + 1) (m + 2) (m + 3) (m + 4) (m + 5)} (\sum_{j = 1}^{m + 4} \frac{1}{{(2 j - 1)}^{2}}) = - \frac{\sqrt{2}}{945} π^{2} + \frac{8369}{567000}$ 。

3.2. 数值级数

在公式(1)~(20)，令 $x = 1$ ， $\arcsin x = \frac{π}{2}$ ；计算公式中的项 $(\frac{a_{1}}{b_{1}} + \frac{a_{2}}{b_{2}} + \dots + \frac{a_{t}}{b_{t}}) B = 0$ ，

因为 $(\frac{a_{1}}{b_{1}} + \frac{a_{2}}{b_{2}} + \dots + \frac{a_{t}}{b_{t}}) = 0$ ，而 $B = \frac{π^{2} / 8}{\sqrt{1 - x^{2}}} \to \infty$ ， $(x \to 1)$

这时定理计算公式(1)~(20)中， $(\frac{a_{1}}{b_{1}} + \frac{a_{2}}{b_{2}} + \dots + \frac{a_{t}}{b_{t}})$ 与B相乘的项的均为0

则分母为奇偶性不定线性因子乘积的二项式系数级数连带奇数倒数平方和数值恒等式

1) $\sum_{m = 1}^{\infty} \frac{2 m!}{2^{2 m} {(m!)}^{2} (m + 1)} (\sum_{j = 1}^{m} \frac{1}{{(2 j - 1)}^{2}}) = 1$

2) $\sum_{m = 1}^{\infty} \frac{2 m!}{2^{2 m} {(m!)}^{2} (m + 2)} (\sum_{j = 1}^{m + 1} \frac{1}{{(2 j - 1)}^{2}}) = \frac{8}{9}$ ；

3) $\sum_{m = 1}^{\infty} \frac{2 m!}{2^{2 m} {(m!)}^{2} (m + 3)} (\sum_{j = 1}^{m + 2} \frac{1}{{(2 j - 1)}^{2}}) = \frac{544}{675}$ ；

4) $\sum_{m = 1}^{\infty} \frac{2 m!}{2^{2 m} {(m!)}^{2} (m + 4)} (\sum_{j = 1}^{m + 3} \frac{1}{{(2 j - 1)}^{2}}) = \frac{8192}{11025}$ ；

5) $\sum_{m = 1}^{\infty} \frac{2 m!}{2^{2 m} {(m!)}^{2} (m + 5)} (\sum_{j = 1}^{m + 4} \frac{1}{{(2 j - 1)}^{2}}) = \frac{16384}{23625}$ ；

6) $\sum_{m = 1}^{\infty} \frac{2 m!}{2^{2 m} {(m!)}^{2} (m + 1) (m + 2)} (\sum_{j = 1}^{m + 1} \frac{1}{{(2 j - 1)}^{2}}) = \frac{1}{9}$ ；

7) $\sum_{m = 1}^{\infty} \frac{2 m!}{2^{2 m} {(m!)}^{2} (m + 1) (m + 3)} (\sum_{j = 1}^{m + 2} \frac{1}{{(2 j - 1)}^{2}}) = \frac{131}{1350}$ ；

8) $\sum_{m = 1}^{\infty} \frac{2 m!}{2^{2 m} {(m!)}^{2} (m + 2) (m + 3)} (\sum_{j = 1}^{m + 2} \frac{1}{{(2 j - 1)}^{2}}) = \frac{56}{675}$ ；

9) $\sum_{m = 1}^{\infty} \frac{2 m!}{2^{2 m} {(m!)}^{2} (m + 1) (m + 4)} (\sum_{j = 1}^{m + 3} \frac{1}{{(2 j - 1)}^{2}}) = \frac{2833}{33075}$ ；

10) $\sum_{m = 1}^{\infty} \frac{2 m!}{2^{2 m} {(m!)}^{2} (m + 2) (m + 4)} (\sum_{j = 1}^{m + 3} \frac{1}{{(2 j - 1)}^{2}}) = \frac{268}{3675}$ ；

11) $\sum_{m = 1}^{\infty} \frac{2 m!}{2^{2 m} {(m!)}^{2} (m + 3) (m + 4)} (\sum_{j = 1}^{m + 3} \frac{1}{{(2 j - 1)}^{2}}) = \frac{416}{6615}$ ；

12) $\sum_{m = 1}^{\infty} \frac{2 m!}{2^{2 m} {(m!)}^{2} (m + 1) (m + 2) (m + 3)} (\sum_{j = 1}^{m + 2} \frac{1}{{(2 j - 1)}^{2}}) = \frac{19}{1350}$ ；

13) $\sum_{m = 1}^{\infty} \frac{2 m!}{2^{2 m} {(m!)}^{2} (m + 1) (m + 2) (m + 4)} (\sum_{j = 1}^{m + 3} \frac{1}{{(2 j - 1)}^{2}}) = \frac{178501}{33075}$ ；

14) $\sum_{m = 1}^{\infty} \frac{2 m!}{2^{2 m} {(m!)}^{2} (m + 1) (m + 3) (m + 4)} (\sum_{j = 1}^{m + 3} \frac{1}{{(2 j - 1)}^{2}}) = \frac{251}{22050}$ ；

15) $\sum_{m = 1}^{\infty} \frac{2 m!}{2^{2 m} {(m!)}^{2} (m + 2) (m + 3) (m + 4)} (\sum_{j = 1}^{m + 3} \frac{1}{{(2 j - 1)}^{2}}) = \frac{332}{33075}$ ；

16) $\sum_{m = 1}^{\infty} \frac{2 m!}{2^{2 m} {(m!)}^{2} (m + 1) (m + 2) (m + 3) (m + 4)} (\sum_{j = 1}^{m + 3} \frac{1}{{(2 j - 1)}^{2}}) = \frac{89}{66150}$ ；

17) $\sum_{m = 1}^{\infty} \frac{(2 m)!}{2^{2 m} {(m!)}^{2} (m + 1) (m + 2) (m + 4) (m + 5)} (\sum_{j = 1}^{m + 4} \frac{1}{{(2 j - 1)}^{2}}) = \frac{780701}{5953500}$ ；

18) $\sum_{m = 1}^{\infty} \frac{(2 m)!}{2^{2 m} {(m!)}^{2} (m + 1) (m + 3) (m + 4) (m + 5)} (\sum_{j = 1}^{m + 4} \frac{1}{{(2 j - 1)}^{2}}) = \frac{13}{147000}$ ；

19) $\sum_{m = 1}^{\infty} \frac{(2 m)!}{2^{2 m} {(m!)}^{2} (m + 2) (m + 3) (m + 4) (m + 5)} (\sum_{j = 1}^{m + 4} \frac{1}{{(2 j - 1)}^{2}}) = \frac{556}{496125}$ ；

20) $\sum_{m = 1}^{\infty} \frac{(2 m)!}{2^{2 m} {(m!)}^{2} (m + 1) (m + 2) (m + 3) (m + 4) (m + 5)} (\sum_{j = 1}^{m + 4} \frac{1}{{(2 j - 1)}^{2}}) = \frac{223}{3969000}$ 。

3.3. 数值级数

在推论公式(1)~(5)令 $x = 1$ ， $b = - \frac{\sqrt{2}}{4} \ln^{2} (1 + \sqrt{2})$

则交错的分母为奇偶性不定因子乘积二项式系数级数连带奇数倒数平方和数值恒等式

1) $\sum_{m = 1}^{\infty} \frac{{(- 1)}^{m} (2 m)!}{2^{2 m} {(m!)}^{2} (m + 1)} (\sum_{j = 1}^{m} \frac{1}{{(2 j - 1)}^{2}}) = - \sqrt{2} \ln^{2} (1 + \sqrt{2}) + 1$ ；

2) $\sum_{m = 1}^{\infty} \frac{{(- 1)}^{m} (2 m)!}{2^{2 m} {(m!)}^{2} (m + 2)} (\sum_{j = 1}^{m + 1} \frac{1}{{(2 j - 1)}^{2}}) = 3 \sqrt{2} \ln^{2} (1 + \sqrt{2}) - \frac{4}{9}$ ；

3) $\sum_{m = 1}^{\infty} \frac{{(- 1)}^{m} (2 m)!}{2^{2 m} {(m!)}^{2} (m + 3)} (\sum_{j = 1}^{m + 2} \frac{1}{{(2 j - 1)}^{2}}) = - \frac{7}{15} \sqrt{2} \ln^{2} (1 + \sqrt{2}) + \frac{304}{675}$ ；

4) $\sum_{m = 1}^{\infty} \frac{{(- 1)}^{m} (2 m)!}{2^{2 m} {(m!)}^{2} (m + 4)} (\sum_{j = 1}^{m + 3} \frac{1}{{(2 j - 1)}^{2}}) = \frac{9}{35} \sqrt{2} \ln^{2} (1 + \sqrt{2}) - \frac{736}{2205}$ ；

5) $\sum_{m = 1}^{\infty} \frac{{(- 1)}^{m} (2 m)!}{2^{2 m} {(m!)}^{2} (m + 5)} (\sum_{j = 1}^{m + 4} \frac{1}{{(2 j - 1)}^{2}}) = - \frac{107}{315} \sqrt{2} \ln^{2} (1 + \sqrt{2}) + \frac{18176}{55125}$ 。

文章引用

薛坐远,薛雪,及万会. 二项式系数级数连带奇数倒数平方和
The Binomial Coefficient Series Is Associated with a Sum of Odd Reciprocal Squares[J]. 理论数学, 2019, 09(07): 818-835. https://doi.org/10.12677/PM.2019.97108

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