摩訶不思議な級数を作る方法（小ネタ）

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あいさつ

んちゃ！
今回は次の文献をもとに級数で遊んでみます。
おもちゃで遊ぶ子どもの気持ちになって読んで頂けると助かります。
単発ネタです。かなり短いのでその点はご了承お願いいたします。

今回使う公式

まずは今回の記事を書くきっかけになった公式を証明します。参考文献👈

$\forall {a_{n}}_{n \in N} \subset C : \sum_{n = 1}^{N} \frac{1}{a_{n}} \prod_{m = 1}^{n} \frac{a_{m}}{x + a_{m}} = \frac{1}{x} (1 - \prod_{m = 1}^{N} \frac{a_{m}}{x + a_{m}})$

[1]下記の様に差分により書き直す。
$\frac{1}{a_{n}} \prod_{m = 1}^{n} \frac{a_{m}}{x + a_{m}} = \frac{1}{x} (\prod_{m = 1}^{n - 1} \frac{a_{m}}{x + a_{m}} - \prod_{m = 1}^{n} \frac{a_{m}}{x + a_{m}})$
[2][1]により
$\begin{array}{rcl} \sum_{n = 1}^{N} \frac{1}{a_{n}} \prod_{m = 0}^{n} \frac{a_{m}}{x + a_{m}} & = & \frac{1}{x} (1 - \prod_{m = 1}^{N} \frac{a_{m}}{x + a_{m}}) \end{array}$

上記公式を魔改造してみます。

$\begin{array}{rcl} \forall & {a_{n}}_{n \in N} (a_{n} - b_{n} \neq 0), {b_{n}}_{n \in N} \subset C : \forall {γ_{n} (x)} \subset {γ : C \to C | γ は有理関数} \\ \sum_{n = 1}^{N} & \frac{γ_{n - 1} (x) (x + a_{n}) (x + b_{n}) - γ_{n} (x) (a_{n} - b_{n})}{a_{n} - b_{n}} \prod_{m = 1}^{n} \frac{a_{m} - b_{m}}{(x + a_{m}) (x + b_{m})} = γ_{0} (x) - γ_{N} (x) \prod_{m = 1}^{N} \frac{a_{m} - b_{m}}{(x + a_{m}) (x + b_{m})} \end{array}$

[1]適当な有理関数列 ${γ_{n} (x)}_{n \in N_{0}}$ を用いて差分で書き直す。
$\begin{array}{rcl} γ_{n - 1} (x) \prod_{m = 1}^{n - 1} \frac{a_{m} - b_{m}}{(x + a_{m}) (x + b_{m})} - γ_{n} (x) \prod_{m = 1}^{n} \frac{a_{m} - b_{m}}{(x + a_{m}) (x + b_{m})} & = & \frac{γ_{n - 1} (x) (x + a_{n}) (x + b_{n}) - γ_{n} (x) (a_{n} - b_{n})}{(x + a_{n}) (x + b_{n})} \prod_{m = 1}^{n - 1} \frac{a_{m} - b_{m}}{(x + a_{m}) (x + b_{m})} \\ = & \frac{γ_{n - 1} (x) (x + a_{n}) (x + b_{n}) - γ_{n} (x) (a_{n} - b_{n})}{a_{n} - b_{n}} \prod_{m = 1}^{n} \frac{a_{m} - b_{m}}{(x + a_{m}) (x + b_{m})} \end{array}$
[2][1]を代入すると以下の式が得られる。
$\begin{array}{r} \sum_{n = 1}^{N} \frac{γ_{n - 1} (x) (x + a_{n}) (x + b_{n}) - γ_{n} (x) (a_{n} - b_{n})}{a_{n} - b_{n}} \prod_{m = 1}^{n} \frac{a_{m} - b_{m}}{(x + a_{m}) (x + b_{m})} = γ_{0} (x) - γ_{N} (x) \prod_{m = 1}^{N} \frac{a_{m} - b_{m}}{(x + a_{m}) (x + b_{m})} \end{array}$

さらに魔改造します。

$\begin{array}{rcl} \forall & {a_{n}}_{n \in N} (a_{n} - b_{n} \neq 0), {b_{n}}_{n \in N} \subset C : \forall {γ_{n} (x)} \subset {γ : C \to C | γ は有理関数} \\ \sum_{n = 1}^{N} & \frac{γ_{n - 1} (x) (x + a_{n})^{K} (x + b_{n})^{L} - γ_{n} (x) (a_{n} - b_{n})^{K + L}}{(a_{n} - b_{n})^{K + L}} \prod_{m = 1}^{n} \frac{(a_{m} - b_{m})^{K + L}}{(x + a_{m})^{K} (x + b_{m})^{M}} = γ_{0} (x) - γ_{N} (x) \prod_{m = 1}^{N} \frac{(a_{m} - b_{m})^{K + L}}{(x + a_{m})^{K} (x + b_{m})^{M}} \end{array}$

[1]
$\begin{array}{rcl} γ_{n - 1} (x) \prod_{m = 1}^{n - 1} \frac{(a_{m} - b_{m})^{K + L}}{(x + a_{m})^{K} (x + b_{m})^{M}} - γ_{n} (x) \prod_{m = 1}^{n} \frac{(a_{m} - b_{m})^{K + L}}{(x + a_{m})^{K} (x + b_{m})^{M}} & = & \frac{γ_{n - 1} (x) (x + a_{n})^{K} (x + b_{n})^{L} - γ_{n} (x) (a_{n} - b_{n})^{K + L}}{(x + a_{n})^{K} (x + b_{n})^{L}} \prod_{m = 1}^{n - 1} \frac{(a_{m} - b_{m})^{K + L}}{(x + a_{m})^{K} (x + b_{m})^{M}} \\ = & \frac{γ_{n - 1} (x) (x + a_{n})^{K} (x + b_{n})^{L} - γ_{n} (x) (a_{n} - b_{n})^{K + L}}{(a_{n} - b_{n})^{K + L}} \prod_{m = 1}^{n} \frac{(a_{m} - b_{m})^{K + L}}{(x + a_{m})^{K} (x + b_{m})^{M}} \end{array}$
[2]代入して
$\begin{array}{r} \sum_{n = 1}^{N} \frac{γ_{n - 1} (x) (x + a_{n})^{K} (x + b_{n})^{L} - γ_{n} (x) (a_{n} - b_{n})^{K + L}}{(a_{n} - b_{n})^{K + L}} \prod_{m = 1}^{n} \frac{(a_{m} - b_{m})^{K + L}}{(x + a_{m})^{K} (x + b_{m})^{M}} = γ_{0} (x) - γ_{N} (x) \prod_{m = 1}^{N} \frac{(a_{m} - b_{m})^{K + L}}{(x + a_{m})^{K} (x + b_{m})^{M}} \end{array}$

応用

では応用してみましょう。
途中計算ミスがある可能性があります。
くどいですが必ず自分で計算して確認してみてください。

$γ_{n} (x) = \frac{1}{(x + n + 1)^{5}}, a_{n} = (\begin{matrix} 2 n + 2 \\ n + 1 \end{matrix}), b_{n} = (\begin{matrix} 2 n \\ n \end{matrix})$ とすると下記の式を得る。
$\sum_{n = 1}^{\infty} \frac{1}{n^{5}} \prod_{m = 1}^{n - 1} \frac{3 m - 1}{(4 m + 2) (\begin{matrix} 2 m \\ m \end{matrix})} = 2 \sum_{n = 1}^{\infty} \frac{2 n + 1}{n^{5} (3 n - 1)} (\begin{matrix} 2 n \\ n \end{matrix}) \prod_{m = 1}^{n} \frac{3 m - 1}{(4 m + 2) (\begin{matrix} 2 m \\ m \end{matrix})}$

[1]
$\begin{array}{rcl} a_{n} - b_{n} & = & \frac{3 n - 1}{n + 1} (\begin{array}{c} 2 n \\ n \end{array}) \end{array}$
[2]
$\begin{array}{rcl} \sum_{n = 1}^{N} \frac{\frac{1}{(x + n)^{5}} {x + \frac{2 (2 n + 1)}{n + 1} (\begin{array}{c} 2 n \\ n \end{array})} {x + (\begin{array}{c} 2 n \\ n \end{array})} - \frac{1}{(x + n + 1)^{5}} \frac{3 n - 1}{n + 1} (\begin{array}{c} 2 n \\ n \end{array})}{\frac{3 n - 1}{n + 1} (\begin{array}{c} 2 n \\ n \end{array})} \prod_{m = 1}^{n} \frac{\frac{3 m - 1}{m + 1} (\begin{array}{c} 2 m \\ m \end{array})}{{x + \frac{2 (2 m + 1)}{m + 1} (\begin{array}{c} 2 m \\ m \end{array})} {x + (\begin{array}{c} 2 m \\ m \end{array})}} & = & \frac{1}{(x + 1)^{5}} - \frac{1}{(x + N + 1)^{5}} \prod_{m = 1}^{N} \frac{\frac{3 m - 1}{m + 1} (\begin{array}{c} 2 m \\ m \end{array})}{{x + \frac{2 (2 m + 1)}{m + 1} (\begin{array}{c} 2 m \\ m \end{array})} {x + (\begin{array}{c} 2 m \\ m \end{array})}} \end{array}$
[3] $x = 0$ を代入して
$\begin{array}{r} 2 \sum_{n = 1}^{\infty} \frac{2 n + 1}{n^{5} (3 n - 1)} (\begin{array}{c} 2 n \\ n \end{array}) \prod_{m = 1}^{n} \frac{3 m - 1}{(4 m + 2) (\begin{array}{c} 2 m \\ m \end{array})} - \sum_{n = 1}^{\infty} \frac{1}{(n + 1)^{5}} \prod_{m = 1}^{n} \frac{3 m - 1}{(4 m + 2) (\begin{array}{c} 2 m \\ m \end{array})} = 1 \end{array}$
[4]
$\sum_{n = 1}^{\infty} \frac{1}{n^{5}} \prod_{m = 1}^{n - 1} \frac{3 m - 1}{(4 m + 2) (\begin{matrix} 2 m \\ m \end{matrix})} = 2 \sum_{n = 1}^{\infty} \frac{2 n + 1}{n^{5} (3 n - 1)} (\begin{matrix} 2 n \\ n \end{matrix}) \prod_{m = 1}^{n} \frac{3 m - 1}{(4 m + 2) (\begin{matrix} 2 m \\ m \end{matrix})}$

$C (n, m)$ をコネクターとする。この時、 $a_{n} = C (n - 1, m), b_{n, m} = C (n, m)$ としましょう。
コネクターの定義より
$\exists R (n, m) \in C [n, m] s . t . a_{n} - b_{n} = R (n, m) C (n, m)$
を満たしますので、以下の式が成り立つ。
$\sum_{n = 1}^{N} \frac{γ_{n - 1} (x) [x + {1 + R (n, m)} C (n, m)]^{K} {x + C (n, m)}^{L} - γ_{n} (x) {R (n, m) C (n, m)}^{K + L}}{{R (n, m) C (n, m)}^{K + L}} \prod_{k = 1}^{n} \frac{{R (k, m) C (k, m)}^{K + L}}{[x + {1 + R (k, m)} C (k, m)]^{K} {x + C (k, m)}^{L}} = γ_{0} (x) - γ_{N} (x) \prod_{k = 1}^{N} \frac{{R (k, m) C (k, m)}^{K + L}}{[x + {1 + R (k, m)} C (k, m)]^{K} {x + C (k, m)}^{L}}$
特に右辺の第二項が収束する場合
$\begin{array}{r} \sum_{n = 1}^{\infty} \frac{γ_{n - 1} (x) [x + {1 + R (n, m)} C (n, m)]^{K} {x + C (n, m)}^{L} - γ_{n} (x) {R (n, m) C (n, m)}^{K + L}}{{R (n, m) C (n, m)}^{K + L}} \prod_{k = 1}^{n} \frac{{R (k, m) C (k, m)}^{K + L}}{[x + {1 + R (k, m)} C (k, m)]^{K} {x + C (k, m)}^{L}} = γ_{0} (x) \end{array}$

$γ_{n} (x) = \frac{1}{(x + n + 1)^{3}}, a_{n} = \frac{1}{{(\begin{matrix} n + m - 1 \\ m \end{matrix})}^{10}}, b_{n} = \frac{1}{{(\begin{matrix} n + m \\ m \end{matrix})}^{10}}$ とすると以下の様に書ける。
$\sum_{n = 1}^{\infty} \frac{1}{n^{3}} \prod_{k = 1}^{n - 1} \frac{{\frac{(k + m)^{10} - k^{10}}{k^{10}} \frac{1}{{(\begin{matrix} k + m \\ m \end{matrix})}^{10}}}^{K + L}}{{x + \frac{(k + m)^{10}}{k^{10} {(\begin{matrix} k + m \\ m \end{matrix})}^{10}}}^{K} {x + \frac{1}{{(\begin{matrix} k + m \\ m \end{matrix})}^{10}}}^{L}} = \sum_{n = 1}^{\infty} \frac{n^{10 L - 3} (n + m)^{10 K}}{{(n + m)^{10} - n^{10}}^{K + L}} \prod_{k = 1}^{n} \frac{{\frac{(k + m)^{10} - k^{10}}{k^{10}} \frac{1}{{(\begin{matrix} k + m \\ m \end{matrix})}^{10}}}^{K + L}}{{x + \frac{(k + m)^{10}}{k^{10} {(\begin{matrix} k + m \\ m \end{matrix})}^{10}}}^{K} {x + \frac{1}{{(\begin{matrix} k + m \\ m \end{matrix})}^{10}}}^{L}}$

$\begin{array}{r} {\begin{cases} R (n, m) = \frac{(n + m)^{10} - n^{10}}{n^{10}} \\ 1 + R (n, m) = \frac{(n + m)^{10}}{n^{10}} \end{cases} \end{array}$
より
$\sum_{n = 1}^{\infty} \frac{\frac{1}{(x + n)^{3}} {x + \frac{(n + m)^{10}}{n^{10} {(\begin{matrix} n + m \\ m \end{matrix})}^{10}}}^{K} {x + \frac{1}{{(\begin{matrix} n + m \\ m \end{matrix})}^{10}}}^{L} - \frac{1}{(x + n + 1)^{3}} {\frac{(n + m)^{10} - n^{10}}{n^{10}} \frac{1}{{(\begin{matrix} n + m \\ m \end{matrix})}^{10}}}^{K + L}}{{\frac{(n + m)^{10} - n^{10}}{n^{10}} \frac{1}{{(\begin{matrix} n + m \\ m \end{matrix})}^{10}}}^{K + L}} \prod_{k = 1}^{n} \frac{{\frac{(k + m)^{10} - k^{10}}{k^{10}} \frac{1}{{(\begin{matrix} k + m \\ m \end{matrix})}^{10}}}^{K + L}}{{x + \frac{(k + m)^{10}}{k^{10} {(\begin{matrix} k + m \\ m \end{matrix})}^{10}}}^{K} {x + \frac{1}{{(\begin{matrix} k + m \\ m \end{matrix})}^{10}}}^{L}} = \frac{1}{(x + 1)^{3}}$
この式から以下の結論を得る。
$\begin{array}{r} \sum_{n = 1}^{\infty} \frac{1}{n^{3}} \prod_{k = 1}^{n - 1} \frac{{\frac{(k + m)^{10} - k^{10}}{k^{10}} \frac{1}{{(\begin{array}{c} k + m \\ m \end{array})}^{10}}}^{K + L}}{{x + \frac{(k + m)^{10}}{k^{10} {(\begin{array}{c} k + m \\ m \end{array})}^{10}}}^{K} {x + \frac{1}{{(\begin{array}{c} k + m \\ m \end{array})}^{10}}}^{L}} = \sum_{n = 1}^{\infty} \frac{n^{10 L - 3} (n + m)^{10 K}}{{(n + m)^{10} - n^{10}}^{K + L}} \prod_{k = 1}^{n} \frac{{\frac{(k + m)^{10} - k^{10}}{k^{10}} \frac{1}{{(\begin{array}{c} k + m \\ m \end{array})}^{10}}}^{K + L}}{{x + \frac{(k + m)^{10}}{k^{10} {(\begin{array}{c} k + m \\ m \end{array})}^{10}}}^{K} {x + \frac{1}{{(\begin{array}{c} k + m \\ m \end{array})}^{10}}}^{L}} \end{array}$