文献あり

級数を求めたい[2]

167

導入

前回の続きです。今回は前回の級数の分母にいろいろ飾りつけをします。

本題

$C_{n} := \frac{1}{2^{2 n}} (\binom{2 n}{n}) = (- 1)^{n} (\binom{- 1 / 2}{n}) = \frac{(2 n - 1)!!}{(2 n)!!}$
$H (s; s) := \sum_{0 < n_{1} \leq \dots \leq n_{i} \leq n} \frac{C_{n}}{n_{1}^{s_{1}} \dots n_{i}^{s_{i}} n^{s}}$
$H (s; s) [x] := \sum_{0 < n_{1} \leq \dots \leq n_{i} \leq n} \frac{C_{n}}{n_{1}^{s_{1}} \dots n_{i}^{s_{i}} n^{s}} x^{n}$
$H (s; s) ⟨ x ⟩ := \sum_{0 < n_{1} \leq \dots \leq n_{i} \leq n} \frac{C_{n}}{n_{1}^{s_{1}} \dots n_{i}^{s_{i}} n^{s}} x^{n_{1}}$
$a, b_{0}, b_{1}, c$ :それぞれ $\frac{d t}{t}, \frac{d t}{1 - t}, \frac{d t}{- 1 - t}, \frac{((1 - t)^{- 1 / 2} - 1) d t}{t}$ に対応する微分形式の語
$D := Q ⟨ a, b_{0}, b_{1}, c ⟩$ (非可換多項式環)

既知の結果

$H (\emptyset; k) = - 2 \sum_{i_{1}, \dots, i_{k - 1} \in {1, \overset{―}{1}}} ζ (i_{1}, \dots, i_{k - 1}, \overset{―}{1})$

$\begin{aligned} H (\emptyset; k) [x] & = \sum_{0 < n} \frac{C_{n}}{n^{k}} x^{n} \\ = \sum_{0 < n} \frac{C_{n}}{n^{k - 1}} \int_{0}^{x} t^{n - 1} d x \\ = \int_{0}^{x} \frac{d t}{t} H (\emptyset; k - 1) [t] \\ = \dots \\ = \int_{1 > t_{1} > \dots > t_{k - 1} > 0} \frac{d t_{1}}{t_{1}} \dots \frac{d t_{k - 1}}{t_{k - 1}} \sum_{0 < n} \frac{C_{n}}{n} t_{k - 1}^{n} \\ = \int_{1 > t_{1} > \dots > t_{k - 1} > 0} \frac{d t_{1}}{t_{1}} \dots \frac{d t_{k - 1}}{t_{k - 1}} \sum_{0 < n} C_{n} \int_{0}^{t_{k - 1}} t_{k}^{n} d t_{k} \\ = \int_{1 > t_{1} > \dots > t_{k} > 0} \frac{d t_{1}}{t_{1}} \dots \frac{d t_{k - 1}}{t_{k - 1}} \sum_{0 < n} C_{n} t_{k}^{n} d t_{k} \\ = \int_{1 > t_{1} > \dots > t_{k} > 0} \frac{d t_{1}}{t_{1}} \dots \frac{d t_{k - 1}}{t_{k - 1}} \frac{((1 - t_{k})^{- 1 / 2} - 1) d t_{k}}{t_{k}} \\ = \int_{0}^{x} a^{k - 1} c \end{aligned}$
ここで、置換 $g : x \mapsto 1 - x^{2}$ によって $\int_{0}^{1} a^{k - 1} c$ は
$\begin{aligned} \int_{0}^{1} a^{k - 1} c & = \int_{0}^{1} \frac{d t_{1}}{t_{1}} \int_{0}^{t_{1}} \frac{d t_{2}}{t_{2}} \dots \int_{0}^{t_{k - 1}} \frac{((1 - t_{k})^{- 1 / 2} - 1) d t_{k}}{t_{k}} \\ = \int_{0}^{1} \frac{d t_{1}}{t_{1}} \int_{0}^{t_{1}} \frac{d t_{2}}{t_{2}} \dots \int_{\sqrt{1 - t_{k - 1}}}^{1} \frac{- 2 d t_{k}}{- 1 - t_{k}} \\ = \dots \\ = \int_{0}^{1} (\frac{d t_{1}}{1 - t_{1}} + \frac{d t_{1}}{- 1 - t_{1}}) \dots \int_{t_{k - 2}}^{1} (\frac{d t_{k - 1}}{1 - t_{k - 1}} + \frac{d t_{k - 1}}{- 1 - t_{k - 1}}) \int_{t_{k - 1}}^{1} \frac{- 2 d t_{k}}{- 1 - t_{k}} \\ = - 2 \int_{0}^{1} b_{1} (b_{0} + b_{1})^{k - 1} \\ = - 2 \sum_{i_{1}, \dots, i_{k - 1} \in {1, \overset{―}{1}}} ζ (i_{1}, \dots, i_{k - 1}, \overset{―}{1}) \end{aligned}$

主定理

$H (s; s)$ はAMZVが $Q$ 上張る空間に含まれる。

$k = (k_{1}, \dots, k_{r})$ のとき
$k_{\to} := (k_{1}, \dots, k_{r}, 1)$
$k_{\leftarrow} := (k_{1}, \dots, k_{r - 1})$
$k_{↑} := (k_{1}, \dots, k_{r} + 1)$
$k_{↓} := (k_{1}, \dots, k_{r} - 1)$
$_{\leftarrow} k := (1, k_{1}, \dots, k_{r})$
$_{\to} k := (k_{2}, \dots, k_{r})$
$_{↑} k := (k_{1} + 1, \dots, k_{r})$
$_{↓} k := (k_{1} - 1, \dots, k_{r})$
とする。

$H (_{↑} s; k) ⟨ x ⟩ = \int_{0}^{x} \frac{d t}{t} H (s; k) ⟨ t ⟩$
$H (_{\leftarrow} s; k) ⟨ x ⟩ = \int_{0}^{x} \frac{d t}{1 - t} (H (s; k) - H (s; k) ⟨ t ⟩)$

$\begin{aligned} H (_{↑} s; k) ⟨ x ⟩ & = \sum_{0 < n_{1} \leq \dots \leq n_{i} \leq n} \frac{C_{n}}{n_{1}^{s_{1} + 1} \dots n_{i}^{s_{i}} n^{k}} x^{n_{1}} \\ = \sum_{0 < n_{1} \leq \dots \leq n_{i} \leq n} \frac{C_{n}}{n_{1}^{s_{1}} \dots n_{i}^{s_{i}} n^{k}} \int_{0}^{x} t^{n_{1} - 1} d t \\ = \int_{0}^{x} \sum_{0 < n_{1} \leq \dots \leq n_{i} \leq n} \frac{C_{n}}{n_{1}^{s_{1}} \dots n_{i}^{s_{i}} n^{k}} \frac{t^{n_{1}}}{t} d t \\ = \int_{0}^{x} \frac{d t}{t} H (s; k) ⟨ t ⟩ \end{aligned}$
$\begin{aligned} H (_{\leftarrow} s; k) ⟨ x ⟩ & = \sum_{0 < m \leq n_{1} \leq \dots \leq n_{i} \leq n} \frac{C_{n}}{m n_{1}^{s_{1}} \dots n_{i}^{s_{i}} n^{k}} x^{m} \\ = \sum_{0 < m \leq n_{1} \leq \dots \leq n_{i} \leq n} \frac{C_{n}}{n_{1}^{s_{1}} \dots n_{i}^{s_{i}} n^{k}} \int_{0}^{x} t^{m - 1} d t \\ = \int_{0}^{x} \sum_{0 < n_{1} \leq \dots \leq n_{i} \leq n} \frac{C_{n}}{n_{1}^{s_{1}} \dots n_{i}^{s_{i}} n^{k}} \frac{1 - t^{n_{1}}}{1 - t} d t \\ = \int_{0}^{x} \frac{d t}{1 - t} (H (s; k) - H (s; k) ⟨ t ⟩) \end{aligned}$

ここで、任意のインデックスはweight $1$ ,depth $1$ のインデックス $(1)$ から $_{↑} *$ と $_{\leftarrow} *$ の繰り返しで作ることができる。 $s$ の一番右の要素が $1$ の時、 $(H (s; k) - H (s; k) ⟨ t ⟩)$ は
$\begin{aligned} H (s; k) - H (s; k) ⟨ t ⟩ & = \sum_{0 < n_{1} \leq \dots \leq n_{i} \leq n} \frac{C_{n}}{n_{1} n_{2}^{s_{2}} \dots n_{i}^{s_{i}} n^{k}} (1 - t^{n_{1}}) \\ = \sum_{0 < n_{1} \leq \dots \leq n_{i} \leq n} \frac{C_{n}}{n_{2}^{s_{2}} \dots n_{i}^{s_{i}} n^{k}} \int_{t}^{1} u^{n_{1} - 1} d u \\ = \int_{t}^{1} \sum_{0 < n_{1} \leq \dots \leq n_{i} \leq n} \frac{C_{n}}{n_{2}^{s_{2}} \dots n_{i}^{s_{i}} n^{k}} u^{n_{1} - 1} d u \\ = \int_{t}^{1} \sum_{0 < n_{2} \leq \dots \leq n_{i} \leq n} \frac{C_{n}}{n_{2}^{s_{2}} \dots n_{i}^{s_{i}} n^{k}} \frac{1 - u^{n_{2}}}{1 - u} d u \\ = \int_{t}^{1} \frac{d u}{1 - u} \sum_{0 < n_{2} \leq \dots \leq n_{i} \leq n} \frac{C_{n}}{n_{2}^{s_{2}} \dots n_{i}^{s_{i}} n^{k}} (1 - u^{n_{2}}) \\ = \int_{t}^{1} \frac{d u}{1 - u} (H (_{\to} s; k) - H (_{\to} s; k) ⟨ u ⟩) \end{aligned}$
また、 $s$ の一番右の要素が $1$ より大きい時、 $(H (s; k) - H (s; k) ⟨ t ⟩)$ は
$\begin{aligned} H (s; k) - H (s; k) ⟨ t ⟩ & = \sum_{0 < n_{1} \leq \dots \leq n_{i} \leq n} \frac{C_{n}}{n_{1}^{s_{1}} \dots n_{i}^{s_{i}} n^{k}} (1 - t^{n_{1}}) \\ = \sum_{0 < n_{1} \leq \dots \leq n_{i} \leq n} \frac{C_{n}}{n_{1}^{s_{1} - 1} n_{2}^{s_{2}} \dots n_{i}^{s_{i}} n^{k}} \int_{t}^{1} u^{n_{1} - 1} d u \\ = \int_{t}^{1} \sum_{0 < n_{1} \leq \dots \leq n_{i} \leq n} \frac{C_{n}}{n_{1}^{s_{1} - 1} n_{2}^{s_{2}} \dots n_{i}^{s_{i}} n^{k}} u^{n_{1} - 1} d u \\ = \int_{t}^{1} \sum_{0 < n_{1} \leq \dots \leq n_{i} \leq n} \frac{C_{n}}{n_{1}^{s_{1} - 1} n_{2}^{s_{2}} \dots n_{i}^{s_{i}} n^{k}} \frac{u^{n_{1}}}{u} d u \\ = \int_{t}^{1} \frac{d u}{u} \sum_{0 < n_{1} \leq \dots \leq n_{i} \leq n} \frac{C_{n}}{n_{1}^{s_{1} - 1} n_{2}^{s_{2}} \dots n_{i}^{s_{i}} n^{k}} u^{n_{1}} \\ = \int_{t}^{1} \frac{d u}{u} H (_{↓} s; k) ⟨ u ⟩ \end{aligned}$
となるので上の操作を繰り返して $H (1, k) ⟨ x ⟩$ もしくは $H (1, k) - H (1, k) ⟨ x ⟩$ を $\int_{0}^{t}, \int_{t}^{1}, \frac{d t}{t}, \frac{d t}{1 - t}$ を用いて反復積分した形に直すことができる。
$\begin{aligned} H (1; k) ⟨ t ⟩ & = \sum_{0 < m \leq n} \frac{C_{n}}{m n^{k}} t^{m} \\ = \sum_{0 < m \leq n} \frac{C_{n}}{n^{k}} \int_{0}^{t} u^{m - 1} d u \\ = \int_{0}^{t} \sum_{0 < m \leq n} \frac{C_{n}}{n^{k}} u^{m - 1} d u \\ = \int_{0}^{t} \sum_{0 < n} \frac{C_{n}}{n^{k}} \frac{1 - u^{n}}{1 - u} d u \\ = \int_{0}^{t} \frac{d u}{1 - u} \sum_{0 < n} \frac{C_{n}}{n^{k}} (1 - u^{n}) \\ = \int_{0}^{t} \frac{d u}{1 - u} (H (\emptyset; k) - H (\emptyset; k) ⟨ u ⟩) \\ = \int_{0}^{t} \frac{d u}{1 - u} (H (\emptyset; k) - H (\emptyset; k) [u]) \\ = \int_{0}^{t} \frac{d u}{1 - u} \int_{u}^{1} a^{k - 1} c \end{aligned}$
$\begin{aligned} H (1; k) - H (1; k) ⟨ t ⟩ & = \sum_{0 < m \leq n} \frac{C_{n}}{m n^{k}} (1 - t^{m}) \\ = \sum_{0 < m \leq n} \frac{C_{n}}{n^{k}} \int_{t}^{1} u^{m - 1} d u \\ = \int_{t}^{1} \sum_{0 < m \leq n} \frac{C_{n}}{n^{k}} u^{m - 1} d u \\ = \int_{t}^{1} \sum_{0 < n} \frac{C_{n}}{n^{k}} \frac{1 - u^{n}}{1 - u} d u \\ = \int_{t}^{1} \frac{d u}{1 - u} \sum_{0 < n} \frac{C_{n}}{n^{k}} (1 - u^{n}) \\ = \int_{t}^{1} \frac{d u}{1 - u} (H (\emptyset; k) - H (\emptyset; k) ⟨ u ⟩) \\ = \int_{t}^{1} \frac{d u}{1 - u} (H (\emptyset; k) - H (\emptyset; k) [u]) \\ = \int_{t}^{1} \frac{d u}{1 - u} \int_{u}^{1} a^{k - 1} c \end{aligned}$
なので、 $H (s; k)$ は $\int_{0}^{t}, \int_{t}^{1}, \frac{d t}{t}, \frac{d t}{1 - t}, \frac{((1 - t)^{- 1 / 2} - 1) d t}{t}$ を用いた反復積分で表せるから、この記事や上の定理1でも用いたpull-back $g : t \mapsto 1 - t^{2}$ により $H (s; k)$ がAMZVの $Q$ 線形結合で表されることがわかる。

計算例

$\begin{aligned} \sum_{0 < m \leq n} \frac{C_{n}}{m n} & = H (1; 1) \\ = H (1; 1) ⟨ 1 ⟩ \\ = \int_{0}^{1} \frac{d t}{1 - t} \int_{t}^{1} \frac{(1 - u)^{- 1 / 2} - 1}{u} d u \\ = \int_{0}^{1} \frac{((1 - u)^{- 1 / 2} - 1) d u}{u} \int_{0}^{u} \frac{d t}{1 - t} \\ = 2 \int_{0}^{1} \frac{((1 - u)^{- 1 / 2} - 1) d u}{u} \int_{\sqrt{1 - u}}^{1} \frac{d t}{t} \\ = 4 \int_{0}^{1} \frac{(1 - u) d u}{1 - u^{2}} \int_{u}^{1} \frac{d t}{t} \\ = - 4 \int_{0}^{1} \frac{d t}{t} \int_{0}^{t} \frac{d u}{- 1 - u} \\ = - 4 ζ (\overset{―}{2}) \\ = \frac{π^{2}}{3} \end{aligned}$
この記事にある通り、
$\sum_{0 < m \leq n} \frac{C_{n}}{m^{2} n} = H (2; 1) = \frac{3}{2} ζ (3)$
です。