伯努利数

一.伯努利公式

伯努利数是一个用于解决 $n$ 次方和的数列。

它的递归定义公式如下：

\sum_{i = 0}^{n} (\binom{n + 1}{i}) B_{i} = [n = 0] (1.1)

$\sum_{i=0}^n \binom {n+1} i B_i=[n=0] ~~~~~~~~ (1.1)$

通过这个定义可以得到伯努利数的前几项： $1,-\frac{1}{2},\frac{1}{6},0...$

令 $S_m(n)=\sum_{i=0}^{n-1} i^m$ ，伯努利通过找规律发现了伯努利公式：

S_{m} (n) = \frac{1}{m + 1} \sum_{i = 0}^{m} (\binom{m + 1}{i}) B_{i} n^{m + 1 - i} (1.2)

$S_m(n)=\frac{1}{m+1} \sum_{i=0}^m \binom {m+1} i B_i n^{m+1-i} ~~~~~~~~ (1.2)$

《具体数学》上给出的证明如下：

\begin{aligned} S_{m + 1} (n) + n^{m + 1} & = \sum_{i = 0}^{n} i^{m + 1} \\ = \sum_{i = 0}^{n - 1} (i + 1)^{m + 1} \\ = \sum_{i = 0}^{n - 1} \sum_{j = 0}^{m + 1} (\binom{m + 1}{j}) i^{j} \\ = \sum_{j = 0}^{m + 1} (\binom{m + 1}{j}) \sum_{i = 0}^{n - 1} i^{j} \\ = \sum_{j = 0}^{m + 1} (\binom{m + 1}{j}) S_{j} (n) & (1.3) \end{aligned}

$\begin{aligned} S_{m+1}(n)+n^{m+1}&=\sum_{i=0}^ni^{m+1}\\ &=\sum_{i=0}^{n-1} (i+1)^{m+1} \\ &=\sum_{i=0}^{n-1}\sum_{j=0}^{m+1} \binom {m+1}{j} i^j \\ &=\sum_{j=0}^{m+1}\binom {m+1}{j} \sum_{i=0}^{n-1} i^j \\ &=\sum_{j=0}^{m+1}\binom {m+1}{j} S_j(n) & (1.3) \end{aligned}$

由 $(1.3)$ 两边同时减去 $n^{m+1}$ 得，

S_{m + 1} (n) = \sum_{j = 0}^{m} (\binom{m + 1}{j}) S_{j} (n) (1.4)

$S_{m+1}(n)=\sum_{j=0}^{m}\binom {m+1}{j} S_j(n) ~~~~~~~~ (1.4)$

设 $S'_ m(n)$ 为 $(1.2)$ 右式， $\Delta=S_m(n)-S'_m(n)$ 。

归纳证明 $S_m(n)=S'_m(n)$ ：

1.当 $m=0$ 时成立。

2.设对于 $\forall i\in[0,m),S_i(n)=S'_ i(n)$ , 由 $(3.4)$ 得：

\begin{aligned} n^{m + 1} & = \sum_{j = 0}^{m} (\binom{m + 1}{j}) S_{j}^{'} (n) + (\binom{m + 1}{m}) Δ & (只有j=m时有差异) \\ = \sum_{j = 0}^{m} (\binom{m + 1}{j}) \frac{1}{j + 1} \sum_{k = 0}^{j} (\binom{j + 1}{k}) B_{k} n^{j + 1 - k} + (m + 1) Δ & (化简代入) \\ = \sum_{j = 0}^{m} \sum_{k = 0}^{j} (\binom{m + 1}{j}) (\binom{j + 1}{k}) \frac{B_{k}}{j + 1} n^{j + 1 - k} + (m + 1) Δ \\ = \sum_{j = 0}^{m} \sum_{k = 0}^{j} (\binom{m + 1}{j}) (\binom{j + 1}{j - k}) \frac{B_{j - k}}{j + 1} n^{k + 1} + (m + 1) Δ & (将k换为j-k) \\ = \sum_{j = 0}^{m} \sum_{k = 0}^{j} (\binom{m + 1}{j}) (\binom{j + 1}{k + 1}) \frac{B_{j - k}}{j + 1} n^{k + 1} + (m + 1) Δ & ((\binom{n}{m}) = (\binom{n}{n - m})) \\ = \sum_{k = 0}^{m} \sum_{j = k}^{m} (\binom{m + 1}{j}) (\binom{j + 1}{k + 1}) \frac{B_{j - k}}{j + 1} n^{k + 1} + (m + 1) Δ \\ = \sum_{k = 0}^{m} \sum_{j = k}^{m} (\binom{m + 1}{j}) (\binom{j}{k}) * \frac{j + 1}{k + 1} * \frac{B_{j - k}}{j + 1} n^{k + 1} + (m + 1) Δ & ((\binom{n + 1}{m + 1}) = (\binom{n}{m}) \times \frac{n + 1}{m + 1}) \\ = \sum_{k = 0}^{m} \frac{n^{k + 1}}{k + 1} \sum_{j = k}^{m} (\binom{m + 1}{j}) (\binom{j}{k}) B_{j - k} + (m + 1) Δ \\ = \sum_{k = 0}^{m} \frac{n^{k + 1}}{k + 1} (\binom{m + 1}{k}) \sum_{j = k}^{m} (\binom{m + 1 - k}{j - k}) B_{j - k} + (m + 1) Δ \\ = \sum_{k = 0}^{m} \frac{n^{k + 1}}{k + 1} (\binom{m + 1}{k}) \sum_{j = 0}^{m - k} (\binom{m + 1 - k}{j}) B_{j} + (m + 1) Δ \\ = \sum_{k = 0}^{m} \frac{n^{k + 1}}{k + 1} (\binom{m + 1}{k}) [m - k = 0] + (m + 1) Δ & (将(1.1)带入) \\ = \frac{n^{m + 1}}{m + 1} (\binom{m + 1}{m}) + (m + 1) Δ \\ = n^{m + 1} + (m + 1) Δ \end{aligned}

$\begin{aligned} n^{m+1}&=\sum_{j=0}^{m} \binom{m+1}{j} S'_ {j}(n)+ \binom{m+1}{m} \Delta & \text{(只有j=m时有差异)} \\ &=\sum_{j=0}^{m} \binom{m+1}{j} \frac{1}{j+1} \sum_{k=0}^j \binom {j+1} k B_k n^{j+1-k} + (m+1) \Delta & \text{(化简代入)} \\ &=\sum_{j=0}^{m} \sum_{k=0}^j \binom{m+1}{j} \binom {j+1} k \frac{B_k}{j+1} n^{j+1-k} + (m+1) \Delta \\ &=\sum_{j=0}^{m} \sum_{k=0}^j \binom{m+1}{j} \binom {j+1} {j-k} \frac{B_{j-k}}{j+1} n^{k+1} + (m+1) \Delta & \text{(将k换为j-k)} \\ &=\sum_{j=0}^{m} \sum_{k=0}^j \binom{m+1}{j} \binom {j+1} {k+1} \frac{B_{j-k}}{j+1} n^{k+1} + (m+1) \Delta & (\binom {n}{m} = \binom {n}{n-m}) \\ &=\sum_{k=0}^{m} \sum_{j=k}^{m} \binom{m+1}{j} \binom {j+1} {k+1} \frac{B_{j-k}}{j+1} n^{k+1} + (m+1) \Delta \\ &=\sum_{k=0}^{m} \sum_{j=k}^{m} \binom{m+1}{j} \binom {j} {k} * \frac{j+1}{k+1} * \frac{B_{j-k}}{j+1} n^{k+1} + (m+1) \Delta & (\binom{n+1}{m+1}=\binom{n}{m} \times \frac{n+1}{m+1}) \\ &=\sum_{k=0}^{m} \frac{n^{k+1}}{k+1} \sum_{j=k}^{m} \binom{m+1}{j} \binom {j} {k} B_{j-k} + (m+1) \Delta \\ &=\sum_{k=0}^{m} \frac{n^{k+1}}{k+1} \binom{m+1}{k} \sum_{j=k}^{m} \binom{m+1-k}{j-k} B_{j-k} + (m+1) \Delta \\ &=\sum_{k=0}^{m} \frac{n^{k+1}}{k+1} \binom{m+1}{k} \sum_{j=0}^{m-k} \binom{m+1-k}{j} B_{j} + (m+1) \Delta \\ &=\sum_{k=0}^{m} \frac{n^{k+1}}{k+1} \binom{m+1}{k} [m-k=0] + (m+1) \Delta & (\text{将(1.1)带入)} \\ &=\frac{n^{m+1}}{m+1} \binom{m+1}{m} + (m+1) \Delta \\ &=n^{m+1} + (m+1) \Delta \\ \end{aligned}$

所以 $(m+1)\Delta=0$ , 又 $m \ge 1$ , 所以 $\Delta=0$ ，证毕。