23 | 3 月 | 2017 | FreeSandal

數學證明追求邏輯嚴謹，因此常常讀來定義不斷符號滿篇。偶兒間讀到簡明扼要之定理推導，應當會樂於分享乎？特別介紹證明維基網頁給有興趣的讀者︰

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借著分析比較兩種維基版本上的證明︰

Faulhaber’s formula

Theorem

Let n and p be positive integers.

Then:

$\sum \limits_{k = 1}^n k^p = \frac 1 {p + 1} \sum \limits_{i = 0}^p \left({-1}\right)^i \binom {p + 1} i B_i n^{p + 1 - i}$

where Bn denotes the nth Bernoulli number.

Proof

Let $x \ge 0$ .

$\sum \limits_{k = 0}^{n - 1} e^{k x} = \sum \limits_{k = 0}^{n - 1} \sum \limits_{p = 0}^\infty \frac {\left({k x}\right)^p} {p!}$ Power Series Expansion for Exponential Function

$= \sum \limits_{p = 0}^\infty \left({\sum \limits_{k = 0}^{n - 1} k^p}\right) \frac {x^p} {p!}$ rearrangement is valid by Tonelli’s Theorem

We also have:

$\sum \limits_{k = 0}^{n - 1} e^{k x} = \frac {1 - e^{n x} } {1 - e^x}$ Geometric Series

$= \frac {e^{n x} - 1} x \frac x {e^x - 1}$

$= \sum \limits_{p = 0}^\infty \frac {n^{p + 1} x^p} {\left({p + 1}\right)!} \sum \limits_{p = 0}^\infty \frac {B_p x^p} {p!}$ by definition of Bernoulli Numbers

$= \sum \limits_{p = 0}^\infty \sum \limits_{i = 0}^p \frac {n^{p + 1 - i} x^{p - i} } {\left({p + 1 - i}\right)!} \frac {B_i x^i} {i!}$ Cauchy Product

$= \sum \limits_{p = 0}^\infty \left({\frac 1 {p + 1} \sum \limits_{i = 0}^p \binom {p + 1} i B_i n^{p + 1 - i} }\right) \frac {x^p} {p!}$

By equating coefficients, we find that:

$\sum \limits_{k = 0}^{n - 1} k^p = \frac 1 {p + 1} \sum \limits_{i = 0}^p \binom {p + 1} i B_i n^{p + 1 - i}$

$\implies \ \ \sum \limits_{k = 1}^n k^p = \frac 1 {p + 1} \sum \limits_{i = 0}^p \left({-1}\right)^i \binom {p + 1} i B_i n^{p + 1 - i}$ since $B_1 = - \frac{1}{2}$ and Odd Bernoulli Numbers Vanish

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Proof

Let

S_{p}(n)=\sum_{k=1}^{n} k^p,

denote the sum under consideration for integer $p\ge 0.$

Define the following exponential generating function with (initially) indeterminate $z$

G(z,n)=\sum_{p=0}^{\infty} S_{p}(n) \frac{1}{p!}z^p.

We find

</span

This is an entire function in $z$ so that $z$ can be taken to be any complex number.

We next recall the exponential generating function for the Bernoulli polynomials $B_j(x)$

\frac{ze^{zx}}{e^{z}-1}=\sum_{j=0}^{\infty} B_j(x) \frac{z^j}{j!},

where $B_j=B_j(0)$ denotes the Bernoulli number (with the convention $B_{1}=-\frac{1}{2}$ ). We obtain the Faulhaber formula by expanding the generating function as follows: