巴塞尔问题(Basel problem)的多种解法

（PS：本文会不断更新）

如何计算

\sum_{k = 1}^{\infty} \frac{1}{k^{2}} = \frac{π^{2}}{6}

解决这个问题的方法在近代不断涌现。这里我从各处摘抄到一些方法，列举在此，仅供大家参考。

如有错误，请向我指出，谢谢！（PS：最近发现忻州师范学院某网页抄了我博客后不给Reference，希望大家明辨是非）

首先，我们需要知道这个问题的等价形式,将这个数列除以4，我们自然得到

\sum_{k = 1}^{\infty} \frac{1}{(2 k - 1)^{2}} = \frac{π^{2}}{8}

而以下某些证明会用到这一点。

证明1：欧拉的证明

欧拉的证明是十分聪明的。他只是将幂级数同有限的多项式联系到了一起，就得到了答案。首先注意到

\sin (x) = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} + \dots

从而

\frac{\sin (x)}{x} = 1 - \frac{x^{2}}{3!} + \frac{x^{4}}{5!} - \frac{x^{6}}{7!} + \dots

x = n \cdot π, (n = \pm 1, \pm 2, \pm 3, \dots) .

\begin{aligned} \frac{\sin (x)}{x} & = (1 - \frac{x}{π}) (1 + \frac{x}{π}) (1 - \frac{x}{2 π}) (1 + \frac{x}{2 π}) (1 - \frac{x}{3 π}) (1 + \frac{x}{3 π}) \dots \\ = (1 - \frac{x^{2}}{π^{2}}) (1 - \frac{x^{2}}{4 π^{2}}) (1 - \frac{x^{2}}{9 π^{2}}) \dots . \end{aligned}

从而我们对这个无穷乘积的

- (\frac{1}{π^{2}} + \frac{1}{4 π^{2}} + \frac{1}{9 π^{2}} + \dots) = - \frac{1}{π^{2}} \sum_{n = 1}^{\infty} \frac{1}{n^{2}} .

- \frac{1}{6} = - \frac{1}{π^{2}} \sum_{n = 1}^{\infty} \frac{1}{n^{2}} .

注：欧拉给出过严谨的证明，但是由于他的第一个证明太广为人知，所以有时候会认为他没给出真正的证明。不过贴吧里的 tq唐乾吧友提醒了我，实际上，欧拉有他真正的证明。是通过如下方式：首先令

z^{n} - a^{n} = (z - 1) \prod_{k = 1}^{(n - 1) / 2} (z^{2} - 2 a z \cos \frac{2 k π}{n} + a^{2})

\begin{aligned} {(1 + \frac{x}{N})}^{N} - {(1 + \frac{x}{N})}^{N} & = \frac{2 x}{N} \prod_{k = 1}^{(N - 1) / 2} (2 + \frac{2 x^{2}}{N^{2}} - 2 (1 - \frac{x^{2}}{N^{2}}) \cos \frac{2 k π}{N}) \\ = \frac{2 x}{N} \prod_{k = 1}^{(N - 1) / 2} ((1 - \cos \frac{2 k π}{N}) + \frac{x^{2}}{N^{2}} (1 + \cos \frac{2 k π}{N})) \\ = C_{N} x \prod_{k = 1}^{(N - 1) / 2} (1 + \frac{x^{2}}{N^{2}} \frac{1 + \cos (2 k π / N)}{1 - \cos (2 k π / N)}) \end{aligned}

考虑一次项系数知道

\frac{e^{x} - e^{- x}}{2} = x \prod_{k = 1}^{\infty} (1 + \frac{x^{2}}{k^{2} π^{2}})

比较三次项系数可知答案

证明2：一个初等的证明
以下证明第一次来自Ioannis Papadimitriou于1973年在American Math Monthly 80(4):424-425页发表的。Apostol在同一份杂志425-430发表了用这个方法计算

这似乎是这个问题最“初等”的一个证明了，只需要知道三角函数相应知识就能够完成。我们先证明一个恒等式：

Lemma: 令

$\cot^{2} ω_{m} + \cot^{2} (2 ω_{m}) + \dots \cot^{2} (m ω_{m}) = \frac{m (2 m - 1)}{3} .$

证明：由于

\begin{aligned} \sin n θ & = (\binom{n}{1}) \sin θ \cos^{n - 1} θ - (\binom{n}{3}) \sin^{3} θ \cos^{n - 3} θ + \dots \pm \sin^{n} θ \\ = \sin^{n} θ ((\binom{n}{1}) \cot^{n - 1} θ - (\binom{n}{3}) \cot^{n - 3} θ + \dots \pm 1) \end{aligned}

(\binom{n}{1}) x^{m} - (\binom{n}{3}) x^{m - 1} + \dots \pm 1

由于三角不等式

\sum_{k = 1}^{m} \cot^{2} (k ω_{m}) < \sum_{k = 1}^{m} \frac{1}{k^{2} ω_{m}^{2}} < m + \sum_{k = 1}^{m} \cot^{2} (k ω_{m})

\frac{m (2 m - 1) π^{2}}{3 (2 m + 1)^{2}} < \sum_{k = 1}^{m} \frac{1}{k^{2}} < \frac{m (2 m - 1) π^{2}}{3 (2 m + 1)^{2}} + \frac{m π^{2}}{(2 m + 1)^{2}}

证明3：数学分析的证明
这个证明来自Apostol在1983年的“Mathematical Intelligencer”,只需要简单的高数知识。

注意到恒等式

\frac{1}{n^{2}} = \int_{0}^{1} \int_{0}^{1} x^{n - 1} y^{n - 1} d x d y

\sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \int_{0}^{1} \int_{0}^{1} (\sum_{n = 1}^{\infty} (x y)^{n - 1}) d x d y = \int_{0}^{1} \int_{0}^{1} \frac{1}{1 - x y} d x d y

\sum_{n = 1}^{\infty} \frac{1}{n^{2}} = 2 \iint_{S} \frac{1}{1 - u^{2} + v^{2}} d u d v

\begin{aligned} 2 \iint_{S} \frac{1}{1 - u^{2} + v^{2}} d u d v & = 4 \int_{0}^{1 / 2} \int_{0}^{u} \frac{1}{1 - u^{2} + v^{2}} d v d u + 4 \int_{1 / 2}^{1} \int_{0}^{1 - u} \frac{1}{1 - u^{2} + v^{2}} d v d u \\ = 4 \int_{0}^{1 / 2} \frac{1}{\sqrt{1 - u^{2}}} \arctan (\frac{u}{\sqrt{1 - u^{2}}}) d u \\ + 4 \int_{1 / 2}^{1} \frac{1}{\sqrt{1 - u^{2}}} \arctan (\frac{1 - u}{\sqrt{1 - u^{2}}}) d u \end{aligned}

\begin{aligned} \sum_{n = 1}^{\infty} \frac{1}{n^{2}} & = 4 \int_{0}^{1 / 2} \frac{\arcsin u}{\sqrt{1 - u^{2}}} d u + 4 \int_{1 / 2}^{1} \frac{1}{\sqrt{1 - u^{2}}} (\frac{π}{4} - \frac{\arcsin u}{2}) d u \\ = [2 \arcsin u^{2}]_{0}^{1 / 2} + [π \arcsin u - \arcsin u^{2}]_{1 / 2}^{1} \\ = \frac{π^{2}}{18} + \frac{π^{2}}{2} - \frac{π^{2}}{4} - \frac{π^{2}}{6} + \frac{π^{2}}{36} \\ = \frac{π^{2}}{6} \end{aligned}

证明4:数学分析的证明

(Calabi, Beukers & Kock.)同样利用上一问的结论,不过这次我们计算的是：

\sum_{n = 0}^{\infty} \frac{1}{(2 n + 1)^{2}} = \int_{0}^{1} \int_{0}^{1} \frac{d x d y}{1 - x^{2} y^{2}}

做代换

(u, v) = (\arctan x \sqrt{\frac{1 - y^{2}}{1 - x^{2}}}, \arctan x \sqrt{\frac{1 - x^{2}}{1 - y^{2}}})

从而有

雅可比行列式即为

\begin{aligned} \frac{\partial (x, y)}{\partial (u, v)} & = | \begin{matrix} \cos u / \cos v & \sin u \sin v / \cos v^{2} \\ \sin u \sin v / \cos u^{2} & \cos v / \cos u \end{matrix} | \\ = 1 - \frac{\sin^{2} u \sin^{2} v}{\cos^{2} u \cos^{2} v} = 1 - x^{2} y^{2} \end{aligned}

从而

\frac{3}{4} ζ (2) = \sum_{n = 0}^{\infty} \frac{1}{(2 n + 1)^{2}} = \iint_{A} d u d v

其中

证明5:复分析的证明

这个证明在很多复分析书上都有。我们同样可以利用留数计算该结果,考虑

实轴交点为

| \cot π z |^{2} = \frac{\cos^{2} x + \sinh^{2} y}{\sin^{2} x + \sinh^{2} y}

| \oint_{P_{n}} z^{- 2} \cot π z | \leq \frac{2}{n^{2}} (8 n + 2)

成立，在

利用留数定理，则有

2 π i \sum_{k = - \infty}^{\infty} Res (z^{- 2} \cot π z, k) = lim_{n \to \infty} \oint_{P_{n}} z^{- 2} \cot π z d z = 0

而每一点的留数，计算有

\sum_{k = 1}^{\infty} \frac{2}{π k^{2}} = \frac{π}{3}

答案显而易见了。

证明6:复数积分的证明

本证明由Dennis C.Russell给出。考虑积分

I = \int_{0}^{π / 2} \ln (2 \cos x) d x

那么利用

\begin{aligned} I & = \int_{0}^{π / 2} i x + \ln (1 + e^{- 2 i x}) d x \\ = i \frac{π^{2}}{8} + \int_{0}^{π / 2} l n (1 + e^{- 2 i x}) d x \end{aligned}

\ln (1 + x) = x - x^{2} / 2 + x^{3} / 3 - x^{4} / 4 + \dots

\ln (1 + e^{- 2 i x}) = e^{2 i x} - e^{- 4 i x} / 2 + e^{- 6 i x} / 3 + \dots

\int_{0}^{π / 2} \ln (1 + e^{- 2 i x}) d x = - \frac{1}{2 i} (e^{- i π} - 1 - \frac{e^{- 2 i π} - 1}{2^{2}} + \frac{e^{- 3 i π} - 1}{3^{2}} - \frac{e^{- 4 i π} - 1}{4^{2}} + \dots)

\int_{0}^{π / 2} \ln (1 + e^{- 2 i x}) d x = \frac{1}{i} \sum_{k = 1}^{\infty} \frac{1}{(2 k - 1)^{2}} = \frac{- 3 i}{4} ζ (2)

I = i (\frac{π^{2}}{8} + \frac{- 3}{4} ζ (2))

\int_{0}^{π / 2} \ln (\cos x) d x = - \frac{π}{2} \ln 2

证明7:泰勒公式证明

(Boo Rim Choe 在1987 American Mathematical Monthly上发表)利用反三角函数

\arcsin x = \sum_{n = 0}^{\infty} \frac{1 \cdot 3 \dots (2 n - 1)}{2 \cdot 4 \dots (2 n)} \frac{x^{2 n + 1}}{2 n + 1}

t = \sum_{n = 0}^{\infty} \frac{1 \cdot 3 \dots (2 n - 1)}{2 \cdot 4 \dots (2 n)} \frac{\sin^{2 n + 1} t}{2 n + 1}

\int_{0}^{π / 2} \sin^{2 n + 1} x d x = \frac{2 \cdot 4 \dots (2 n)}{3 \cdot 5 \dots (2 n + 1)}

\frac{π^{2}}{8} = \int_{0}^{π / 2} t d t = \sum_{n = 0}^{\infty} \frac{1}{(2 n + 1)^{2}}

证明8:复分析证明

(T. Marshall 在American Math Monthly,2010)对于

R (z) = \sum \frac{1}{\log^{2} z}

这个和是对于每一个

这里有几个Claim:

由于 1.和 3.有

R (z) = \frac{z}{(z - 1)^{2}} .

现在令

\sum_{n = - \infty}^{\infty} \frac{1}{(w - n)^{2}} = \frac{π^{2}}{\sin^{2} (π w)}

\sum_{k = 0}^{\infty} \frac{1}{(2 k + 1)^{2}} = \frac{π^{2}}{8},

证明9:傅立叶分析证明

考虑函数

f (x) = \frac{π^{2}}{3} + \sum_{n = 1}^{\infty} ((- 1)^{n} \frac{4}{n^{2}} \cos n x)

显而易见，代入

证明10:傅立叶分析证明

考虑函数

f (x) = 2 \sum_{n = 1}^{\infty} (\frac{(- 1)^{n + 1}}{n} \sin n x)

利用Parseval等式

\sum_{n = 1}^{\infty} | a_{n} |^{2} = \frac{1}{2 π} \int_{- π}^{π} x^{2} d x

其中

那么有

2 \sum_{k = 1}^{\infty} \frac{1}{k^{2}} = \frac{1}{2 π} \int_{- π}^{π} x^{2} d x

可得答案

证明11:傅立叶分析证明

考虑

f (t) = \sum_{n = 1}^{\infty} \frac{\cos n t}{n^{2}}

\sum_{n = 1}^{N} \sin n t = \frac{e^{i t} - e^{i (N + 1) t}}{2 i (1 - e^{i t})} + \frac{1 - e^{i N) t}}{2 i (1 - e^{i t})}

\frac{2}{| 1 - e^{i t} |} = \frac{1}{\sin t / 2}

\sum_{n = 1}^{\infty} \frac{\sin t}{n}

f^{'} (t) = - \sum_{n = 1}^{\infty} \frac{\sin n t}{n} = ℑ (\log (1 - e^{i t})) = \arg (1 - e^{i t}) = \frac{t - π}{2}

- ζ (2) / 2 - ζ (2) = f (π) - f (0) = \int_{0}^{π} \frac{t - π}{2} d t = - \frac{π^{2}}{4}

证明12:泊松公式证明

(Richard Troll)由泊松求和公式

\sum_{n = - \infty}^{\infty} f (n) = \sum_{k = - \infty}^{\infty} \hat{f} (k)

其中

那么有

\hat{f} (ξ) = \frac{2 a}{a^{2} + 4 π^{2} ξ^{2}}

也就是说

\frac{1}{2 a} \sum_{n \in Z} e^{- a | n |} - \frac{1}{a^{2}} = \sum_{k = 1}^{\infty} \frac{2}{a^{2} + 4 π^{2} k^{2}}

则

lim_{a \to 0} \sum_{k = 1}^{\infty} \frac{2}{a^{2} + 4 π^{2} k^{2}} = lim_{a \to 0} {\frac{1}{2 a} (\frac{e^{a} + 1}{e^{a} - 1}) - \frac{1}{a^{2}}} = \frac{1}{12}

从而就有

证明13:概率论证明

(Luigi Pace 发表于2011 American Math Monthly)

设

令随机变量

\begin{aligned} p_{Y} (y) & = \int_{0}^{\infty} x p_{X_{1}} (x y) p_{X_{2}} (x) d x = \frac{4}{π^{2}} \int_{0}^{\infty} \frac{x}{(1 + x^{2} y^{2}) (1 + x^{2})} d x \\ = \frac{2}{π^{2} (y^{2} - 1)} {[\log (\frac{1 + x^{2} y^{2}}{1 + x^{2}})]}_{x = 0}^{\infty} = \frac{2}{π^{2}} \frac{\log (y^{2})}{y^{2} - 1} = \frac{4}{π^{2}} \frac{\log (y)}{y^{2} - 1} . \end{aligned}

由于

\frac{1}{2} = \int_{0}^{1} \frac{4}{π^{2}} \frac{\log (y)}{y^{2} - 1} d y

也就是说

\frac{π^{2}}{8} = \int_{0}^{1} \frac{- \log (y)}{1 - y^{2}} d y = - \int_{0}^{1} \log (y) (1 + y^{2} + y^{4} + \dots) d y = \sum_{k = 0}^{\infty} \frac{1}{(2 k + 1)^{2}}

那么答案显而易见。

证明14:积分+函数方程证明

(H Haruki,S Haruki在1983年 American Mathematical Monthly发表)

由于

\sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \sum_{n = 1}^{\infty} \frac{1}{n} \int_{0}^{1} x^{n - 1} d x = \int_{0}^{1} \frac{\log (1 - x)}{x} d x

只需要算出这个积分值即可，我们令

f (a) = \int_{0}^{1} \frac{\log (x^{2} - 2 x \cos a + 1)}{x} d x

要证明

利用等式

f (a / 2) + f (π - a / 2) = \int_{0}^{1} \frac{\log (x^{4} - 2 x^{2} \cos a + 1)}{x} = \frac{1}{2} \frac{\log (t^{2} - 2 t \cos a + 1)}{t} d t = \frac{1}{2} f (a)

中间是令

f^{″} (a_{0} / 2) + f^{″} (π - a_{0} / 2) = 2 f^{″} (a_{0}) = 2 M

lim_{n \to \infty} f^{″} (a_{0} / 2^{n}) = f^{″} (0) = M

f (a) = α \frac{a^{2}}{2} + β a + γ

f^{'} (a) = \int_{0}^{1} \frac{2 \sin a}{1 + x^{2} - 2 x \cos a} d x

\int_{0}^{1} \frac{\log (1 - x)}{x} d x = - \frac{π^{2}}{6}

证明15:三角恒等式的初等证明

(Josef Hofbauer发表于2002年American Mathematical Monthly)

\frac{1}{\sin^{2} x} = \frac{1}{4 \sin^{2} \frac{x}{2} \cos^{2} \frac{x}{2}} = \frac{1}{4} [\frac{1}{\sin^{2} \frac{x}{2}} + \frac{1}{\sin^{2} \frac{π + x}{2}}]

1 = \frac{1}{\sin^{2} \frac{π}{2}} = \frac{1}{4 [\frac{1}{\sin^{2} \frac{π}{4}} + \frac{1}{\sin^{2} \frac{3 π}{4}}]} = \dots = \frac{1}{4^{n}} \sum_{k = 0}^{2^{n} - 1} \frac{1}{\sin^{2} \frac{(2 k + 1) π}{2^{n + 1}}} = \frac{2}{4^{n}} \sum_{k = 0}^{2^{n - 1} - 1} \frac{1}{\sin^{2} \frac{(2 k + 1) π}{2^{n + 1}}}

又由于

令

1 > \frac{8}{π^{2}} \sum_{k = 0}^{2^{n} - 1} \frac{1}{(2 k + 1)^{2}} > 1 - \frac{1}{N}

证明16:三角多项式的证明

(Kortram发表于1996年 Mathematics Magazine)

对于奇数

F_{n} (y) = n y \prod_{j = 1}^{m} (1 - \frac{y^{2}}{\sin^{2} (j π / n)})

\sin n x = n \sin x \prod_{j = 1}^{m} (1 - \frac{\sin^{2} x}{\sin^{2} (j π / n)})

- \frac{n^{3}}{6} = - \frac{n}{6} - n \sum_{j = 1}^{m} \frac{1}{\sin^{2} (j π / n)}

\frac{1}{6} - \sum_{j = 1}^{m} \frac{1}{n^{2} \sin^{2} (j π / n)} = \frac{1}{6 n^{2}}

\frac{1}{6} - \sum_{j = 1}^{M} \frac{1}{n^{2} \sin^{2} (j π / n)} = \frac{1}{6 n^{2}} + \sum_{j = M + 1}^{m} \frac{1}{n^{2} \sin^{2} (j π / n)}

0 < \frac{1}{6} - \sum_{j = 1}^{M} \frac{1}{n^{2} \sin^{2} (j π / n)} = \frac{1}{6 n^{2}} + \sum_{j = M + 1}^{m} \frac{1}{4 j^{2}}

0 \leq \frac{1}{6} - \sum_{j = 1}^{M} \frac{1}{π^{2} j^{2}} \leq \sum_{j = M + 1}^{m} \frac{1}{4 j^{2}}

\sum_{j = 1}^{\infty} \frac{1}{π^{2} j^{2}} = \frac{1}{6}

证明17:积分证明

(Matsuoka发表于1961年American Mathematical Montly)

考虑积分

I_{n} = \int_{0}^{π / 2} \cos^{2 n} x d x and J_{n} = \int_{0}^{π / 2} x^{2} \cos^{2 n} x d x

I_{n} = \frac{1 \cdot 3 \cdot 5 \dots (2 n - 1)}{2 \cdot 4 \cdot 6 \dots 2 n} \frac{π}{2} = \frac{(2 n)!}{4^{n} (n!)^{2}} \frac{π}{2}

\begin{aligned} I_{n} & = [x \cos^{2 n} x]_{0}^{π / 2} + 2 n \int_{0}^{π / 2} x \sin x \cos^{2 n - 1} x d x \\ = n (2 n - 1) J_{n - 1} - 2 n^{2} J_{n} \end{aligned}

\frac{(2 n)!}{4^{n} (n!)^{2}} \frac{π}{2} = n (2 n - 1) J_{n - 1} - 2 n^{2} J_{n}

\frac{π}{4 n^{2}} = \frac{4^{n - 1} (n - 1)!^{2}}{(2 n - 2)!} J_{n - 1} - \frac{4^{n} n!^{2}}{(2 n)!} J_{n}

\frac{π}{4} \sum_{n = 1}^{N} \frac{1}{n^{2}} = J_{0} - \frac{4^{N} N!^{2}}{(2 N)!} J_{N}

J_{N} < \frac{π^{2}}{4} \int_{0}^{π / 2} \sin^{2} x \cos^{2 N} x d x = \frac{π^{2}}{4} (I_{N} - I_{N + 1}) = \frac{π^{2} I_{N}}{8 (N + 1)}

0 < \frac{4^{N} N!^{2}}{(2 N)!} J_{N} < \frac{π^{3}}{16 (N + 1)}

证明18:Fejér核的证明

(Stark在1969年American Mathematical Monthly上的证明)

对于Fejér核有如下等式：

{(\frac{\sin n x / 2}{\sin x / 2})}^{2} = \sum_{k = - n}^{n} (n - | k |) e^{i k x} = n + 2 \sum_{k = 1}^{n} (n - k) \cos k x

\begin{aligned} \int_{0}^{π} x {(\frac{\sin n x / 2}{\sin x / 2})}^{2} & = \frac{n π^{2}}{2} + 2 \sum_{k = 1}^{n} (n - k) \int_{0}^{π} x \cos k x d x \\ = \frac{n π^{2}}{2} - 2 \sum_{k = 1}^{n} (n - k) \frac{1 - (- 1)^{k}}{k^{2}} \\ = \frac{n π^{2}}{2} - 4 n \sum_{1 \leq k \leq n, 2 ∤ k} \frac{1}{k^{2}} + 4 \sum_{1 \leq k \leq n, 2 ∤ k} \frac{1}{k} \end{aligned}

\int_{0}^{π} \frac{x}{8 N} {(\frac{\sin N x}{\sin x / 2})}^{2} d x = \frac{π^{2}}{8} - \sum_{r = 0}^{N - 1} \frac{1}{(2 r + 1)^{2}} + O (\frac{\log N}{N})

\int_{0}^{π} \frac{x}{8 N} {(\frac{\sin N x}{\sin x / 2})}^{2} d x < \frac{π^{2}}{8 N} \int_{0}^{π} \sin^{2} N x \frac{d x}{x} = \frac{π^{2}}{8 N} \int_{0}^{N π} \sin^{2} y \frac{d y}{y} = O (\frac{\log N}{N})

\frac{π^{2}}{8} = \sum_{r = 0}^{\infty} \frac{1}{(2 r + 1)^{2}}

证明19:Gregory定理证明

证明来自Borwein & Borwein的著作"Pi and the AGM"

以下公式是著名的Gregory定理：

\frac{π}{4} = \sum_{n = 0}^{\infty} \frac{(- 1)^{n}}{2 n + 1}

令

a_{N} = \sum_{n = - N}^{N} \frac{(- 1)^{n}}{2 n + 1}, b_{N} = \sum_{n = - N}^{N} \frac{1}{(2 n + 1)^{2}}

我们需要证明

如果

\frac{1}{(2 n + 1) (2 m + 1)} = \frac{1}{2 (m - n)} (\frac{1}{2 n + 1} - \frac{1}{2 m + 1})

\begin{aligned} a_{N}^{2} - b_{N} & = \sum_{n = - N}^{N} \sum_{m = - N, m \neq n}^{N} \frac{(- 1)^{m + n}}{2 (m - n)} (\frac{1}{2 n + 1} - \frac{1}{2 m + 1}) \\ = \sum_{n = - N}^{N} \sum_{m = - N, m \neq n}^{N} \frac{(- 1)^{m + n}}{(m - n)} \frac{1}{(m - n) (2 n + 1)} = \sum_{n = - N}^{N} \frac{(- 1)^{n} c_{n, N}}{2 n + 1} \end{aligned}

c_{n, N} = \sum_{m = - N, m \neq n}^{N} \frac{(- 1)^{m}}{m - n}

c_{n, N} = (- 1)^{n + 1} \sum_{j = N - n + 1}^{N + n} \frac{(- 1)^{j}}{j}

\begin{aligned} | a_{N}^{2} - b_{N} | & \leq \sum (\frac{1}{(2 n - 1) (N - n + 1)} + \frac{1}{(2 n + 1) (N - n + 1)}) \\ = \sum_{n = 1}^{N} \frac{1}{2 N + 1} (\frac{2}{2 n - 1} + \frac{1}{N - n + 1}) + \sum_{n = 1}^{N} \frac{1}{2 N + 3} (\frac{2}{2 n + 1} + \frac{1}{N - n + 1}) \\ \leq \frac{1}{2 N + 1} (2 + 4 \log (2 N + 1) + 2 + 2 \log (N + 1)) \end{aligned}

证明20:数论的证明

(本证明来自华罗庚的数论)

需要用到整数能被表示为四个平方的和。令

r (n) = 8 \sum_{m | n, 4 ∤ m} m

R (N) = 1 + 8 \sum_{n = 1}^{N} \sum_{m | n, 4 ∤ m} m = 1 + 8 \sum_{m \leq N, 4 ∤ m} m ⌊ \frac{N}{m} ⌋ = 1 + 8 (θ (N) - 4 θ (N / 4))

θ (x) = \sum_{m \leq x} m ⌊ \frac{x}{m} ⌋

\begin{aligned} θ (x) & = \sum_{m r \leq x} m = \sum_{r \leq x} \sum_{m = 1}^{⌊ x / r ⌋} m = \frac{1}{2} \sum_{r \leq x} ({⌊ \frac{x}{r} ⌋}^{2} + ⌊ \frac{x}{r} ⌋) = \frac{1}{2} \sum_{r \leq x} ({⌊ \frac{x}{r} ⌋}^{2} + O (\frac{x}{r})) \\ = \frac{x^{2}}{2} (ζ (2) + O (1 / x)) + O (x \log x) = \frac{ζ (2) x^{2}}{2} + O (x \log x) \end{aligned}

R (N) \sim \frac{π^{2}}{2} N^{2} \sim 4 ζ (2) (N^{2} - \frac{N^{2}}{4})

证明21:类似的初等证明

首先我们要证明这个等式：

$\sum_{k = 1}^{n} \cot^{2} (\frac{2 k - 1}{2 n} \frac{π}{2}) = 2 n^{2} - n$

是由于注意到

\cos 2 n θ = Re (\cos θ + i \sin θ)^{2 n} = \sum_{k = 0}^{n} (- 1)^{k} (\binom{2 n}{2 k}) \cos^{2 n - 2 k} θ \sin^{2 k} θ

就立即可得

\frac{\cos 2 n θ}{\sin^{2 n} θ} = \sum_{k = 0}^{n} (- 1)^{k} (\binom{2 n}{2 k}) \cot^{2 n - 2 k} θ

令

f (x) = \sum_{k = 0}^{n} (- 1)^{k} (\binom{2 n}{2 k}) x^{n - k}

有根

有了这个等式，我们类似初等证明中的方法进行证明

现在

1 / θ^{2} - 2 / 3 < \cot^{2} θ < 1 / θ^{2}

2 n^{2} - n < \sum_{k = 1}^{n} {(\frac{2 n}{2 k - 1} \frac{2}{π})}^{2} < 2 n^{2} - n + 2 n / 3

\frac{π^{2}}{16} \frac{2 n^{2} - n}{n^{2}} < \sum_{k = 1}^{n} \frac{1}{(2 k - 1)^{2}} < \frac{π^{2}}{16} \frac{2 n^{2} - n / 3}{n^{2}}

\sum_{k = 1}^{\infty} \frac{1}{(2 k - 1)^{2}} = \frac{π^{2}}{8}

证明22:伯努利数的证明

函数

\frac{x}{e^{x} - 1} = \sum_{n \in Z} \frac{2 π i n}{x - 2 π i n} = \sum_{n \in Z} - (\frac{1}{1 - \frac{x}{2 π i n}}) .

而注意到后者又可以展开为几何级数相加：

\frac{x}{e^{x} - 1} = - \sum_{n \in Z} \sum_{k \geq 0} {(\frac{x}{2 π i n})}^{k} = \sum_{k \geq 0} (- 1)^{n + 1} \frac{2 ζ (2 n)}{(2 π)^{2 n}} x^{2 n}

是由于在重排级数的同时，奇数项消去了而偶数项留下了，所以我们就得到如下式子：

B_{2 n} = (- 1)^{n + 1} \frac{2 ζ (2 n)}{(2 π)^{2 n}}

也就是要求计算

B_{2} = lim_{x \to 0} \frac{1}{x^{2}} {\frac{x}{e^{x} - 1} - 1 + \frac{x}{2}} = \frac{1}{12}

那么

证明23:超几何正切分布的证明

（本证明来自Lars Holst于2013年Journal of Applied Probability的证明）

注意到超几何正切函数

\int_{- \infty}^{x} \frac{2}{π (e^{y} - e^{- y})} d y = \frac{2}{π} \arctan (e^{x}) .

这样可以知道

$f_{2} (x) = \frac{4 x}{π^{2} (e^{x} - e^{- x})} .$

这是因为

\begin{aligned} \int_{- \infty}^{\infty} & \frac{2}{π (e^{y} + e^{- y})} \frac{2}{π (e^{x - y} + e^{y - x})} d y \\ = \frac{4}{π^{2}} \int_{0}^{\infty} \frac{u e^{- x}}{(1 + u^{2}) (1 + u^{2} e^{- 2 x})} d u \\ = \frac{4}{π (e^{x} - e^{- x})} \int_{0}^{\infty} (\frac{u}{1 + u^{2}} - \frac{u e^{- 2 x}}{1 + u^{2} e^{- 2 x}}) d u \\ = \frac{4 x}{π (e^{x} - e^{- x})} \end{aligned}

\begin{aligned} \sum_{k = 0}^{\infty} \frac{1}{(2 k + 1)^{2}} & = \sum_{k = 0}^{\infty} \int_{0}^{\infty} x e^{- (2 k + 1) x} d x \\ = \int_{0}^{\infty} x e^{- x} \sum_{k = 0}^{\infty} e^{- 2 k x} d x = \int_{0}^{\infty} \frac{x e^{- x}}{1 - e^{- 2 x}} d x \\ = \frac{π^{2}}{8} \int_{- \infty}^{\infty} f_{2} (x) d x = \frac{π^{2}}{8} \end{aligned}

这样可以得到结论。

http://www.cnblogs.com/misaka01034/p/BaselProof.html#title01

posted on 2017-10-11 00:12 Eufisky 阅读(8811) 评论(1) 编辑收藏举报

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