转载自http://www.ivy-end.com/archives/1098

今天我们来介绍一个曾经在数列题中做过无数遍的数列\(a_{n}=\frac{1}{n^{2}}\)，为了方便下文的讨论，我们记\(S_{n}\)为数列\(a_{n}\)的前\(n\)项和。我们做过最熟悉的题目莫过于证明\(S_{n}<2\)，这个证明对于大家来说应该是非常容易的\[S_{n}=\sum_{i=1}^{n}\frac{1}{i^{2}}=1+\sum_{i=2}^{n}\frac{1}{i\cdot\left(i-1 \right )}=2-\frac{1}{n}.\]

那么大家是否考虑过，\(\lim_{n \to \infty }S_{n}\)究竟等于多少？我记得严桂华讲过欧拉曾经计算过这个问题，今天在国外一个数学讨论网站上翻到了一篇关于如何计算\(\lim_{n \to \infty }S_{n}\)的问答（原文地址）。在此仅介绍几种方法。

首先给出结果，再用多种方法进行证明\[\lim_{n \to \infty }S_{n}=\frac{\pi^{2}}{6}.\]不得不说，这一结果是非常不可思议的，因为它居然包含了\(\pi\)。读者如果不相信的话，可以自行计算一下，当\(n\)趋向于正无穷时，答案便越接近\(\frac{\pi^{2}}{6}\)。

好了，废话不说，我们进入证明部分。

法一：

当\(0<x<\frac{\pi}{2}\)时，有\(0<\sin{x}<x<\tan{x}\)。至于这个命题的证明，请参考下图：

这样，我们可以得到\[\frac{1}{\tan^{2}{x}}<\frac{1}{x^{2}}<\frac{1}{\sin^{2}{x}}.\]我们知道，\(\frac{1}{\tan^{2}{x}}=\frac{1}{\sin^{2}{x}}-1\)（很容易由三角恒等式导出）。将区间\(\left(0,\frac{\pi}{2}\right)\)等分成\(2^{n}\)个部分，令\(x_{k}=\frac{\pi}{2}\cdot\frac{k}{2^{n}}\)，并将不等式两边对\(x_{k}\)进行求和，可以得到\[\sum_{k=1}^{2^n-1} \frac{1}{\sin^2 x_k} – \sum_{k=1}^{2^n-1} 1 < \sum_{k=1}^{2^n-1} \frac{1}{x_k^2} < \sum_{k=1}^{2^n-1} \frac{1}{\sin^2 x_k}.\]我们不妨将不等式右边的表达式记为\(T_{n}\)，那么原不等式可以化简为\[T_{n} – \left(2^n – 1\right) < \sum_{k=1}^{2^n-1} \left( \frac{2 \cdot 2^n}{\pi} \right)^2 \frac{1}{k^2} < T_n.\]接下来我们主要考察\(T_{n}\)。它看起来是一个非常复杂的求和，实际上可以非常简单的计算出来。首先，我们考虑下面的式子\[\frac{1}{\sin^2 x} + \frac{1}{\sin^2 (\frac{\pi}{2}-x)} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x \cdot \sin^2 x} = \frac{4}{\sin^2 2x}.\]因此，如果我们将\(T_{n}\)以\(\frac{\pi}{4}\)为中点进行配对（选取区间\(\left(0,\frac{\pi}{2}\right)\)左边的点\(x_{k}\)，与右边的点\(\frac{\pi}{2}-x_{k}\)），考虑递推，再加上中点\(\frac{1}{\sin^{2}{\frac{\pi}{4}}}=2\)，则可以得到\[T_{n}=4T_{n-1}+2.\]又\(T_{1}=2\)，我们可以很容易的得到\(T_{n}\)的通项公式\[T_n = \frac{2(4^n-1)}{3}.\]于是，我们得到了这样的不等式\[\frac{2(4^n-1)}{3} – (2^n-1) \leq \frac{4^{n+1}}{\pi^2} \sum_{k=1}^{2^n-1} \frac{1}{k^2} \leq \frac{2(4^n-1)}{3}.\]两边同时乘以\(\frac{\pi^{2}}{4^{n+1}}\)使得不等式的中间与我们的目标形式更加相似，这样我们又得到一个不等式\[\frac{\pi^{2}}{4^{n+1}} \cdot \frac{2(4^n-1)}{3} – (2^n-1) \leq \sum_{k=1}^{2^n-1} \frac{1}{k^2} \leq\frac{\pi^{2}}{4^{n+1}} \cdot \frac{2(4^n-1)}{3}.\]当\(n\rightarrow\infty\)时，不等式左右两边同时趋向于\(\frac{\pi^{2}}{6}\)，由夹逼准则得\(\lim_{n \to \infty }S_{n}=\frac{\pi^{2}}{6}\)。证毕。

法二：

我们可以借助函数\(f\left(x\right)=x^{2},x\in\left[-\pi,\pi\right]\)来找到它对应的三角傅立叶级数\[\frac{a_{0}}{2}+\sum_{n=1}^{\infty }(a_{n}\cos nx+b_{n}\sin x),\]且它是周期收敛的。

又\(f\left(x\right)\)为偶函数，这足够让我们得到它的系数\[a_{n}=\frac{1}{\pi }\int_{-\pi }^{\pi }f(x)\cos nx\;dx\qquad n=0,1,2,3,\cdots,\]因为\[b_{n}=\frac{1}{\pi }\int_{-\pi }^{\pi }f(x)\sin nx\;dx=0\qquad n=1,2,3,\cdots.\]令\(n=0\)得\[a_{0}=\frac{1}{\pi }\int_{-\pi }^{\pi }x^{2}dx=\frac{2}{\pi }\int_{0}^{\pi}x^{2}dx=\frac{2\pi ^{2}}{3}.\]当\(n=1,2,3,\cdots\)时，我们有\[a_{n}=\frac{1}{\pi }\int_{-\pi }^{\pi }x^{2}\cos nx\;dx\\=\frac{2}{\pi }\int_{0}^{\pi }x^{2}\cos nx\;dx=\frac{2}{\pi }\times \frac{2\pi }{n^{2}}(-1)^{n}=(-1)^{n}\frac{4}{n^{2}},\]因为\[\int x^2\cos nx\;dx=\frac{2x}{n^{2}}\cos nx+\left( \frac{x^{2}}{n}-\frac{2}{n^{3}}\right) \sin nx.\]因此\[f(x)=\frac{\pi ^{2}}{3}+\sum_{n=1}^{\infty }\left( (-1)^{n}\frac{4}{n^{2}}\cos nx\right).\]又\(f\left(\pi\right)=\pi^{2}\)，我们可以得到\[\pi ^{2}=\frac{\pi ^{2}}{3}+\sum_{n=1}^{\infty }\left( (-1)^{n}\frac{4}{n^{2}}\cos \left( n\pi \right) \right)\]即\[\pi ^{2}=\frac{\pi ^{2}}{3}+4\sum_{n=1}^{\infty }\left( (-1)^{n}(-1)^{n}\frac{1}{n^{2}}\right)\]进一步化简得\[\pi ^{2}=\frac{\pi ^{2}}{3}+4\sum_{n=1}^{\infty }\frac{1}{n^{2}}.\]因此\[\sum_{n=1}^{\infty }\frac{1}{n^{2}}=\frac{\pi ^{2}}{4}-\frac{\pi ^{2}}{12}=\frac{\pi ^{2}}{6}.\]证毕。

法三：

当\(x=0\)时\(\sin{x}\)可以使用泰勒级数展开\[\sin x = x – \frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots.\]于是我们可以得到\[\frac{x^3}{3!}=x\left(\frac{x^2}{\pi^2} + \frac{x^2}{2^2\pi^2}+ \frac{x^2}{3^2\pi^2}+\cdots\right)=x^3\sum_{n=1}^{\infty}\frac{1}{n^2\pi^2},\]即\[\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}.\]证毕。

法四：

设\(z\)为复数，考虑\[\frac{\pi^2}{\sin^2\pi z}=\sum_{n=-\infty}^{\infty}\frac{1}{(z-n)^2}\]由复变函数分析可得\[\frac{\pi^2}{\sin^2\pi z}-\frac{1}{z^2}=\sum_{n=1}^{\infty}\frac{1}{(z-n)^2}+\sum_{n=1}^{\infty}\frac{1}{(z+n)^2}.\]这样，当右边\(z=0\)时，我们就可以得到\[\lim_{z\to 0}\left(\frac{\pi^2}{\sin^2\pi z}-\frac{1}{z^2}\right)=2\sum_{n=1}^{\infty}\frac{1}{n^2}.\]又\[\lim_{z\to 0}\left(\frac{\pi^2}{\sin^2\pi z}-\frac{1}{z^2}\right)=\frac{\pi^2}{3}.\]因此\[\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}.\]证毕。

法五：

这是欧拉的方法。令\[s=\sin^{-1}{x}.\]那么\[\int_{0}^{\frac{\pi}{2}}sds=\frac{\pi^{2}}{8}.\]但是\[\int\limits_0^1 {\frac{{{{\sin }^{ – 1}}x}}{{\sqrt {1 – {x^2}} }}dx} = \frac{{{\pi ^2}}}{8}.\]又\[{\sin ^{ – 1}}x = \int {\frac{{dx}}{{\sqrt {1 – {x^2}} }}} = x + \frac{1}{2}\frac{{{x^3}}}{3} + \frac{{1 \cdot 3}}{{2 \cdot 4}}\frac{{{x^5}}}{5} + \frac{{1 \cdot 3 \cdot 5}}{{2 \cdot 4 \cdot 6}}\frac{{{x^7}}}{7} + \cdots.\]我们可以得到\[\int\limits_0^1 {\left\{ {\frac{{dx}}{{\sqrt {1 – {x^2}} }}\int {\frac{{dx}}{{\sqrt {1 – {x^2}} }}} } \right\}} = \int\limits_0^1 {\left\{ {x + \frac{1}{2}\frac{{{x^3}}}{3}\frac{{dx}}{{\sqrt {1 – {x^2}} }} + \frac{{1 \cdot 3}}{{2 \cdot 4}}\frac{{{x^5}}}{5}\frac{{dx}}{{\sqrt {1 – {x^2}} }} + \cdots } \right\}}.\]但是\[\int\limits_0^1 {\frac{{{x^{2n + 1}}}}{{\sqrt {1 – {x^2}} }}dx} = \frac{{2n}}{{2n + 1}}\int\limits_0^1 {\frac{{{x^{2n – 1}}}}{{\sqrt {1 – {x^2}} }}dx}.\]这样，可以得到\[\int\limits_0^1 {\frac{{{x^{2n + 1}}}}{{\sqrt {1 – {x^2}} }}dx} = \frac{{\left( {2n} \right)!!}}{{\left( {2n + 1} \right)!!}}.\]又所有的次数都是奇数，所以最终的结果为\[\frac{{{\pi ^2}}}{8} = 1 + \frac{1}{2}\frac{1}{3}\left( {\frac{2}{3}} \right) + \frac{{1 \cdot 3}}{{2 \cdot 4}}\frac{1}{5}\left( {\frac{{2 \cdot 4}}{{3 \cdot 5}}} \right) + \frac{{1 \cdot 3 \cdot 5}}{{2 \cdot 4 \cdot 6}}\frac{1}{7}\left( {\frac{{2 \cdot 4 \cdot 6}}{{3 \cdot 5 \cdot 7}}} \right) \cdots,\]\[\frac{{{\pi ^2}}}{8} = 1 + \frac{1}{{{3^2}}} + \frac{1}{{{5^2}}} + \frac{1}{{{7^2}}} + \cdots.\]令\[1 + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + \cdots = \omega.\]则\[\frac{1}{{{2^2}}} + \frac{1}{{{4^2}}} + \frac{1}{{{6^2}}} + \frac{1}{{{8^2}}} + \cdots = \frac{\omega }{4}.\]这样，我们可以得到\[\frac{\omega }{4} + \frac{{{\pi ^2}}}{8} = \omega.\]即\[\omega = \frac{{{\pi ^2}}}{6}.\]证毕。

法六：

\[\sum_{i=1}^{n}\frac{1}{i^{2}}=\zeta(2)=\frac{4}{3}\sum_{n=0}^{+\infty}\frac{1}{(2n+1)^2}=\frac{4}{3}\int_{0}^{1}\frac{\log y}{y^2-1}dy\]\[=\frac{2}{3}\int_{0}^{1}\frac{1}{y^2-1}\left[\log\left(\frac{1+x^2 y^2}{1+x^2}\right)\right]_{x=0}^{+\infty}dy=\frac{4}{3}\int_{0}^{1}\int_{0}^{+\infty}\frac{x}{(1+x^2)(1+x^2 y^2)}dx\,dy\]
\[=\frac{4}{3}\int_{0}^{1}\int_{0}^{+\infty}\frac{dx\, dz}{(1+x^2)(1+z^2)}=\frac{4}{3}\cdot\frac{\pi}{4}\cdot\frac{\pi}{2}=\frac{\pi^2}{6}.\]证毕。

就介绍这几种方法吧，大家可以有选择的进行阅读。以后数学作业中再有证明\(S_{n} < 2\)的时候，不妨这样写\(\lim_{n \to \infty }S_{n}=\frac{\pi}{6} < 2\)，怕老师不承认，可以顺便证明一下，至于给不给分，就不好说了。

In mathematics, the Iverson bracket, named after Kenneth E. Iverson, is a notation that denotes a number that is 1 if the condition in square brackets is satisfied, and 0 otherwise. More exactly,

\( [P] = \begin{cases} 1 & \text{if } P \text{ is true;} \\ 0 & \text{otherwise.} \end{cases} \)

where P is a statement that can be true or false. This notation was introduced by Kenneth E. Iverson in his programming language APL,while the specific restriction to square brackets was advocated by Donald Knuth to avoid ambiguity in parenthesized logical expressions.

The Iverson bracket converts a Boolean value to an integer value through the natural map \( \textbf{false}\mapsto 0; \textbf{true}\mapsto1 \), which allows counting to be represented as summation. For instance, the Euler phi function that counts the number of positive integers up to n which are coprime to n can be expressed by

\( \phi(n)=\sum_{i=1}^{n}[\gcd(i,n)=1],\qquad\text{for }n\in\mathbb N^+. \)

More generally the notation allows moving boundary conditions of summations (or integrals) as a separate factor into the summand, freeing up space around the summation operator, but more importantly allowing it to be manipulated algebraically. For example

\( \sum_{1\le i \le 10} i^2 = \sum_{i} i^2[1 \le i \le 10]. \)

In the first sum, the index \( i \) is limited to be in the range 1 to 10. The second sum is allowed to range over all integers, but where i is strictly less than 1 or strictly greater than 10, the summand is 0, contributing nothing to the sum. Such use of the Iverson bracket can permit easier manipulation of these expressions.

Another use of the Iverson bracket is to simplify equations with special cases. For example, the formula

\( \sum_{1\le k\le n \atop \gcd(k,n)=1}\!\!k = \frac{1}{2}n\varphi(n) \)

which is valid for n > 1 but which is off by 1/2 for n = 1. To get an identity valid for all positive integers n (i.e., all values for which \(\phi(n) \) is defined), a correction term involving the Iverson bracket may be added:

\( \sum_{1\le k\le n \atop \gcd(k,n)=1}\!\!k = \frac{1}{2}n(\varphi(n)+[n=1]) \)

 -By Wikipedia https://en.wikipedia.org/wiki/Iverson_bracket

Via Ivy – End  http://www.ivy-end.com/archives/1064

闲来无事，逛逛Mathematics。看到一道关于数列的题目，不妨拿来研究一下。原题地址

题目是这么说的，计算\[\left ( 1+\frac{1}{2}+\cdots + \frac{1}{n} \right )^{2}+\cdots + \left ( \frac{1}{n-1}+\frac{1}{n} \right )^{2}+\left ( \frac{1}{n} \right )^{2}.\]各位可以自己动笔计算一下。

下面我们主要讨论一下给出的两种种解法。

法一：

令\(S_{n}=\left ( 1+\frac{1}{2}+\cdots + \frac{1}{n} \right )^{2}+\cdots + \left ( \frac{1}{n-1}+\frac{1}{n} \right )^{2}+\left ( \frac{1}{n} \right )^{2}\)。

则\[S_{n}-S_{n-1}=\left [\left ( 1+\frac{1}{2}+\cdots + \frac{1}{n} \right )^{2}+\cdots + \left ( \frac{1}{n-1}+\frac{1}{n} \right )^{2}+\left ( \frac{1}{n} \right )^{2} \right ]\\-\left [\left ( 1+\frac{1}{2}+\cdots + \frac{1}{n-1} \right )^{2}+\cdots + \left ( \frac{1}{n-2}+\frac{1}{n-1} \right )^{2}+\left ( \frac{1}{n-1} \right )^{2} \right ].\]整理得\[S_{n}-S_{n-1}=\left [\left ( 1+\frac{1}{2}+\cdots + \frac{1}{n-1} \right )^{2}+2\cdot\left ( 1+\frac{1}{2}+\cdots + \frac{1}{n-1} \right )\cdot\left ( \frac{1}{n} \right )+\left ( \frac{1}{n} \right )^{2} \right ]\\ + \left [ \left ( \frac{1}{2}+\frac{1}{3}+\cdots + \frac{1}{n-1} \right )^{2}+2\cdot\left ( \frac{1}{2}+\frac{1}{3}+\cdots + \frac{1}{n-1} \right )\cdot\left ( \frac{1}{n} \right )+\left ( \frac{1}{n} \right )^{2} \right ]\\ + \cdots + \left [ \left ( \frac{1}{n-1} \right )^{2}+2\cdot\left ( \frac{1}{n-1} \right )\cdot\left ( \frac{1}{n} \right )+\left ( \frac{1}{n} \right )^{2} \right ]+\left [ \left ( \frac{1}{n} \right )^{2} \right ]\\-\left [\left ( 1+\frac{1}{2}+\cdots + \frac{1}{n-1} \right )^{2}+\cdots + \left ( \frac{1}{n-2}+\frac{1}{n-1} \right )^{2}+\left ( \frac{1}{n-1} \right )^{2} \right ].\]化简得\[S_{n}-S_{n-1}=n\cdot\left ( \frac{1}{n} \right )^{2}+2\cdot\frac{1}{n}\cdot\left [ 1+2\cdot\frac{1}{2}+3\cdot\frac{1}{3}+\cdots +\left ( n-1 \right )\cdot\frac{1}{n-1} \right ].\]即\[S_{n}-S_{n-1}=\frac{1}{n}+\frac{2\left ( n-1 \right )}{n}=2-\frac{1}{n}.\]运用累加法求和即可得到\[S_{n}=2n-H_{n}.\]其中\(H_{n}=1+\frac{1}{2}+\frac{1}{2}+\cdots +\frac{1}{n}.\)

法二：

这种方法比第一种方法高明很多，它需要用到一个引理——Iverson Bracket。简单的介绍一下这个引理，它的符号是中括号，当且仅当中括号里的表达式成立时，其值为\(1\)，否则为\(0\)。例如\(\left [ 1 < 2 \right ]=1\)，而\(\left [ 1 > 2 \right ]=0\)。

下面我们就用Iverson Bracket来求解这个问题\[S_{n}=\sum_{k}\left ( \sum_{i}\frac{1}{i}\cdot\left [ k \leq i \right ] \right )^{2}=\sum_{k}\sum_{i,j}\frac{1}{ij}\cdot\left [ k \leq i, k \leq j \right ]=\sum_{i,j}\frac{1}{ij}\sum_{k}\left [ k \leq \min\left ( i,j \right ) \right ].\]即\[S_{n}=\sum_{i,j}\frac{1}{ij}\cdot\min\left ( i,j \right )=\sum_{i,j}\frac{1}{\max\left ( i,j \right )}=\sum_{m}\frac{1}{m}\sum_{i,j}\left [ \max\left ( i,j \right )=m \right ].\]即\[S_{n}=\sum_{m}\frac{1}{m}\cdot\left ( 2m-1 \right )=2n-\sum_{m}\frac{1}{m}=2-H_{n}.\]