$$\prod\limits_{n\ge2}\left(1+\frac{1}{n^2}\right)$$
在这之前，先来介绍一下entire function(整函数，英语中有时也作integral function)，什么是entire function呢？
If a complex function is analytic at all finite points of the complex plane C, then it is said to be entire, sometimes also called "integral" (Knopp 1996, p. 112).
Wolfram MathWorld - Entire Function
由上定义，我们有entire function就是整个复平面上处处解析的函数。那么什么样的复函数可以被称为是“解析的”呢？
A complex function is said to be analytic on a region \(\mathbb R\) if it is complex differentiable at every point in \(\mathbb R\)
Wolfram MathWorld - Analytic Function
即，如果一个复函数对区域\(\mathbb R\)上的每个点都是可微的，那么它就是解析的。这样的处处可解析的复函数又叫做“Holomorphic Function（全纯函数）”
Note, 在区域上某个点可微，即不管以何种方式／路径趋近，极限都要存在。这一点和二元函数在某个点极限存在的定义十分类似。
值得注意的是，二元函数的极限存在，是指点\(P\left(x, y\right)\)以任何方式趋向于\(P_0\left(x_0, y_0\right)\)时，函数都无限趋近于同一常数A。如果点\(P\left(x, y\right)\)以不同方式趋于\(P_0\left(x_0, y_0\right)\)时，函数\(f(x, y)\)趋向于不同的值，则可断定函数\(f(x, y)\)在点\(P_0\left(x_0, y_0\right)\)的极限一定不存在；如果点\(P\left(x, y\right)\)以一种特殊的方式趋于\(P_0\left(x_0, y_0\right)\)，即使函数无限趋近于某一确定的值，我们也不能断定函数的极限存在。
《高等数学》（下册）P54
例如
\(f\;'(0) = \lim_{z\to0}\frac{f(z)-f(0)}{z-0} = \lim_{z\to0}\frac{f(z)}{z} \qquad...\qquad (1)\)
有上述定理可知，它与如下极限等价，
$$\lim_{r\to0} \frac{f(re^{i\theta})}{re^{i\theta}}$$
即(1)中\(z\)沿着某一条确定的直线趋于0，该直线与\(x\)轴的角度就是\(\theta\)。
Proof
整函数\(\sin(\pi z)\)的零点集合是\(\mathbb Z\)，使得如下级数收敛的最小的整数h为1。
$$\sum_{n \in \mathbb Z^*} \frac{1}{n^{h+1}}$$
（这也是整函数\(\sin(\pi z)\)的亏格(genus)，应用Weierstrass factorization定理需要它）
那么什么是Weierstrass factorization定理呢？
The Weierstrass factorization theorem asserts that entire functions can be represented by a product involving their zeroes. In addition, every sequence tending to infinity has an associated entire function with zeroes at precisely the points of that sequence.
Wikipedia - Weierstrass factorization theorem
即，Weierstrass factorization定理断言整函数可以用包含它们的零点的积表示。此外，每个趋于无穷的序列都伴随一个零点恰好在这个序列上的整函数。
由Weierstrass factorization定理可得，
$$\sin(\pi z)=z e^{g(z)} \prod_{n \in \mathbb Z^*} \left(1-\frac{z}{n} \right) e^{z/n}$$
\(g(z)\)是一个待定的整函数，取对数导数[1]（这样我们就能将积写为和）
\(\pi \cot (\pi z)=\frac{1}{z}+g'(z)+ \sum_{n \in \mathbb Z^*} \left(\frac{1}{z-n}+\frac{1}{n}\right)\qquad...\qquad(2)\)
考虑这个等式，
\(2\pi \cot (2\pi z) = \pi \frac{\cos^2(\pi z) - \sin^2(\pi z)}{\sin (\pi z)\cos (\pi z)} = \pi \cot (\pi z) + \pi\cot \left(\pi \left(z+\tfrac{1}{2}\right)\right)\)
令函数
\(s(z) = \frac{1}{z} + \sum_{n\neq 0} \left(\frac{1}{z-n} + \frac{1}{n}\right) = \lim_{N\to\infty} \sum_{n=-N}^N \frac{1}{z-n}\)
令\(z=z和z=z+\frac{1}{2}\)带入\(s(z)\)求和有，
\(\begin{align}s(z) + s\left(z+\tfrac{1}{2}\right) &= \lim_{N\to\infty} \left(\sum_{n=-N}^N \frac{1}{z-n} + \sum_{n=-N}^N\frac{1}{z+\frac{1}{2}-n}\right)\\
&= 2\lim_{N\to\infty} \left(\sum_{n=-N}^N\frac{1}{2z-2n} + \sum_{n=-N}^N \frac{1}{2z+1-2n}\right)\\
&= 2\lim_{N\to\infty} \sum_{k=-(2N+1)}^{2N} \frac{1}{2z-k}\\
&= 2s(2z)\end{align}\)
故整函数\(g(z)\)满足
$$g(z) + g\left(z+\tfrac{1}{2}\right) = 2g(2z)$$
一个满足上式的连续的1周期函数\(f\)在\(\mathbb R\)上为常数，分别考虑\(f\)的实部和虚部，假定\(f\)是实值函数（\(g\)也是\(\mathbb R\)上的实值函数），根据连续性和周期性可知，\(\exists{x_0}\in [0,1]\)使得\(f(x) \leqslant f(x_0)\)（\(\forall{x}\in\mathbb R\)）。令\(x_k = 2^{-k}x_0\)（\(k\in\mathbb N\)）则有，
\(2 f(x_0) = 2f(x_k) = 2f(2x_{k+1}) = f(x_{k+1}) + f\left(x_{k+1}+\tfrac{1}{2}\right)\)
显然\(f(x_k) = f(x_0)\)成立，由连续性有\(f(0) = \max \{ f(x) : x\in\mathbb{R}\}\)，类似的也有\(f(0) = \min \{ f(x) : x\in\mathbb{R}\}\)。故\(f \equiv f(0)\)，为常数。由此，因为\(g\)是整函数，那么\(g\)在\(\mathbb{C}\)上也为常数，有\(g(z) \equiv g(0) = 0\)，自然地，\(g'(z)=0\)。
于是(2)可写为$$\pi \cot(\pi z) = \frac{1}{z}+\sum_{n \in \mathbb Z^*} \frac{1}{z-n}+\frac{1}{n}$$
取\(z \to 0\)时\(\frac{\sin(\pi z)}{z}\)的极限，有\(e^{g(z)} \equiv \pi\)，于是(1)可写为
$$\begin{align}\sin(\pi z) &= \pi z \prod_{n \in \mathbb Z^*} \left( 1-\frac{z}{n} \right) e^{z/n} \\&=\pi z \prod_{n=1}^\infty \left(1-\frac{z^2}{n^2}\right)\end{align}$$
令\(z=\frac{z}{\pi}\)，于是有
\(\sin(z)=z \prod_{n=1}^\infty \left(1-\frac{z^2}{n^2 \pi^2} \right)\qquad...\qquad(3)\)
由正弦双曲函数与正弦函数的关系，$$\sinh(z)=-i \sin(iz)$$
可由(3)式变形为
\(\begin{align}\sinh(z)=-i \sin(iz) &= -i \cdot i z \prod_{n=1}^\infty \left(1+\frac{z^2}{\pi^2 n^2} \right) \\&= z \prod_{n=1}^\infty \left(1+\frac{z^2}{\pi^2 n^2} \right)\end{align}\)
令\(z=\pi x\)，则有$$\begin{align}\sinh(\pi x) &= \pi x\prod_{n=1}^\infty \left(1+\frac{\pi^2 x^2}{\pi^2 n^2} \right) \\&= \pi x\prod_{n=1}^\infty \left(1+\frac{x^2}{n^2} \right)\end{align}$$
也就是
\(\frac{\sinh(\pi x)}{\pi x}=\prod_{n=1}^\infty \left(1+\frac{x^2}{n^2} \right)\qquad...\qquad(4)\)
可以观察到我们要求的极限\(\prod\limits_{n\ge2}\left(1+\frac{1}{n^2}\right)\)几乎就是\(x=1\)时的(4)式，那么，令\(x=1\)有
$$\prod_{n=1}^\infty \left(1+\frac{1}{n^2} \right)=\frac{\sinh(\pi)}{\pi}$$
因为\(n\)从2开始，所以有
$$\begin{align}\prod_{n=2}^\infty \left(1+\frac{1}{n^2} \right)&=\frac{\prod_{n=1}^\infty \left(1+\frac{1}{n^2} \right)}{1+\tfrac{1}{1}}\\&=\frac{\sinh \pi}{2\pi}\end{align}$$
[1] Wikipedia - Logarithmic derivative
[2] Wolfram MathWorld - Entire Function
[3] Wolfram MathWorld - Analytic Function
[4] Wikipedia - Weierstrass factorization theorem