sinn/n和cosn/n类型的若干问题

解：取$\displaystyle f\left( x \right) = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\left| {\cos \left( {k + x} \right)} \right|}$，则

$$f\left( {x + m} \right) = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\left| {\cos \left( {k + x + m} \right)} \right|} \left( {m \in {N^*}} \right).$$

$$\begin{align*}f\left( x \right) &= \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\left| {\cos \left( {k + x} \right)} \right|}  \\&= \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^m {\left| {\cos \left( {k + x} \right)} \right|}  + \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = m + 1}^n {\left| {\cos \left( {k + x} \right)} \right|}  \\&= \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = m + 1}^n {\left| {\cos \left( {k + x} \right)} \right|}  \\&= \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{s = 1}^n {\left| {\cos \left( {s + m + x} \right)} \right|}  = f\left( {x + m} \right).\end{align*}$$

又

$$f\left( {x + 2\pi } \right) = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\left| {\cos \left( {k + x + 2\pi } \right)} \right|}  = f\left( x \right).$$

所以

$$\begin{align}\label{MA1}f\left( x \right) = f\left( {x + m} \right) = f\left( {x + 2\pi } \right).\qquad(1)\end{align}$$

对于$\displaystyle f\left( x \right) = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\left| {\cos \left( {k + x} \right)} \right|}$，其中$\displaystyle \frac{1}{n}$单调且$\displaystyle\lim\limits_{n\rightarrow \infty}{\frac{1}{n}}=0,|\cos(k+x)|$在$x\in[0,1]$上一致有界，由狄利克雷判别法可知：函数$\displaystyle f(x)$一致收敛.同时（1）式说明周期可为任意整数，又可为$\displaystyle 2\pi$，那么必为常数函数，$\displaystyle f(x)\equiv C.$所以

$$\begin{align*}f\left( 0 \right) &= \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\left| {\cos k} \right|}  = \int_0^1 {f\left( x \right)dx}  \\&= \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\int_0^1 {\sum\limits_{k = 1}^n {\left| {\cos \left( {k + x} \right)} \right|} dx}  \\&= \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\int_0^1 {\left| {\cos \left( {k + x} \right)} \right|dx} }  \\&= \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\int_k^{k + 1} {\left| {\cos t} \right|dt} }  \\&= \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\int_1^{n + 1} {\left| {\cos t} \right|dt}  \\&= 2\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {\frac{n}{\pi }} \right] \\&= \frac{2}{\pi }\mathop {\lim }\limits_{n \to \infty } \frac{\pi }{n}\left[ {\frac{n}{\pi }} \right] = \frac{2}{\pi }\end{align*}$$

注：在区间$[1,n+1]$中，含有函数$\omega(t)=|\cos t|$的周期个数是$\left[ {\frac{n}{\pi }} \right]$，每个周期的积分值都是$2$，所以得到上述结果.

$$\mathop {\lim }\limits_{n \to \infty } \frac{\pi }{n}\left[ {\frac{n}{\pi }} \right] = \mathop {\lim }\limits_{n \to \infty } \frac{{\frac{n}{\pi } - \left( {\frac{n}{\pi }} \right)}}{{\frac{n}{\pi }}} = \mathop {\lim }\limits_{n \to \infty } \left( {1 - \frac{{\left( {\frac{n}{\pi }} \right)}}{{\frac{n}{\pi }}}} \right) = 1$$

其中$\left( {\frac{n}{\pi }} \right)\in(0,1).$

另解一：构建$|\cos k|,k\in\mathbb{N}^*$在$|\cos x|$上的随机分布，所以$|\cos k|$的均值就为

$$\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\left| {\cos k} \right|}  = E\left( {\left| {\cos k} \right|} \right) = \frac{{\int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\left| {\cos x} \right|dx} }}{\pi } = \frac{2}{\pi }.$$

另解二：令

$$f(x)=\int_0^x{|\cos t|dt}.$$

当$x\in[n\pi,(n+1)\pi)$时，

$$\int_0^{n\pi } {\left| {\cos t} \right|dt}  \le f\left( x \right) < \int_0^{\left( {n + 1} \right)\pi } {\left| {\cos t} \right|dt} .$$

$$\Rightarrow 2n\leq f(x)<2(n+1).$$

由此得

$$\frac{{2n}}{{\left( {n + 1} \right)\pi }} < \frac{{f\left( x \right)}}{x} < \frac{{2\left( {n + 1} \right)}}{{n\pi }}.$$

记$S_n=\sum\limits_{k = 1}^n {\left| {\cos k} \right|}$，则有

$$\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\left| {\cos k} \right|}  = \mathop {\lim }\limits_{n \to \infty } \frac{{{S_n}}}{n} = \mathop {\lim }\limits_{n \to \infty } \frac{{f\left( x \right)}}{x} = \frac{2}{\pi }.$$