$\sin x_0+\frac{\cos x_0}{1!}(x-x_0)+\cdots +\frac{\sin (x_0+n\frac{\pi}{2})}{n!}(x-x_0)^n+\cdots$

级数
\begin{equation}\label{eq:1.fuck}
\sin x_0+\frac{\cos x_0}{1!}(x-x_0)+\cdots +\frac{\sin (x_0+n\frac{\pi}{2})}{n!}(x-x_0)^n+\cdots
\end{equation}
绝对收敛.

证明:
\begin{align*}
\lim_{n\to\infty}|\frac{\frac{\sin(x_0+(n+1)
\frac{\pi}{2})(x-x_0)^{n+1}}{(n+1)!}}{\frac{\sin(x_0+n
\frac{\pi}{2})(x-x_0)^n}{n!}}|&=\lim_{n\to\infty}|\frac{x-x_0}{n+1}\frac{\sin(x_0+n
\frac{\pi}{2}+ \frac{\pi}{2})}{\sin(x_0+n
\frac{\pi}{2})}|\\&=\lim_{n\to\infty}|\frac{x-x_0}{n+1}\frac{1}{\tan
(x_0+\frac{n\pi}{2})}|=0
\end{align*}
可见达朗贝尔比值判别法行不通.我们寻找另外的方法.要证明级数\ref{eq:1.fuck}绝对收敛，只用证明

\begin{equation}
|\sin x_0|+|\frac{\cos x_0}{1!}||x-x_0|+\cdots+|\frac{\sin(x_0+n \frac{\pi}{2})}{n!}||x-x_0|^n+\cdots
\end{equation}绝对收敛.为此，只用证明
\begin{equation}
1+\frac{1}{1!}|x-x_0|+\cdots+\frac{1}{n!}|x-x_0|^n+\cdots
\end{equation}绝对收敛.我们采用达朗贝尔比例判别法.
\begin{equation}
\lim_{n\to\infty}\frac{\frac{1}{(n+1)!}|x-x_0|^{n+1}}{\frac{1}{n!}|x-x_0|^n}=\lim_{n\to\infty}\frac{|x-x_0|}{n+1}=0
\end{equation}
因此级数绝对收敛.

 

 

 

注:根据泰勒公式，我们知道该级数绝对收敛至$\sin x$.