$\sum\frac{n!}{(x+1)(x+2)(x+3)(x+4)\cdots(x+n+2)}$

hbghlyj

$\forall n\inN,x>-1,$
\begin{align*}\frac1{(x+1)^2}&>\frac{1}{(x+1)(x+2)}\\&+\frac{1!}{(x+1)(x+2)(x+3)}\\&+\frac{2!}{(x+1)(x+2)(x+3)(x+4)}\\&+\ldots\\&+\frac{n!}{(x+1)(x+2)(x+3)(x+4)\cdots(x+n+2)}\end{align*}

ZCos666

即证
\[ \begin{aligned}
\dfrac{1}{x+1}=\dfrac{x+2}{x+1}-1&>\dfrac{1!}{x+3}\\
&+\dfrac{2!}{(x+3)(x+4)}\\
&+\cdots\\
&+\dfrac{n!}{(x+3)(x+4)\cdots(x+n+2)}
\end{aligned} \]
即证
\[ \begin{aligned}
\dfrac{2}{x+1}=\dfrac{x+3}{x+1}-1!&>\dfrac{2!}{x+4}\\
&+\dfrac{3!}{(x+4)(x+5)}\\
&+\cdots\\
&+\dfrac{n!}{(x+4)(x+5)\cdots(x+n+2)}
\end{aligned} \]
即证
\[ \begin{aligned}
\dfrac{6}{x+1}=\dfrac{2(x+4)}{x+1}-2!&>\dfrac{3!}{x+5}\\
&+\dfrac{3!}{(x+5)(x+6)}\\
&+\cdots\\
&+\dfrac{n!}{(x+6)(x+7)\cdots(x+n+2)}
\end{aligned} \]
以此类推，即证
\[ \dfrac{n!}{x+1}=\dfrac{(n-1)!(x+n+1)}{x+1}-(n-1)!>\dfrac{n!}{x+n+2} \]
显然成立

战巡

右边这太难看，首先把每一项改成这样
\[\frac{n!}{(x+1)(x+2)...(x+n+2)}=\frac{n!\Gamma(x+1)}{\Gamma(x+n+3)}=\frac{\Gamma(n+1)\Gamma(x+1)}{\Gamma(x+n+3)}=\frac{\Gamma(x+1)B(n+1,x+2)}{\Gamma(x+2)}\]

然后
\[\sum_{n=0}^\infty\frac{n!\Gamma(x+1)}{\Gamma(x+n+3)}=\sum_{n=0}^\infty\frac{\Gamma(x+1)B(n+1,x+2)}{\Gamma(x+2)}\]
\[=\frac{\Gamma(x+1)}{\Gamma(x+2)}\sum_{n=0}^\infty B(n+1,x+2)\]
\[=\frac{1}{x+1}\sum_{n=0}^\infty\int_0^1t^n(1-t)^{x+1}dt\]
\[=\frac{1}{x+1}\int_0^1(1-t)^xdt\]
\[=\frac{1}{(x+1)^2}\]

kuing

注意到
\[\frac1{x+n+2}=\frac1{x+1}-\frac1{x+1}\cdot\frac{n+1}{x+n+2},\]
所以有裂项式
\[\frac{n!}{(x+1)(x+2)\cdots(x+n+2)}=\frac1{(x+1)^2}\left(\frac{n!}{(x+2)\cdots(x+n+1)}-\frac{(n+1)!}{(x+2)\cdots(x+n+2)}\right),\]
因此原不等式有
\[\RHS=\frac1{(x+1)^2}\left(1-\frac{(n+1)!}{(x+2)\cdots(x+n+2)}\right)<\LHS.\]