已知函数 f(x)=(1x+12)ln(x+1)f\left( x\right) = \left( {\frac{1}{x} + \frac{1}{2}}\right) \ln \left( {x + 1}\right)f(x)=(x1+21)ln(x+1) .(1)求曲线 y=f(x)y = f\left( x\right)y=f(x) 在 x=2x = 2x=2 处的切线斜率;(2)求证:当 x>0x > 0x>0 时, f(x)>1f\left( x\right) > 1f(x)>1 ;(3)证明: 56<ln(n!)−(n+12)lnn+n≤1\frac{5}{6} < \ln \left( {n!}\right) - \left( {n + \frac{1}{2}}\right) \ln n + n \leq 165<ln(n!)−(n+21)lnn+n≤1 .