设函数 f(x)=(x−2)eaxf\left( x\right) = \left( {x - 2}\right) {\mathrm{e}}^{ax}f(x)=(x−2)eax .(1)若曲线 y=f(x)y = f\left( x\right)y=f(x) 在点 (0,f(0))\left( {0,f\left( 0\right) }\right)(0,f(0)) 处的切线方程为 y−3x+b=0y - {3x} + b = 0y−3x+b=0 ,求 a,ba,ba,b 的值;(2)若当 x>0x > 0x>0 时,恒有 f(x)>−x−2f\left( x\right) > - x - 2f(x)>−x−2 ,求实数 aaa 的取值范围;(3) 设 n∈N∗n \in {\mathbf{N}}^{ * }n∈N∗ 时,求证: 312+22+522+32+⋯+2n+1n2+(n+1)2<ln(n+1)\frac{3}{{1}^{2} + {2}^{2}} + \frac{5}{{2}^{2} + {3}^{2}} + \cdots + \frac{{2n} + 1}{{n}^{2} + {\left( n + 1\right) }^{2}} < \ln \left( {n + 1}\right)12+223+22+325+⋯+n2+(n+1)22n+1<ln(n+1) .