((=)=a(x+1).(a-k+1)2-y g(0) ()(0)a(a+1).(a-k+1) k 于是ch+)=1+z+ a d 1)a-k+1)k + 由于e是()=e.一个分支( ) ( ) ( ) ( ) k ( k )f (z ) g z k e - = + - + a a a 1 K a 1 \ g (0 ) = 1 ( ) ( ) ( ) ( ) ! 1 1 ! 0 k k k g k + - + = a a K a ( ) ( ) ... 2! 1 1 ln 1 2 + - = + + + e z z z a a a 于是 a ( )( ) ..., 1 ! 1 1 + < - - + + z z k a a a k k ( ) ( ) 由于e ln 1+ z 是 1+ z = e a ln (1+ z)的一个分支 a a