x>y>1 (y!)^(x-1) 和(x!)^(y-1)的大小
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(y!)^(x-1) 和(x!)^(y-1)
题目出现阶乘“!”,所以x、y为自然数.
又x＞y＞1,所以y≥2,x≥3
不妨设x=y+k,k为自然数
即:
[(y!)^(x-1) ]/[(x!)^(y-1)]=[(y!)^(y+k-1)]/{[(y+k)!]^(y-1)}
=[(y!)^(y-1+k)]/{[(y!(y+1)(y+2)……(y+k)]^(y-1)}
=[(y!)^(y-1+k)]/{[(y!)^(y-1)][(y+1)(y+2)……(y+k)]^(y-1)}
=[(y!)^(y-1)][(y!)^k]/{[(y!)^(y-1)][(y+1)(y+2)……(y+k)]^(y-1)}
=[(y!)^k]/{[(y+1)(y+2)……(y+k)]^(y-1)}
＜[(y!)^k]/{[(y+1)^k]^(y-1)}
=[(y!)^k]/{[(y+1)^k]^(y-1)}
＜[(y!)^k]/[(y^k)^(y-1)]
=[(y!)^k]/[y^(y-1)]^k
=[(y!/y^(y-1)]^k
={(y/y)[(y-1)/y][(y-2)/y]……[3/y][2/y]*1}^k
＜1^k=1
所以[(y!)^(x-1) ]/[(x!)^(y-1)]＜1
则(y!)^(x-1)＜(x!)^(y-1).
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