x>y>1 (y!)^(x-1) 和(x!)^(y-1)的大小

x>y>1 (y!)^(x-1) 和(x!)^(y-1)的大小
x>y>1 (y!)^(x-1) 和(x!)^(y-1)的大小

x>y>1 (y!)^(x-1) 和(x!)^(y-1)的大小
(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).

可以举个特例说明一下情况：x=3,y=2
x!=6,y!=2
(y!)^(x-1)=2^2=4
(x!)^(y-1)=6^1=6
所以：
(x!)^(y-1)>(y!)^(x-1)

作商法
(x!)^(y-1)/(y!)^(x-1)
=[x*.....*(x-y+1)]^(y-x)
因为底数x*.....*(x-y+1)>1
指数y-x<0
所以[x*.....*(x-y+1)]^(y-x)<1
可得(x!)^(y-1)/(y!)^(x-1)<1
可得(x!)^(y-1)<(y!)^(x-1)
即(y!)^(x-1)>(x!)^(y-1)
希望能帮到你，祝学习进步O(∩_∩)O,也别忘了采纳！

(y!)^(x-1) <(x!)^(y-1)
设x=3,y=2……

设k>1，首先证明不等式ln(k!)/(k-1)等价展开,k(ln2+...+lnk)即kln(k+1)-ln(k+1)!>0
即(ln(k+1)-lnk)+(ln(k+1)-ln(k-1))+...+(ln(k+1)-ln2)+(ln(k+1)-ln1)>0
这个式子在k>1成...

全部展开

设k>1，首先证明不等式ln(k!)/(k-1)等价展开,k(ln2+...+lnk)即kln(k+1)-ln(k+1)!>0
即(ln(k+1)-lnk)+(ln(k+1)-ln(k-1))+...+(ln(k+1)-ln2)+(ln(k+1)-ln1)>0
这个式子在k>1成立,那么ln(k!)/(k-1)用归纳法可以得到数列Ak=ln(k!)/(k-1)是递增数列
对于x>y>1得到不等式ln(y!)/(y-1)取指数操作就得到(y!)^(x-1)<(x!)^(y-1)

收起

过程在图片里，很详细了

两边取对数
就是比较loga[(y!)^(x-1)] =(x-1)logay! 与 loga[(x!)^(y-1)]=(y-1)loga(X!)大小
两边同时除以(x-1)(y-1)
即比较loga[y!/(y-1)]与loga[X!/(x-1)]的大小
只需判定f(n)=loga[n!/(n-1) ] (n>1)(令a=n+1)的单调性即可

全部展开

两边取对数
就是比较loga[(y!)^(x-1)] =(x-1)logay! 与 loga[(x!)^(y-1)]=(y-1)loga(X!)大小
两边同时除以(x-1)(y-1)
即比较loga[y!/(y-1)]与loga[X!/(x-1)]的大小
只需判定f(n)=loga[n!/(n-1) ] (n>1)(令a=n+1)的单调性即可
f(n+1)-f(n)=loga[(n+1)!/n ] -loga[n!/(n-1)]={loga[(n+1)+lgn!]/n}-loga[n!/(n-1)]
而logan!-(n-1)=logan+loga(n-1)+loga(n-2)+…+loga2+loga1-(n-1)
=[(logan)-1]+[loga(n-1)-1]+…+(loga2-1)
=loga(n/a)+loga[(n-1)/a]+…+loga(2/a)
∵a=n+1
∴n/a, (n-1)/a, …2/a都小于1
则各自对数都小于零
即logan!<(n-1)
则 logan!/(n-1)<[logan! +1]/[（n-1)+1]=(logan! +1)/n
【a/b<(a+1)/(b+1) (0即log(n+1)[n!/(n-1]<[log(n+1)n!+1]/n=log(n+1)[(n+1)!/n]
所以f(n+1)>f(n)函数递增
得(x!)^(y-1)>(y!)^(x-1)

收起