证明e的x-1次方>x的n次方/n!x属于1到正无穷
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证明e的x-1次方>x的n次方/n!x属于1到正无穷
证明e的x-1次方>x的n次方/n!x属于1到正无穷
证明e的x-1次方>x的n次方/n!x属于1到正无穷
e^(x-1) > (x^n)/n!
n=1
L.S = e^(x-1)
R.S = x
consider
f(x) = e^(x-1) - x
f'(x) = e^(x-1) - 1 > 0 ( for x ∈（1,+∞)
f(x) > f(1) = 0
=> f(x) >0
e^(x-1) > x
for n=1 is true
Assum n = k is true, ie
e^(x-1) > (x^k)/k!
for n = (k+1)
LS = e^(x-1)
RS = (x^(k+1))/(k+1)!
consider
g(x)= e^(x-1) - (x^(k+1))/(k+1)!
g'(x) = e^(x-1) - (k+1)x^k/(k+1)!
= e^(x-1) - x^k/k!
> e^(x-1) - e^(x-1) =0
g(x)> g(1) = 1 - 1/(k+1)! > 0
g(x) > 0
=> e^(x-1) > (x^(k+1))/(k+1)!
By MI, it is true for all n ∈N
1 当n=1时 e^(x-1)>x 因为对于函数f(x)= e^(x-1)-x,f'(x)=e^(x-1)-1， 当x>1时,f'(x)>0,所以当x>1时f(x)= e^(x-1)-x为增函数，f(1)=0 所以 当x>1时f(x)>0 e^(x-1)>x
2 设n=k时， e^(x-1)>x ^k/k!
因为x^k/k！-x^（k+1）/（k+1）！=x^k/k！-x^k/...
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1 当n=1时 e^(x-1)>x 因为对于函数f(x)= e^(x-1)-x,f'(x)=e^(x-1)-1， 当x>1时,f'(x)>0,所以当x>1时f(x)= e^(x-1)-x为增函数，f(1)=0 所以 当x>1时f(x)>0 e^(x-1)>x
2 设n=k时， e^(x-1)>x ^k/k!
因为x^k/k！-x^（k+1）/（k+1）！=x^k/k！-x^k/k！[x/(k+1)]=x^k/k！[1-x/(k+1)]=x^k/k！[(k+1-x)/k+1)]
因为x可看作常数，当k足够大时必有k>x所以(k+1-x)/(k+1)>0,所以x^k/k！-x^（k+1）/（k+1）！>0，所以x^k/k！>x^（k+1）/（k+1）！所以 e^(x-1)>x^（k+1）/（k+1）！
所以当n=k+1时原式成立，所以 e^(x-1)>x ^n/n! n>=1,n是整数 x>1
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