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Topic: Factorial Inequality (Read 974 times) |
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SWF
Uberpuzzler
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Re: Factorial Inequality
« Reply #1 on: Apr 2nd, 2003, 7:04pm » |
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Testing it out for n=1,2, and 3 shows it to be true at least for those values of n. Multiplying the left side of the inequality by x(n)=(2n+2)2(2n+1)2 and the right side by y(n)=4(n+1)n+4/nn results in the original inequality with n replaced by n+1. If x(n)>=y(n) for n>=3, then by induction the orginal inequality is true for all positive integers, n.
Factoring out the 2, gives x(n)=4(n+1)2(2n+1)2.
Dividing out n, gives y(n)=4(n+1)4(1+1/n)n. This is less than 12(n+1)4 because (1+1/n)n<3. That last inequality can be verified by using binomial expansion and comparing to the series for e=1+1+1/2!+1/3!+..., and noting that e<3.
Removing the common factor of 4(n+1)2 in the expressions for x(n) and y(n) shows that x(n) is greater than y(n) for n>=3 because (2n+1)2>3(n+1)2 for n>=3.
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NickH
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Re: Factorial Inequality
« Reply #2 on: Apr 3rd, 2003, 2:56pm » |
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A characteristically ingenious solution, if I may say so!
I derived the result by repeated use of the arithmetic-geometric mean inequality:
[(n+1)/2]2 >= n.1
...
[(n+n)/2]2 >= n.n
Multiplying...
[(2n)!]2/[(n!)2 22n] >= nn n!, from which the result immediately follows.
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SWF
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Re: Factorial Inequality
« Reply #3 on: Apr 3rd, 2003, 5:16pm » |
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Thanks, NickH, but I never would have considered doing it the way you did. Those two terms are not very close together for large n. How about this for something with similar form that is much tighter, and the exponent on n! in the lower term was increased to 4:
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NickH
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Re: Factorial Inequality
« Reply #4 on: Apr 4th, 2003, 9:33pm » |
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Excellent! That comes from Stirling's approximation, I assume, though I haven't worked out the details. Having increased the (n!) exponent to 4, the expression can be rearranged to give a lower bound for 2nCn.
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