How do you get $\displaystyle \frac{1}{n!}\leq\frac{1}{2^{n-1}}$ for $\displaystyle n>1$?

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- Jan 18th 2011, 03:34 AMcrossboneFactorial question
How do you get $\displaystyle \frac{1}{n!}\leq\frac{1}{2^{n-1}}$ for $\displaystyle n>1$?

- Jan 18th 2011, 03:46 AMProve It
$\displaystyle \displaystyle n \cdot (n-1) \cdot (n-2) \cdot \dots \cdot 3 \cdot 2 > 2 \cdot 2 \cdot 2 \cdot \dots \cdot 2 \cdot 2$ (with $\displaystyle \displaystyle n-1$ terms on each side of the inequality)

$\displaystyle \displaystyle n! > 2^{n-1}$

$\displaystyle \displaystyle \frac{1}{n!} < \frac{1}{2^{n-1}}$. - Jan 18th 2011, 04:28 AMcrossbone
I can see the proof, but how do you get the first line ie what idea comes prior to that? why not 2.1, 2.4 or pi?

nevermind. I got it. - Jan 18th 2011, 04:08 PMHallsofIvy
Fpr those who didn't "get it", both n! and $\displaystyle 2^{n-1}$ have n- 1 factors (not counting the "1" in n!). Just compare the size of the factors. n! is the product of all numbers from n down to 2 while $\displaystyle 2^{n-1}$ has all "2"s. 2= 2 while 3> 2 so 3(2)> 2(2), 4> 2 so 4(3)(2)> 2(2)(2), 5> 2 so 5(4)(3)(2)> 2(2)(2)(2), etc.