hi guys,
I'm almost certain that this inequality holds, however for s , n > 0 , n being an integer, s being a real value
(s + n) ! > s^(s + n) as s --> infinity (fixed n)
(s + n)! > s^(s + n) as n --> infinity (fixed s)
Thanks in Advance,
Dave
hi guys,
I'm almost certain that this inequality holds, however for s , n > 0 , n being an integer, s being a real value
(s + n) ! > s^(s + n) as s --> infinity (fixed n)
(s + n)! > s^(s + n) as n --> infinity (fixed s)
Thanks in Advance,
Dave
$\displaystyle \frac{(s+n)!}{s^{s+n}}=\frac{1\cdot 2\cdot\ldots\cdot n\cdot (n+1)\cdot\ldots\cdot (n+s)}{s\cdot s\cdot\ldots\cdot s}$ , with $\displaystyle n+s$ factors in the denominator.
It seems pretty obvious that the denominator is way bigger than the numerator when $\displaystyle s\to\infty$ , isn't it? So the inequality is exactly the other way than what you wrote!
Tonio
If you look at Tonio's ratio for n and n+1,
$\displaystyle \frac{(s+n+1)!}{s^{s+n+1}}$ divided by $\displaystyle \frac{(s+n)!}{s^{s+n}}=\frac{s+n+1}{s}$
is always greater than 1, and much greater as n increases, so the ratio goes to infinity. That proves your second inequality.
But if you look at the ratio for a fixed n as s goes to s+1,
$\displaystyle \frac{(s+n+1)!}{(s+1)^{s+n+1}}$ divided by $\displaystyle \frac{(s+n)!}{s^{s+n}}=\frac{s+n+1}{s+1} * \left(\frac{s}{s+1}\right)^{s} * \left(\frac{s}{s+1}\right)^{n}$.
The first term goes to 1, the second term goes to $\displaystyle \frac{1}{e}$, and the third term goes to 1. So the ratio goes to zero, which proves the opposite of your first inequality.
The difference, of course, is that in the first inequality, the base is growing as well as the exponent, while in the second inequality, only the exponent grows.
- Hollywood