Prove $\log x!$ is $\Omega (xlogx)$

Question

Find a positive real number $C$ and a nonnegative real number $x_o$ such that

$Cx$$\log x$ $\leq$ $\log x!$ for all real numbers $x > x_o$.

I tried to expand $\log x!$ into $\log 1 + \log2 +\log3 +....\log x$. But how do I choose $C$ and $x_o$ so the above inequality hold?

Any hints would be appreciated.

Well... $\log n < \log x$ for $n<x$. Using that inequality should make the choice of $C$ apparent. — abnry, May 26 '13 at 19:31
@nayrb You're thinking about a different direction of inequality: that implies $\log x! \le x \log x$. — A.S, May 26 '13 at 19:32
Closely related to http://math.stackexchange.com/questions/46892/how-do-you-prove-that-nn-is-on2. — lhf, May 26 '13 at 20:00

score 3 · Answer 1 · answered May 26 '13 at 19:38

Note that $$e^n = \sum_{k=1}^{\infty} \dfrac{n^k}{k!} \geq \dfrac{n^n}{n!} \implies \log(n!) \geq n \log(n) - n$$ Also, $$\log(n!) = \sum_{k=1}^n \log(k) \leq \sum_{k=1}^n \log(n) = n \log(n)$$ We hence have $$n \log(n)-n \leq \log(n!) \leq n \log(n)$$

score 0 · Accepted Answer · answered May 26 '13 at 19:55

0

For $k\ge 2$, we have $\log k\ge \int_{k-1}^{k}\log t\,dt$. Summing from $k=2$ to $n$, we find that $$\log 2 +\log 3 + \log 4+ \cdots +\log n \ge \int_1^n \log t\,dt=n\log n-n.$$
If $n\ge 9$, then $n\log n-n \gt \frac{1}{2}n\log n$.

answered May 26 '13 at 19:55

André Nicolas

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Prove $\log x!$ is $\Omega (xlogx)$

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