1. ## A proof

lg(n!)=O(nlgn)

2. Originally Posted by scofield

lg(n!)=O(nlgn)

I'm guessing you wanted to write $\log(n!)=O(n\log n)$, and this follows from $\log(n!)=\sum\limits_{k=1}^n\log n\le n\log n$ ...

Tonio

3. No, actually lg n is correct. lg n is log2 (n).I mean log base 2 of n. However this proof is the same for lg n too.Isn't it?

4. Originally Posted by scofield
No, actually lg n is correct. lg n is log2 (n).I mean log base 2 of n. However this proof is the same for lg n too.Isn't it?

Of course: the above works for any basis...and lg is not standard notation for $\log_2$, is it?

Tonio

5. I know that;

In mathematics, (log2 n) is called the binary logarithm (the logarithm for base 2).It can be written like lg n.The binary logarithm is often used in computer science.Hence, maybe u didn't see this notation.

6. By the way,

Thank you so much for your solution...