How can I prove that $n \log_2(n) ∈ O(log(n!))$ is true?
We start by supposing that $f(n)< c g(n)$ is true,
which means that $n \log_2(n) > c \log(n!)$ for all $n>n_0$ and $c>0$.
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1Not that it matters, but in $\log(n!)$, what is the base? And you will end up wanting to show that $n\log_2(n)$ is less than $c\log(n!)$ for sufficiently large $n$. – André Nicolas Apr 13 '16 at 18:49
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There are a lot of related questions, but you could first start with the usual trick: $\log(n!) = \log \prod_{j=1}^n j = \sum_{j=1}^n \log j > \sum_{j=n/2}^n \log j > \sum_{j=n/2}^n \log \frac{n}{2} > \dots$... – Clement C. Apr 13 '16 at 18:54
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Related: see http://math.stackexchange.com/questions/1489770/proving-asymptotic-barrier-o-notation?lq=1 and linked posts therein. – Clement C. Apr 13 '16 at 18:55
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They don't specify a base for log (n!).. I suppose it is not important when calculating the big O – Natalie Apr 13 '16 at 19:37
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Could you please explain me why do you make the transition from ∑ j=n/2 to n (log j) >∑ j=n/2 to n (log n/2) ? – Natalie Apr 13 '16 at 19:38
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Regarding my comment above: $\sum_{j=1}^n \log j > \sum_{j=n/2}^n \log j$ because each term in the sum is positive, and the LHS has more terms than the RHS. Then $ \sum_{j=n/2}^n \log j > \sum_{j=n/2}^n \log \frac{n}{2} $ because $\log j > \log \frac{n}{2}$ for all $j>\frac{n}{2}$, since $\log$ is increasing. – Clement C. Apr 13 '16 at 19:41