Two sided inequality with logarithm and factorial

Question

Can someone help me to prove this inequality?

$\frac {1}{5}n*log_2 n \leq log_2 n! \leq 5*n*log_2 n $

for $ n \geq 5 $

Answer 1

Hint:

$$\log_2x-1<\lfloor\log_2x\rfloor\le\log_2x$$

so that

$$\int_1^{n+1}(\log_2x-1)\,dx<\int_1^{n+1}\lfloor\log_2 x\rfloor\,dx\le\int_1^{n+1}\log_2x\,dx$$

and

$$\left.\left(x\frac{\ln(x)-1-\ln2}{\ln2}\right)\right|_1^{n+1}<\sum_1^n\log_2n=\log_2n!\le\left.x\frac{\ln(x)-1}{\ln2}\right|_1^{n+1}.$$

Answer 2

Upper bound

We have the upper bound $$\log_2 n!=\log_2 n+\cdots +\log_22\le (n-1)\log_2 n<5n\log_2 n.$$