Need help in proving a version of Stirling Formula

Question

I am trying to prove this inequality but unable to do so ->

$log(n!) \leq \int_2^{n+1} log(x) dx $ .

My attempt -> using Stirling Formula LHS = n log(n) - n + O ( log(n) ) and Integral is n( log(n+1) ) + log( n+1) - n +1-2log 2 .

But I am unable to think now on how to prove that RHS is greater than Or equal to LHS.

Answer 1

$$ \log n! = \sum\limits_{k = 2}^n {\log k} \le \sum\limits_{k = 2}^n {\int_k^{k + 1} {\log xdx} } = \int_2^{n + 1} {\log xdx} $$

Answer 2

Stirling's formula is not useful. Note that $\int_2^{3} \log x dx>\log 2$, $\int_3^{4} \log x dx>\log 3$, ...,$\int_n^{n+1} \log x dx>\log n$. Just add these inequalities.