Show that lnn n ln nSolutionAs n nn1n221 lnn lnnn121 lnn

Show that ln(n!) n ln n.

As n! = n(n-1)(n-2)...2.1,

ln(n!) = ln(n(n1)21)

= ln(n) + ln(n1) ++ ln(2) + ln(1)

n ln(n)

Using a multiplicative variant of Gauss\'s trick, we get

(n!)² = (1n)(2(n1))(3(n2))((n2)3)((n1)2)(n1) nⁿ

This implies that

ln(n!) (1 / 2) n ln n

As n! nⁿ and so ln(n!) n ln n

So (1 / 2) n ln n ln(n!) n ln n and ln(n!) = (n ln (n)).

Therefore, ln(n!) n ln n.