Logarithm inequality with partial sum

Question

How to show that $ \forall n \ge 2 $ : $$ \ln(n+1) \ge \frac{\sum_{k=1}^n ln~k}{n-1} $$

I tried to use integral comparison without success but with the hint I see that I need to do an induction on $(n+1)^{n−1}\geq n!$

Answer 1

By your work we need to prove that: $$(n+1)^{n-1}\geq n!.$$

Induction.

For $n=2$ it's true and from $(n+1)^{n-1}\geq n!$ we obtain: $$(n+1)!=(n+1)n!\leq(n+1)(n+1)^{n-1}=(n+1)^n<(n+2)^n$$ and we are done!

Answer 2

We have as you stated

$$\ln(n+1) \ge \frac{\sum_{k=1}^n \ln k}{n-1} \iff \ln(n+1) \ge \frac{\ln n!}{n-1} $$

$$\iff \ln(n+1)^{n-1}\ge \ln n! \iff (n+1)^{n-1}\ge n!$$

which is trivially true indeed it is true term by term

$$(n+1)^{n-1}=(n+1)(n+1)\ldots(n+1)\cdot 1 \ge n(n-1)\ldots 2\cdot 1=n!$$