Is $\frac{\log(1+k)}{k}\le\frac{\log n}{n-2}$ when $k>n^2+n$?

Question

As in the title: is $$ \frac{\log(1+k)}{k}\le\frac{\log n}{n-2} $$ when $k>n^2+n$ and $n\ge2$? I'd say yes, but can't prove this. Can someone help me please?

Answer 1

$f(x) = \dfrac{\ln(1+x)}{x}\implies f'(x) = \dfrac{1}{x^2}\left(\dfrac{x}{1+x} - \ln(1+x)\right)< 0$ because $g(x) = \dfrac{x}{1+x} - \ln(1+x)$ has $g'(x) = -\dfrac{1}{1+x} + \dfrac{1}{(1+x)^2} < 0\implies g(x) < g(0) = 0\implies f'(x) < 0\implies f(k) \le f(n-1)= \dfrac{\ln(n)}{n-1}$ for if $k \ge n-1$. That is as close as you can get.