Is this logically valid? $$1+\frac{1}{2}+\frac{1}{3}.....+\frac{1}{n-1} > ln(n)$$
and so, necessarily, $$1+\frac{1}{2}+\frac{1}{3}.....+\frac{1}{n-1}+\frac{1}{n} = 1+\frac{1}{2}+\frac{1}{3}.....+\frac{1}{n} > ln(n)$$
assuming that $n \in \mathbb {N}$.

Imagine an inequality $$\frac{1}{n}>0.$$ We can add the same quantity on both sides $$\left(1+\frac{1}{2}+...+\frac{1}{n-1}\right)+\frac{1}{n}>0+\left(1+\frac{1}{2}+...+\frac{1}{n-1}\right)$$
Now we use your first inequality and the transitivity of $>$, which says that if $a>b$ and $b>c$, then $a>c$, where $a:=\left(1+\frac{1}{2}+...+\frac{1}{n-1}\right)+\frac{1}{n}$, $b:=\left(1+\frac{1}{2}+...+\frac{1}{n-1}\right)$ and $c:=\ln(n)$.
