$\displaystyle
\sum_{n=1}^{\infty}\frac{\ln n}{n}> \sum_{n=1}^{\infty}\frac{1}{n}
$
ln is not always bigger then 1
so when i am doing the comparing test
i cant use that
because ln 1 =0
??
Nevermind that.
ln(n)/n may initally be smaller, but when n > e it overtakes 1/n.
Seeing as both terms are over some number n, the determining factor is the size of the numerator. 1 will always be 1, but ln(n) is always increasing.