Yes, the difference between $b_n$ and $a_n$ is always at least $1$. Let
so that $a_n=(n+1)^2f(n)$ and $b_n=(n+1)^2f(n+1).$ Then by the mean value theorem we have that
for some $c\in (n,n+1)$. Next we calculate
and after substituting in the power series expansions for the functions in the parenthesis we have
If $x>2$ then the terms in this series are decreasing in absolute value and
Keeping in mind that $f(x)$ is increasing, it is easy to check that this expression is greater than $1/x^2$ whenever $x>7$.
Substituting this back in the above equation gives $b_n-a_n>1$ for all $n>7$, and the rest is verified directly.