The usual form of Raabe's test says that if $a\_n>0$ and if for large $n$, $\frac{a\_{n+1}}{a\_n}\leq 1-\frac{p}{n}$ for $p>1$, then $\sum a\_n < \infty$. A proof I've seen of this relies on the ratio comparison test: $\frac{a\_{n+1}}{a\_n}\leq 1-\frac{p}{n}< \left(1-\frac{1}{n}\right)^p = \frac{(n-1)^p}{n^p}$, which is $\frac{b\_n}{b\_{n-1}}$ for $b\_n = 1/n^p$, which converges for $p>1$. 

A partial converse says that if $\frac{a\_{n+1}}{a\_n}\geq 1 - \frac{p}{n}$ for $p\leq 1$, then $\sum a\_n$ diverges. To proves this we may again use the ratio comparison test, taking $p=1$, then *a fortiori* it holds for $p<1$.

I am trying to prove a strengthening of the converse, which states: if for large $n$, $\frac{a\_{n+1}}{a\_n}\geq 1 - \frac{1}{n}-\frac{A}{n^2}$ where $A>0$, then $\sum a\_n$ diverges. It seems we'd like to use the ratio comparison test again, but I'm having trouble finding a series to compare it to. 

We might write $1 - \frac{1}{n} - \frac{A}{n^2} = \frac{n^2 - n - A}{n^2}= \frac{n-1}{n} - \frac{A}{n^2}$, but that doesn't seem to be helping. Any ideas?