If $|\frac{c_{n+1}}{c_n}|\leq1+\frac{a}{n}$, where $a<-1$, $a$ does not depend on $n$, then the series $\sum_{n=1}^\infty c_n$ converges absolutely Question: If $|\frac{c_{n+1}}{c_n}|\leq1+\frac{a}{n}$, where $a<-1$ and $a$ does not depend on $n$, then the series $\sum_{n=1}^\infty c_n$ converges absolutely.
My attempt: Let $\epsilon>0$.  To show $\sum c_n$ converges absolutely, we want to show that there exists an $N\in\mathbb{N}$ such that $|\frac{c_{n+1}}{c_n}|$ converges uniformly to some constant $n>N$.  Since $a<-1$, there is some $N_0$ such that $||\frac{c_{n+1}}{c_n}|-1|<\epsilon$ whenever $n>N_0$.  Thus, $|\frac{c_{n+1}}{c_n}|$ converges uniformly (to $1^-$) hence $\sum c_n$ converges absolutely.
I actually asked this question a little over a year ago here: Showing a series converges absolutely and got a couple neat answers, but I was wondering if this would be a more "direct" (I don't know if that is the right word) of doing it.   Or, is there something I messed up?  Any help is, as always, greatly appreciated!  Thank you.

If
$$ \left|\frac{c_{n+1}}{c_n}\right| \leq 1-\frac{k}{n}\qquad\text{with }k>1 $$
then
$$|c_{n+1}|\leq |c_n| e^{-k/n}\leq |c_1|\exp\left(-k H_n\right)\leq |c_1|\exp(-k\log n)=\frac{|c_1|}{n^k}$$
and $\sum_{n\geq 1}\frac{1}{n^k} = \zeta(k)< +\infty$ for any $k>1$.
