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.

Let $|a|<n_0\in\Bbb N.$ For $n>n_0$ we have $$|\;\ln |c_{n+1}|\;|=|\;\ln |c_{n_0}\prod_{j=n_0}^n|c_{j+1}/c_j|\;|\ge$$ $$\ge  |\;\ln |c_{n_0}\prod_{j=n_0}^n(1+a/j)|\;|\ge$$ $$\ge  -|\;\ln |c_{n_0}|\;|+\sum_{j=n_0}^n|a|/j \quad (\bullet)$$ because if $-1<a/j<0$ then $\ln (1+a/j)=(a/j)-(a/j)^2/2+(a/j)^3/3-...<(a/j).$
Now $(-\ln n)+\sum_{j=1}^n(1/j)$ converges to the Euler-Macheroni constant $\gamma\approx 0.577$ (a. k.a. Euler's constant) as $n\to\infty$. So $\lim_{n\to\infty}[-\ln n+\sum_{j=n_0}^n(1/j)\,]=K$ exists in $\Bbb R.$
So by $(\bullet)$, for all but finitely many $n>n_0$ we have $|\ln |c_{n+1}|\;|>-|\ln |c_{n_0}|\;|+|a|\ln n+|a|K-1.$ For convenience let $L=-|\ln |c_{n_0}|\;|+|a|K-1.$ So for all but finitely many $n$ we have $$(\bullet\bullet)\quad |\ln |c_{n+1}|\;|>L+|a|\ln n.$$ So $(|c_n|)_{n>n_0}$ is a decreasing positive sequence whose log's tend to $\infty,$ so by $(\bullet\bullet)$ we have $$|c_{n+1}|<\frac {e^{-L}}{n^{|a|}}$$ for all but finitely many $n.$
