Showing a series converges absolutely The goal is to prove that 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 idea: to have the series converge absolutely, then we want to show that there is some $N\in\mathbb{N}$ such that $|\frac{c_{n+1}}{c_n}|\rightarrow n>N$ uniformly.  I was wanting to play with the idea that $n=1$ would work, but then no such $N$ could exist.  I was thinking about maybe using the Ratio test, but I wasn't able to get it to go anywhere.  Moreover, I am assuming the whole "$a<-1$ and $a$ doesn't depend on $n$" bit is pretty important (why I was thinking about playing with $n=1$, but I'm not quite sure how that fits in.  Any thoughts or help would be greatly appreciated :)
 A: One method is to use Raabe's Test, a refinement of the Ratio Test. Let $S=\sum_n c_n$ be a series of positive terms. If it exists, let
$$
\rho =\lim_{n\to\infty} n\left(\frac{c_n}{c_{n+1}}-1\right)
$$Then
$$
\begin{cases}
\rho >0 & S \text{ converges}\\
\rho <0 & S \text{ diverges}\\
\rho =0 & \text{ (inconclusive)}\\
\end{cases}
$$Since we know something about the ratios, we have
$$
\lim_{n\to\infty} n\left(\frac{c_n}{c_{n+1}}-1\right) \geq \lim_{n\to\infty} n\left((1+a/n)^{-1}-1\right)=\lim_{n\to\infty}-\frac{a n}{a+n} = -a>1
$$
So your series converges.

*

*https://mathworld.wolfram.com/RaabesTest.html
A: Let $c_n$ be a sequence such that
$$\left|\frac{c_{n+1}}{c_n}\right|\le \left(1+\frac an\right)\tag1$$
for some number $a<-1$ and $a$ does not depend on $n$.
Let $k$ be a positive integer $k\ge2 $ such that $-k<a<-1$.  Then, we see from $(1)$ that for $n>k$
$$\begin{align}
|c_{n+1}|&\le \left(1+\frac an\right)|c_n|\\\\
&\le \left(1+\frac an\right)\left(1+\frac a{n-1}\right)|c_{n-1}|\\\\
&\vdots\\\\
&\le \prod_{m=0}^{n-k} \left(1+\frac{1}{n-m}\right)|c_k|\\\\
&=|c_k|\exp\left(\sum_{p=k}^{n}\log\left(1+\frac{a}{p}\right)\right)\\\\
&\le |c_k|\exp\left(\sum_{p=k}^{n}\frac{a}{p}\right)\\\\
&\le |c_k|e^{\left(a \log(n/k)\right)}\\\\
&=|c_k|k^{|a|}\frac1{n^{|a|}}
\end{align}$$
Inasmuch as the series $\sum_{n=1}^\infty \frac1{n^{|a|}}$ converges for $|a|>1$, the series $\sum_{n=1}^\infty |c_n|$ converges and hence the series $\sum_{n=1}^\infty c_n$ converges  absolutely.
