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.
 A: This is known as Raabe's test. The criteria is as follows:

*

*If $\frac{|a_{n+1}|}{|a_n|}\leq 1-\frac{a}{n}$  for some $a>1$, then $\sum_na_n$ converges absolutely

*If $\frac{|a_{n+1}|}{|a_n|}\geq 1-\frac{a}{n}$  for some $a\leq1$, then $\sum_n|a_n|$ diverges.

Another version of the Raabe's test is here
Here is one way to prove this.

Proposition: Suppose $a_n, b_n>0$, and that there is $n_0\in\mathbb{N}$ such that $\frac{a_{n+1}}{a_n}\leq \frac{b_{n+1}}{b_n}$ for all $n\geq n_0$.

*

*If $\sum_nb_n$ converges, so does $\sum_na_n$.

*If $\sum_na_n$ diverges, so does $\sum_nb_n$.


Proof: The hypothesis may be rewritten as $\frac{a_{n+1}}{b_{n+1}}\leq \frac{a_n}{b_n}$ for all $n\geq n_0$. Thus, the sequence of ratios $\frac{a_n}{b_n}$ is monotone non increasing for $n\geq n_0$ and so it bounded above. Hence, there is $c>0$ such that $a_n\leq c b_n$ for all $n\geq n_0$. The conclusions (1) and (2) follow immediately.

Lemma: If $a>1$ and $0<x<1$, then $1-ax\leq (1-x)^a$. If $a\leq1$ and $0<x<1$, then $(1-x)^a\leq 1-ax$.

Proof:
Define $g(x)=ax + (1-x)^a$. Then $g'(x)=a\big(1-(1-x)^{a-1}\big)\geq0$. Thus $g$ is monotone nondecreasing ans $g(x)\geq g(0)=1$ for all $x\geq0$. the second statement is prove similarly.
Proof of Raabe's:
$$\frac{a_{n+1}}{a_n}\leq 1-\frac{a}{n}\leq\Big(1-\frac1n\Big)^a=\frac{(n-1)^a}{n^a}$$
Apply proposition above with $b_n=\frac{1}{n^a}$.

Reference: Courant, R. and Fritz, J. Introduction to Calculus and Analysis, Vol. 1, Vol. 1 (Classics in Mathematics) 1999th Edition.

Edit: the method presented in his solution is more about insight than ingenuity, for it is based on the fact that  many ratio type tests of convergence  can be reduced to the setting the proposition above. This was observed by some clever people in the late 1800's, and came up with the systematic treatment I presented above.
The answer provided by Jack D'Aurizio for example  is more ingenious once one makes the connection to the harmonic series: $\frac1n\leq H_n=\sum^n_{k=1}\frac1k\sim\log n$.
Here is another solution that a more pedestrian solution that is more about building things from the ground up. For simplicity assume $a_n>0$ for all $n$,  $a>1$, and  $\frac{a_{n+1}}{a_n}\leq 1-\frac{a}{n}$ for all $n\geq1$. Then
$$na_{n+1}\geq na_n-aa_n=(n-1)a_n-(a-1)a_n$$
and so,
$$(n-1)a_n-na_{n+1}\geq(a-1)a_n>0$$
This means that the sequence $(n-1)a_n$ is monotone non increasing, and adding terms telescopically gives
$$a_2\geq a_2-na_{n+1}\geq (a-1)(a_1+\ldots a_n)$$
This shows that the sequence $\sum^n_{k=1}a_n$ is bounded and so, the series converges.
A similar argument shows that if $a\leq1$, the sum $\sum_na_n$ diverges,
A: 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 $k>1$.
A: 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 $n.$
