Prove convergence of a series given a bound on the $n$th and $n+1$st terms I'm given that $\Bigl | \frac{a_{n+1}}{a_n}\Bigr|\le 1-\frac{L}{n}$, $\forall n > N \in \mathbb{Z}$ where $L>1$ and $a_n$ is a sequence of positive numbers. I am to prove the convergence of $\sum a_n$.
My immediate thought was to write $1-\frac{L}{n} = \frac{n-L}{n}$, and since $n,L>0$, $n-L<n \Rightarrow \frac{n-L}{n} <1$. Then, the ratio test would result in the convergence of the series. However, this seems a bit lacking to say the least. I know I shouldn't assume every problem needs a "trick," but I think I may have overlooked things.
 For example, I just used that $L$ is positive, that is, it could work for values less than $1$, but the problem said it was strictly greater than $1$. I'd appreciate a push in the right direction, thanks.
 A: Use the criterion of Raabe,that derives from the criterion of Dini-Kummer,that says if $a_n>0$
and $b_n=n(1-\frac {a_{n+1}}{a_n})$ then if 
$\underline {\lim} b_n>1$ ,then $\sum a_n$ converges.
Here $1-\frac {a_{n+1}}{a_n}\geq \frac {L}{n}=>n(1-\frac {a_{n+1}}{a_n})\geq L>1$ and thus $\underline {\lim} b_n>1$.
Ok. First we will prove the Dini-Kummer proposition that says:
Let  $(a_n),(b_n)$ be positive sequences and $d_n=b_n-b_{n+1}\cdot \frac {a_{n+1}}{a_n}$ for every $n=1,2,...$.
Then:
$i)$If $\underline {lim} d_n>0=>\sum a_n$ converges.
$ii)$If there exists a natural number $N:d_n\leq 0 $, $\forall n\geq N$ and $\sum \frac {1}{b_n}$ diverges,then $\sum a_n$ diverges too.
What you need here is to prove the first case $i)$.
I'll write down the proof so you don't have any questions(of course if you want to try it by yourself,just see the first two sentences).

Let $0<r<\underline {lim} d_n$. Then there is $n_0\in \Bbb N$ such that $n\geq n_0=>d_n>r=>0<a_n<\frac {1}{r}\cdot (a_nb_n-a_{n+1}b_{n+1})$. $(1)$
This means that $(a_nb_n)$ is a sequence of strictly decreasing positive numbers and thus it converges to $c\geq 0$.
Hence,$\sum_{n=n_0}^{\infty} {\frac {1}{r}\cdot (a_nb_n-a_{n+1}b_{n+1})}=\lim_{N\to \infty} \sum_{n=n_0}^{N} {\frac {1}{r}\cdot (a_nb_n-a_{n+1}b_{n+1})}=\lim_{N\to \infty} \frac {1}{r}\cdot (a_{n_0}b_{n_0}-a_{N+1}b_{N+1})=\frac {1}{r}\cdot (a_{n_0}b_{n_0}-c)$. 
From comparison test and the $(1)$ relation,we have that $\sum a_n$ converges.
Now for $b_n=n-1$ we have the Raabe's criterion.
Question:For $b_n=1$,what criterion do we have?
Also there is something more general.
Gauss's criterion:
Let $(a_n)$ be a positive sequence, $b\in \Bbb R$ and $(c_n)$ be a bounded sequence such that $$\frac {a_{n+1}}{a_n}=1-\frac {b}{n}-\frac {c_n}{n^2}$$. 
Then $\sum a_n$ converges iff $b>1$.
