When the ratio test fails. Suppose $a_n>0$ and, for large enough $n$, $\frac{a_{n+1}}{a_n} \leq 1-\frac{c}{n}$, where $c>1$. Prove the series $\sum a_n$ converges.
I tried looking at the tail of the distribution, at which point we have a decreasing sequence of positive reals, but other tests don't seem to yield anything. Can't see how the particular form of the RHS helps.
 A: This is indeed Raabe's test, and it is a consequence of Kummer's rule:
If $\dfrac{a_{n+1}}{a_n}\le 1-\dfrac cn$ for all $n\ge N$, we have
$$\frac{a_n}{a_{n+1}}\ge \frac 1{1-\cfrac cn}=\frac n{n-c}.$$
Now, set $k_n=n-c$. We have
$$k_n\frac{a_n}{a_{n+1}}-k_{n+1}\ge n-(n+1-c)=c-1>0.$$
By Kummer's rule, the series $\;\sum_n a_n$ is convergent.

Kummer's rule
The series with positive terms $\sum_n u_n$ converges if and only if there exists a postive sequence $k_n$ and a constant $\delta>0$ such that, if $n$ is large enough, one has
$$k_n\frac {u_n}{u_{n+1}}-k_{n+1}\ge \delta$$

Proof of Kummer's rule:
Suppose that, if $n\ge N$, we have $k_n\dfrac {u_n}{u_{n+1}}-k_{n+1}\ge \delta$. This can be written as
$$\delta u_{n+1} \le k_nu_n-k_{n+1}u_{n+1}.$$
Summing these inequalities from $n=N$, we obtain
\begin{align}
&\delta(u_{N+1}+u_{N+2}+\dots +u_{N+m})\le \\
& k_Nu_N-k_{N+1}u_{N+1}+k_{N+1}u_{N+1}-k_{N+2}u_{N+2}+\dots +k_{N+m-1}u_{N+m-1}-k_{N+m}u_{N+m} \\
&=k_Nu_N-k_{N+m}u_{N+m} \le k_Nu_N,
\end{align}
whence
$$\sum_{n>N}u_n \le \frac1{\delta}k_Nu_N$$
converges since it is a series with positive terms and bounded partial sums.
Conversely, if the series converges, setting
$\;k_n=\frac1{u_n}\sum_{k>n}u_k$, one has
$$k_n\frac{u_n}{u_{n+1}}-k_{n+1}=1. $$
A: The tail end of the sequence will be bounded above a geometric progression of ratio less than $1$, which is known to converge.
