Question: Prove that if $u_n>0$ and $\lim_\limits{n\to\infty}n\left(\frac{u_n}{u_{n+1}}-1\right)=l$ , then $\sum_{n=1}^{\infty} u_n$ diverges if $l<1$.
My Attempt: We choose a positive number $\epsilon$ such that $l+\epsilon<1$.
Since $\lim_\limits{n\to\infty}n\left(\frac{u_n}{u_{n+1}}-1\right)=l$, then for any $\epsilon>0$ , there exists a natural number $m$ such that $$\left| \,\ n\left(\frac{u_n}{u_{n+1}}-1\right)-l\,\ \right|< \epsilon \,\ \forall \,\ n\ge m$$
Therefore, we have Any suggestions on how to proceed from here? P.S. I have read other questions here on MathsSE regarding this proof specifically but I am not satisfied with these answers as one uses big O notation and another has a different form of Raabe's test. I don't want answers with big O notation. Proofs with other methods are welcome but the ones following from where I have ended are most preferred.
$$l-\epsilon
Answer
Your first steps are all correct.
You have obtained the inequality
$$
\frac{u_n}{u_{n+1}} < \frac{r}{n} + 1
$$
for $n\geq m$
but have realized that, in this form, it is rather hard to do anything with it.
Here is the classical argument as to how to proceed.
Rewrite this as
$$
n \frac{u_n}{u_{n+1}} -(n+1) < r-1 \leq 0.
$$
Then observe that
$$ nu_n \leq (n+1)u_{n+1} $$
for all $n$ greater than $m$. Thus we have an increasing
sequence $\{nu_n\}$ and, in particular there is some positive constant $c$
for which $nu_n\geq c$.
Thus for $n\geq m$ we have $$u_n \geq \frac{c}{n}$$
and a comparison with the harmonic series establishes
divergence as you desired.
There is a more general version of Rabbe's test known as
Kummer's test with very much the same proof.
For that you assume there is a sequence of positive numbers
$D_n$ and you compute the limit
$$
L =\lim_{n\to\infty} \left[ D_n\frac{u_n}{u_{n+1}} -D_{n+1} \right]
$$
Raabe's test is just Kummer's test with $D_n=n$.
The divergence part of Kummer's test (and hence also Raabe's test) doesn't actually require limits. You simply need that $$ D_n\frac{u_n}{u_{n+1}} -D_{n+1} \leq 0$$
for all sufficiently large $n$ and the divergence of $\sum_{n=1}^\infty 1/D_n$ and you can conclude divergence.
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