Prove that $\sum a_n$ converges if $g > 1$ and diverges otherwise 
Suppose $\sum_{ n \geq 1} a_n $ is a series of positive terms. Suppose
that
$$ \lim_{n \to \infty} n \ln \dfrac{ a_n }{a_{n+1} } = g $$
Prove that $\sum_{ n \geq 1} a_n $ converges if $g > 1$ and diverges if $g < 1$.
What if $g=1$?

Attempt:
My intuition is that this is a consequence of the ratio test. Write $n \ln (a_n / a_{n+1} ) = \ln (a_n / a_{n+1} )^n $ and we know that for any $\epsilon > 0$ one can find $N$ so that $n > N$ implies
$$ |\ln (a_n /a_{n+1} )^n - g | < \epsilon $$
and
$$ | \ln (a_n /a_{n+1})^n | < \epsilon + |g| $$
or that
$$ e^{-\epsilon - |g| } < \left(\dfrac{a_n}{a_{n+1} } \right)^n < e^{\epsilon + |g| } \implies  e^{\epsilon + |g| } > \left(\dfrac{a_{n+1}}{a_{n} } \right)^n > e^{-|g|-\epsilon }$$
we can take $\epsilon = |g|$ and so we have
$$ e^{2 |g| } > \left(\dfrac{a_{n+1}}{a_{n} } \right)^n > e^{-2|g| } \implies e^{ 2|g|/n} > \dfrac{ a_{n+1} }{a_n} > e^{- 2 |g| / n}$$
But, regardless of the choice of $g$, I always have that $\lim \dfrac{ a_{n+1} }{a_n} = 1 $ so we cant tell anything about convergenge/divergence. What am I doing wrong here?
 A: You want to start by breaking into cases where $g > 1$, $g < 1$ and $g = 1$. For $g > 1$, the main point is you want to chose $\epsilon$ so that $g - \epsilon > 1$. You made an error in bounding by $e^{-\epsilon - g}$ when it should be $e^{g - \epsilon}$. Then you get, $$\frac{a_{n+1}}{a_n} \leq e^{\frac{-(g-\epsilon)}{n}} \leq \left(\frac{n+1}{n}\right)^{-(g-\epsilon)}$$ So taking $m$ large enough so for $n \geq m$, the above bound holds, $$a_{n} \leq \frac{a_m m^{g-\epsilon}}{n^{g-\epsilon}}$$ and the sum will converge. 
If $g < 1$ take $\epsilon$ so that $g + 2\epsilon < 1$, then by your calculations, for sufficiently large $n$, $$\frac{a_{n+1}}{a_n} \geq e^{\frac{-(g + \epsilon)}{n}} \geq \left(\frac{n+1}{n}\right)^{-(g+2\epsilon)}$$
and so $\sum a_n$ diverges.
If $g = 1$, then the series can either converge or diverge. Taking $a_n = \frac{1}{n}$ gives an example where the sum diverges. On the other hand, for each $n \geq 2$, let $d(n)$ be a positive integer satisfying $$\sum_{k = d(n)}^\infty \frac{1}{k^{\frac{n+1}{n}}} < \frac{1}{2^n}$$ Then for $1 \leq k < d(2)$, set $$a_k = k^{-2}$$ and for $n \geq 2$, for $d(n) \leq k < d(n+1)$, set $$a_k = k^{\frac{-(n+1)}{n}}$$ Then $\sum a_n$ converges, but the corresponding $g$ is $1$.
