Consider the series $\sum_{n=1}^{\infty }a_{n}$ where $a_{n}> 0$ for all $n\in \mathbb{N}$. Assume that: $\lim_{n \to \infty }n \ln\left ( \frac{a_{n}}{a_{n+1}} \right )=g$.
I need to prove that if $g> 1$, then $\sum_{n=1}^{\infty }a_{n}$ converges. Similarly, if $g< 1$, then $\sum_{n=1}^{\infty }a_{n}$ diverges.
Here is the solution to this problem as given to me, however I couldn't understand the last part of this proof. If someone has an idea, please share. In the solution to this problem, we need the following lemma (which I will not prove):
Lemma: Let $\sum_{n=1}^{\infty }a_{n}$ and $\sum_{n=1}^{\infty }b_{n}$ be two series of positive terms that satisfy the following inequality: $\frac{a_{n+1}}{a_{n}}\leq \frac{b_{n+1}}{b_{n}}$ for $n\geq K$. One can prove that if $\sum_{n=1}^{\infty }b_{n}$ converges, then $\sum_{n=1}^{\infty }a_{n}$ also converges.
If $g>1$ and let $\epsilon > 0$ be so small so that $g-\epsilon>1$. Then, for sufficiently large $n$: $n\ln\left ( \frac{a_{n}}{a_{n+1}} \right )>g-\epsilon$. Also, from the inequality: $\left ( 1+\frac{1}{n} \right )^{n}< e< \left ( 1+\frac{1}{n} \right )^{n+1}$, it follows that: $n\ln\left ( 1+\frac{1}{n} \right )<1$. So, $n\ln\left ( 1+\frac{1}{n} \right )<1
I can understand everything up to this part. The part I couldn't get is the following: The solution says that it follows from the above inequality that $\frac{a_{n+1}}{a_{n}}<\frac{\frac{1}{(n+1)^{g-\epsilon }}}{\frac{1}{n^{g-\epsilon }}}.\tag{*}\label{lab} $
Then we use the previous lemma to conclude the convergence of the series. Can anyone tell me how to derive the inequality \eqref{lab}? Thanks.