# If $\lim_{n \to \infty }n \ln\left ( \frac{a_{n}}{a_{n+1}} \right )=g>1$, then $\sum_{n=1}^{\infty }a_{n}$ is convergent

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.

Let $u=g-\epsilon$, then $n\log(a_n/a_{n+1})\gt u$, hence $a_{n+1}/a_n\lt\mathrm e^{-u/n}$. Since $\mathrm e\gt\left(1+\frac1n\right)^n$, this yields $\mathrm e^{-u/n}\lt\left(1+\frac1n\right)^{-u}$. The RHS is the odd-loking ratio $\frac{\frac1{(n+1)^u}}{\frac1{n^u}}$ in the RHS of $(*)$.
Edit The other case is $g\lt1$, then $n\log(a_n/a_{n+1})\lt1$ for every $n$ large enough, hence $a_{n+1}/a_n\gt\mathrm e^{-1/n}$. Since $\mathrm e\lt\left(1+\frac1{n-1}\right)^{n}$, $\mathrm e^{-1/n}\gt\frac{n-1}{n}$. Thus, the sequence $(na_{n+1})_n$ is nondecreasing for $n$ large enough. In particular, $a_{n+1}\gt C/n$ for every $n$ large enough, for some positive $C$, and this implies that the series $\sum\limits_na_n$ diverges.