proving that a series is divergent

Question

Let $a_j=\bigg(\ln (j+e)\bigg)^{jp}$, where $j\in \mathbb{N}$ and $p>0$ is a real number. I want to show that the series $$\sum_{j=1}^\infty\frac{a_j}{a_{j+1}}$$ is convergent for $p>1$ and divergent for $ 0<p\leq 1$. I did the former case by comparing it with a $p-$ series. I have a difficulty in the latter case. Please give me some hint.

Answer 1

$$\sum\frac{a_j}{a_{j+1}}=\sum\frac{\left(\ln (j+e)\right)^{jp}}{\left(\ln (j+1+e)\right)^{(j+1)p}}=\sum\left(\frac{\ln (j+e)}{\ln (j+1+e)}\right)^{jp}\frac{1}{\left(\ln (j+1+e)\right)^p}\cdot $$ Applying limit comparison test with the divergent series $\sum \frac{1}{\left(\ln (j+1+e)\right)^p}$ we get $$\lim_{j\to\infty}\frac{\left(\frac{\ln (j+e)}{\ln (j+1+e)}\right)^{jp}\frac{1}{\left(\ln (j+1+e)\right)^p}}{\frac{1}{\left(\ln (j+1+e)\right)^p}}=\lim_{j\to\infty}\left(\frac{\ln (j+e)}{\ln (j+1+e)}\right)^{jp}=1,$$ and conclude $\sum\frac{a_j}{a_{j+1}}$ is divergent for all $p>0$.