Proof of a Limit related to Gauss' Convergence test

So this is the question:if the series $\sum_{n=1}^{\infty} a_n$ is such that $$\frac{a_n}{a_{n+1}} = 1 + \frac pn + \alpha_n$$ and the series $\sum_{n=1}^{\infty} \alpha_n$ converges absolutely, then $a_n\sim \frac{c}{n^p}$and that $\sum_{n=1}^{\infty} a_n$ and it converges absolutely for $p>1$ and diverges for $p \leq 1$

Here is my failed attempt

$$\text{let} \:c_n=\ln(\frac{b_n}{b_{n+1}})=\ln(\alpha_n+\frac{p}{n}+1)\le \alpha_n+\frac{p}{n} $$$$\implies \sum_{i=1}^{n}c_i={\ln(b_1)}-{\ln(b_{n+1})}=\ln\prod_{i=1}^{n}(\alpha_i+\frac{p}{i}+1)$$

$$\le\sum_{i=1}^n\alpha_i+p\sum_{i=1}^n \frac 1i$$

Although I applied the inequality, it seems that it doesn't work.And I have absolutely no idea how to continue this? Could you help me?

Edit:

I also noticed that a similar question has already been asked here Gauss' test for Convergence

And nobody answered.