Proving that $a_n\sim \frac c{ n^p}$ given $\frac{a_n }{ a_{n+1 }}=1+\frac pn +b_n$, where $\sum b_n$ is absolutely convergent. $\frac{a_n }{ a_{n+1 }}=1+\frac pn +b_n$, where $\sum b_n$ is absolutely convergent, then it is to be proven that $a_n\sim \frac c{ n^p}$. 
I tried to prove it like this: 
Let's first consider a particular case when $p=0$ so that we have $\ln a_n-\ln a_{n+1 }=\ln (1+b_n)$. It follows that $\ln a_1-\ln a_{n+1 }=\sum_{n=1}^n \ln(1+b_n)\tag 1$ 
Now, we have $||\frac{\ln (1+b_n) }{b_n }|-1|=||\frac{ b_n+o(b_n)}{b_n }|-1|=||1+\frac{o(b_n) }{b_n }|-1\le|\frac{ o(b_n)}{b_n }|\implies \lim |\frac{\ln (1+b_n) }{b_n }|=1\implies |\ln (1+b_n)|\sim |b_n|$. It follows that $\sum \ln (1+b_n)$ is absolutely cgt. if and only if $b_n$ is absolutely cgt. 
So $(1)$ becomes: $\lim (\ln a_1-\ln a_{n+1})=x$, where $x=\sum \ln (1+b_n)$, which exists as $b_n$ is abs. cgt. It follows that $\lim a_{n+1}=a_1 e^{-x}$ that is $\lim a_{n+1}$ exists. 
Now we come back to the general case: 
We multiply both sides by $\frac{n^p}{(n+1)^p}$ to get: 
$\frac{n^p}{(n+1)^p}\frac{a_n }{ a_{n+1 }}=\frac{n^p}{(n+1)^p}(1+\frac pn +b_n)$ and let $c_n:=a_n n^p$ so that we have 
$\frac{c_n }{ c_{n+1 }}=\frac{n^p}{(n+1)^p}(1+\frac pn +b_n)=1+\theta_n$, where $\theta_n=\frac{n^p}{(n+1)^p}(1+\frac pn +b_n)-1$ $\tag 2$
We have (refer Note) for $p\ne -1$: 
$\begin{align}\theta_n&=(1+\frac 1n)^{-p}+\frac pn (1+\frac 1n)^{-p}+b_n(1+\frac 1n)^{-p}-1\\&=1-\frac pn+O(\frac 1{n^2})+\frac pn (1-\frac pn +O(\frac 1{n^2}))+b_n(1-\frac pn+O(\frac 1{n^2}))-1\\&=O(\frac 1{n^2})(1+\frac pn+b_n)-\frac {p^2}{n^2}+b_n-\frac pn b_n\end {align}$ 
It follows that $|\theta_n|\le |O(\frac 1{n^2})(1+\frac pn+b_n)|+|\frac {p^2}{n^2}|+|b_n|+|b_n|$ 
It follows that $\sum \theta_n$ is abs. cgt. by comparison test. 
Now if $p=-1$, then our $\theta_n=-\frac 1 {n^2}+b_n+\frac {b_n}{n}$ and again it follows that $\sum \theta_n$ is abs. cgt.
It follows from $(1)$ and $(2)$ that: $\lim c_n$ converges to some $c\in \mathbb R$ and therefore $a_n\sim \frac c{n^p}$ 
Is my proof correct? Thanks.
Note: Here, I have used $\frac {n^p}{(1+n)^p}=(1+\frac 1n)^{-p}=1-\frac pn +O(\frac 1{n^2})$
 A: Your approach looks okay. Here is a possible simplification.

The Binomial Theorem says that $(1+x)^p=1+px+O\!\left(x^2\right)$. Therefore, if $x$ is bounded and bounded away from $-\frac1p$,
$$
\begin{align}
\frac{(1+x)^p}{1+px}
=1+O\!\left(x^2\right)\tag1
\end{align}
$$
Furthermore, if $x$ is bounded away from $-1$,
$$
\begin{align}
1+x+y
&=(1+x)\left(1+\frac{y}{1+x}\right)\\[3pt]
&=(1+x)(1+O(y))\tag2
\end{align}
$$
Thus,
$$
\begin{align}
\frac{a_n}{a_{n+1}}
&=1+\frac pn+b_n\tag{3a}\\
&=\left(1+\frac pn\right)\left(1+O(b_n)\right)\tag{3b}\\
&=\left(1+\frac1n\right)^p\left(1+O\!\left(\frac1{n^2}\right)\right)^{-1}\left(1+O(b_n)\right)\tag{3c}\\
\end{align}
$$
Taking the product of $(3)$, we have
$$
\begin{align}
a_n
&=a_1\,\prod_{k=1}^{n-1}\left(\frac{k}{k+1}\right)^p\prod_{k=1}^{n-1}\left(1+O\!\left(\frac1{k^2}\right)\right)\prod_{k=1}^{n-1}\left(1+O(b_k)\right)^{-1}\tag{4a}\\
&\sim\frac{a_1c_1/c_2}{n^p}\tag{4b}
\end{align}
$$
since $c_1=\prod\limits_{k=1}^\infty\left(1+O\!\left(\frac1{k^2}\right)\right)$ and $c_2=\prod\limits_{k=1}^\infty\left(1+O(b_k)\right)$ both converge, because $\sum\limits_{k=1}^\infty\frac1{k^2}$ and $\sum\limits_{k=1}^\infty b_k$ both converge absolutely.
