If a sequence divergences to ##+ \infty## then so does every subsequence. For any ##r > 0##, we must have that ##\frac{r}{n} \leq |b_n|## for only a finite number of terms, otherwise ##\infty = \sum^\infty \frac{r}{n} \leq \sum^\infty |b_n|## where the sum is taken over an arbitrary subsequence. Therefore, there exists a ##N## such that ##\frac{r}{n} > |b_n|## for all ##n > N##. Therefore, there exists a ##N## such that ##\frac{r}{n} + b_n > 0## for all ##n > N##.
Define a ##q## such that ##1 < q < p##. There exists an ##N## such that ##\frac{p-q}{n} + b_n > 0## for all ##n > N##. From ##\frac{a_n}{a_{n+1}} = 1 + \frac{p}{n} + b_n## we have
\begin{align*}
\dfrac{|a_n|}{|a_{n+1}|} & = |1 + \frac{q}{n} + \frac{p-q}{n} + b_n|
\nonumber \\
& > 1 + \frac{q}{n}
\end{align*}
for ##n > N##.
Rearranged:
\begin{align*}
n \left( \dfrac{|a_n|}{|a_{n+1}|} - 1 \right) > q \qquad (*)
\end{align*}
The Raabe-Duhamel's test: Let ##\{ c_n \}## be a sequence of positive numbers. Define
\begin{align*}
\rho_n := n \left( \dfrac{c_n}{c_{n+1}} - 1 \right)
\end{align*}
if
\begin{align*}
L = \lim_{n \rightarrow \infty} \rho_n
\end{align*}
exists and ##L > 1## the series converges.
From ##(*)## we have
\begin{align*}
L = \lim_{n \rightarrow \infty} \rho_n = \lim_{n \rightarrow \infty} n \left( \dfrac{|a_n|}{|a_{n+1}|} - 1 \right) > q > 1 .
\end{align*}
Hence, ##\sum |a_n|## converges.
