What can be said about the convergence of $\sum \frac {a_n}{a_nn^p+1} \text {?} $ Suppose that $\sum a_n $ is a divergent series of real numbers.
What can be said about the convergence of $$\sum \frac {a_n}{a_nn^p+1} \hspace {15pt}\text {?} $$
Here $p \in \mathbb {R} $ is unspecified, meaning that different values may lead to different solutions.

Note.
This question (which comes from me, not a teacher) generalizes an exercise from Rudin.  Though the exercise in Rudin (which I solved prior to asking this question) keeps $a_n $ non-negative, I would like to know what happens if this assumption is dropped, in addition to taking arbitrary $p $. I do appreciate the very good answers below for the non-negative case; however, this is why I have not yet accepted an answer.
I did not give my progress because I had not made any. I am asking a question about mathematical analysis, not requesting homework help.
 A: Note:  We assume that $a_n > 0$, in keeping with the exercise in Rudin's "Principles of Mathematical Analysis" that seemed to inspire this question, namely Exercise 3.11(d).  This question is a generalization of that exercise, in which $p = 1$ and $p = 2$ were the only two cases considered.
Note that,
\begin{align*}
\sum_{n=1}^\infty \frac{a_n}{a_n n^p + 1} & = \sum_{n=1}^\infty \frac{1}{n^p + \frac{1}{a_n}} \\
& \leq \sum_{n=1}^\infty \frac{1}{n^p}.
\end{align*}
We have that $\sum \frac{1}{n^p}$ converges if $p > 1$ and diverges if $p \leq 1$ by the following.  If $p \leq 0$, then $\lim_{n \to \infty} \frac{1}{n^p} \neq 0$ for $p \leq 0$, hence $\sum \frac{1}{n^p}$ diverges.  If $p > 0$, then we can use a theorem of Cauchy to get us to the series $$\sum_{k=0}^\infty 2^k \cdot \frac{1}{2^{kp}} = \sum_{k=0}^\infty 2^{(1-p)k}.$$  Given that $2^{1-p} < 1$ if and only if $1 - p < 0$, we have that $\sum \frac{1}{n^p}$ converges for $p > 1$ by comparison with the geometric series $$\sum_{n=0}^\infty (2^{1-p})^n = \frac{1}{1-2^{1-p}}.$$
Therefore, we have that $\sum \frac{a_n}{a_n n^p + 1}$ converges if $p > 1$.
Let $p \leq 1$.  Suppose that we let $a_n = \frac{1}{n}$.  Then $$\sum_{n=1}^\infty \frac{a_n}{n^p a_n + 1} = \sum_{n=1}^\infty \frac{\frac{1}{n}}{n^p \frac{1}{n} + 1} = \sum_{n=1}^\infty \frac{1}{n^p + n} \geq \sum_{n=1}^\infty \frac{1}{2n} = \frac{1}{2} \sum_{n=1}^\infty \frac{1}{n},$$ which diverges.
We therefore have that $\sum \frac{a_n}{a_n n^p + 1}$ converges for any $a_n > 0$ if $p > 1$ and we can choose some $a_n > 0$ such that $\sum \frac{a_n}{a_n n^p + 1}$ diverges for all $p \leq 1$.
A: Here is one observation if $a_n > 0 $, $\sum_{n} a_n $ diverges and $p>1$ then the series 

$$ \sum \frac {a_n}{a_nn^p+1} $$ 

converge.
