If $\sum\limits_{n=1}^{\infty}\frac{1}{p(n)}\in\mathbb{Q}$, is $\sum\limits_{n=1}^{\infty}\frac{n}{p(n)}\in\mathbb{Q}$? Suppose $p(n)$ is a polynomial with rational coefficients and rational roots of degree at least $3$. If we know 
$$\sum_{n=1}^{\infty}\frac{1}{p(n)}\in\mathbb{Q}$$
are we able to infer that
$$\sum_{n=1}^{\infty}\frac{n}{p(n)}\in\mathbb{Q}?$$
I've tried several approaches to proving (or disproving) this to include the following:
-Looking for counterexamples
-Generating functions
-Residues
-Partial fraction decomposition
but nothing has yielded any positive or negative results. Any tips, terms, papers, methods, or generally topics that I could look into would also be welcome.
Edit: As noted by Carl Schildkraut below, if this is true, then we would automatically know that $\zeta(2k+1)$ was irrational. Since this seems to greatly increase the potential difficulty, I offer the following modification in order to simplify it:
Suppose $p(n)$ is a polynomial with rational coefficients, rational roots, $\deg(P)\geq 3$, and every root has order $1$. If we know 
$$\sum_{n=1}^{\infty}\frac{1}{p(n)}\in\mathbb{Q}$$
are we able to infer that
$$\sum_{n=1}^{\infty}\frac{n}{p(n)}\in\mathbb{Q}?$$
 A: Answering my own question as I believe I have finally found a counterexample:
$$\sum_{n=1}^{\infty}\frac{1}{\left(n+\frac{1}{3}\right) \left(n+\frac{5}{6}\right) \left(n+\frac{11}{6}\right) \left(n+\frac{7}{3}\right)}=\frac{9}{154}$$
$$\text{but }\sum_{n=1}^{\infty}\frac{n}{\left(n+\frac{1}{3}\right) \left(n+\frac{5}{6}\right) \left(n+\frac{11}{6}\right) \left(n+\frac{7}{3}\right)}=\frac{1}{90} \left(-261+80 \sqrt{3} \pi -240 \log (2)\right).$$
The reason I said believe is that it is still an open problem whether $\{\pi,\log(r),s\}$ for $r,s\in\mathbb{Q}$ are algebraically independent, but most would agree that they probably are.
I did manage to prove that the conjecture was true if $p(n)$ was degree two or three, but it seems that for degree four and above it is not true. Based on how I constructed this (combining two degree two polynomials in a certain way) it might be the case that counterexamples only exist for polynomials of even degree larger than four (or four might be the only degree with a counterexample), but further study is needed.
A: $\newcommand\Q{\mathbf{Q}}$
I object a little bit to the "accepted" answer. It would require proving that $\sqrt{3} \pi - 3 \log(2)$ is not a rational number. I'm not sure this is so obvious. 
Here is an alternate solution which is less random:
Let~$b > a$ be distinct non-zero  rational numbers such that $2a$ and $2b$ are integers but $a$ and $b$ are not.
Note that $a+b$ and $a-b$ will both be integers.  Let 
$$R= \displaystyle{\frac{a+b}{a-b}},$$
and let
$$f(x) = \frac{1}{a b(a+b)} \cdot x(x+a)(x+b)(x+a+b).$$
Note that
$$ \sum_{n=1}^{\infty} \frac{1}{f(n)} = \sum_{n=1}^{\infty}
\frac{ab(a+b)}{n(n+a)(n+b)(n+a+b)}
=
\sum_{n=1}^{\infty} \frac{1}{n} - \frac{1}{n+a+b} + \frac{R}{n+a} - \frac{R}{n+b} $$
$$= \sum_{n=1}^{b + a } \frac{1}{n} +  R \sum_{n=1}^{b-a} \frac{1}{n+a} \in \Q + \Q = \Q$$
is a telescoping sum and thus rational. On the other hand, if
$$F(a,b) = \frac{1}{R} \sum_{n=1}^{\infty} \frac{n}{f(n)},$$
then
$$F(a,b) =  \frac{1}{R} \sum_{n=1}^{\infty}
\frac{n ab(a+b)}{n(n+a)(n+b)(n+a+b)} = \sum_{n=1}^{\infty}  \frac{b}{n+b} - \frac{a}{n+a}  - \frac{b-a}{n+a+b}$$
$$= \sum_{n=1}^{\infty}  \frac{a}{n+b}  + \frac{b-a}{n+b}  - \frac{a}{n+a}   - \frac{b-a}{n+a+b}$$
$$= \sum_{n=1}^{\infty}  \frac{a}{n+b} - \frac{a}{n+a}  + \frac{b-a}{n+b} - \frac{b-a}{n+a+b}$$
$$= \left(\sum_{n=1}^{b-a} - \frac{a}{n+a} \right)  + (b-a) \sum_{n=1}^{\infty} \frac{1}{n+b} - \frac{1}{n+a+b}$$
$$= \left(\sum_{n=1}^{b-a} - \frac{a}{n+a} \right)  + (b-a)\left( \sum_{n=0}^{\infty} \frac{1}{n+1/2}  - \frac{1}{n+1}
\right) 
- (b-a) \sum_{n=0}^{b-1/2} \frac{1}{n+1/2} + (b-a) \sum_{n=0}^{a+b-1} \frac{1}{n+1}$$
$$ \in \Q + 2(b-a) \log 2 + \Q + \Q = \Q + 2(b-a) \log 2.$$
This is similar to the other solution, except showing that $\log(2)$ is irrational is a direct consequence of the transcendence of $e$. 
