What is the set of all "pseudo-rational" numbers (see details)? Define a “pseudo-rational” number to be a real number $q$ that can be written as
$q=\sum_{n=1}^{\infty} \frac{P(n)}{Q(n)}$
Where $P(x)$ and $Q(x)$ are fixed integer polynomials (independent of n). All rational numbers are pseudo-rational, as is $\pi^2$ using $P(x)=6,Q(x)=x^2$. There must exist numbers that are not pseudo-rational (defined as "pseudo-irrational") because the set of pseudo-rationals is countable. Is $e$ pseudo-rational? Is $\sqrt{2}$?
 A: Also possibly of interest:
$$ \sum_{n=1}^\infty \left(-\frac{2}{b(n-1/b)(n+1/b)} + \frac{b}{n(n+1)}\right) =  \pi \cot(\pi/b)$$
$$ \sum_{n=1}^\infty {\frac {t \left( {t}^{2}{n}^{2}+2\,{n}^{2}+2\,n+1 \right) }{ \left( n+
1 \right) n \left( {t}^{2}{n}^{2}+1 \right) }}
 = \pi \coth(\pi/t) $$
EDIT: 
And, if I'm not mistaken,
$$ \sum_{n=1}^\infty \left(\frac{1-m}{mn} + \sum_{k=1}^{m-1} \frac{1}{mn-k}\right) = \ln(m) $$
for positive integers $m$.
A: One partial answer:
$\pi$ and $\ln(2)$ are both pseudo-rational:
\begin{align}
\ln(2) &= \sum\frac{1}{2n\,(2n-1)} \\
\pi &= \sum\frac{3}{n\,(2n-1)\,(4n-3)} \\
\end{align}
These follow from statements in Wikipedia, including Gauss's digamma theorem, and are also asserted by Mathematica.  By similar manipulations, $\pi\sqrt{3}$ and $\ln(3)$ are pseudo-rational also.
$\ $
Some questions collected from the comments:


*

*Are the pseudo-rationals closed under multiplication?

*Are all pseudo-rationals periods?

*Is there a procedure to decide if the pseudo-rational from $P,Q$ is positive? 


The last one is difficult because there are non-trivial zeroes like $\sum\left(\dfrac{6}{n^2}-\dfrac{8}{(2n-1)^2}\right)$.
