TITLE: What is the set of all "pseudo-rational" numbers (see details)?
QUESTION [21 upvotes]: 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 $\sqrt{2}$?

REPLY [7 votes]: 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 $m$.