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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 $\sqrt{2}$?

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    $\begingroup$ It is an additive subgroup of the reals in which each number is "slowly"approximable. I don't know if all algebraic numbers are in this group. I am pretty sure it is not closed under multiplication and that e is not in this subgroup, but I can't prove it yet. Gerhard "Perhaps It Excludes Liouville Numbers?" Paseman, 2017.01.13. $\endgroup$
    – Gerhard Paseman
    Commented Jan 14, 2017 at 7:17
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    $\begingroup$ Also, you should place certain restrictions on Q and P to avoid division by zero and diverging sums. It includes values of the zeta function in addition to pi^2/6, and likely some more exotic numbers. In addition to checking for algebraic and Liouville numbers, I suggest looking at some mathematical constants that may arise from this, particularly work of Steven Finch. Also, approximation (by rational numbers) theory may help. Gerhard "Try Diverging Away From Nullity" Paseman, 2017.01.13. $\endgroup$
    – Gerhard Paseman
    Commented Jan 14, 2017 at 7:43
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    $\begingroup$ This question might be completely inaccessible. for example $\zeta(7)$ is pseudorational and hence any rational multiple of it is too. But it could maybe be an open problem to even say whether $e$ is a rational multiple of $\zeta(7)$ -- these things are hard. Maybe someone can come up with a clever proof that (some well-known number) is pseudorational, but most numbers won't be and my guess is that you won't be seeing anyone posting proofs that (some well-known number) is not. Here's a question which might be more tractible -- are pseudorational numbers all periods in the sense of Zagier? $\endgroup$
    – Kevin Buzzard
    Commented Jan 14, 2017 at 11:14
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    $\begingroup$ @GerhardPaseman, calculating to within epsilon is routine, but that procedure won't tell you about positivity. How would it work for $P/Q = 6/n^2 - 8/(2n-1)^2$ ? $\endgroup$
    – user44143
    Commented Jan 15, 2017 at 12:12
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    $\begingroup$ It's still an open question whether $e/\pi$ is irrational, so don't expect a proof that $e$ is not pseudorational. $\endgroup$
    – Robert Israel
    Commented Jan 17, 2017 at 1:33

3 Answers 3

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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.

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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 $\sum\left(\dfrac{6}{n^2}-\dfrac{8}{(2n-1)^2}\right)$.

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    $\begingroup$ For $\pi$ there is also the trivial $\sum \frac{8}{(4n-3)(4n-1)}$. $\endgroup$
    – Yaakov Baruch
    Commented Feb 9, 2017 at 7:23
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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 $m$.

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    $\begingroup$ I like the expression for the logs. $\endgroup$
    – user44143
    Commented Jan 18, 2017 at 3:26
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If you just want lots of examples .... note that, if $a$ is rational and not an integer, then $$ \sum_{n=1}^\infty\left(\frac{1}{n} - \frac{1}{n+a-1}\right) = \gamma+\psi(a) $$ is pseudo-rational. ($\psi$ is the digamma) I plugged in $k/12$ for $1 \le k \le 11$ to get these: $$ -2\,\ln \left( 2 \right) ,\\-1/6\,\pi \,\sqrt {3}-3/2\,\ln \left( 3 \right) ,\\-3\,\ln \left( 2 \right) -\pi /2,\\-3/2\,\ln \left( 3 \right) -2\,\ln \left( 2 \right) -1/2\,\pi \,\sqrt {3},\\-1/2\,\sqrt { 3}\ln \left( 2-\sqrt {3} \right) -3/2\,\ln \left( 2-\sqrt {3} \right) -3/4\,\sqrt {3}\ln \left( 3 \right) -9/4\,\ln \left( 3 \right) -3\,\ln \left( 2 \right) -1/2\,\sqrt {2}\pi \,\cos \left( \pi /12 \right) -1/2\,\sqrt {2}\pi \,\cos \left( \pi /12 \right) \sqrt {3}+3/2\,\ln \left( -3+2\,\sqrt {3} \right) +3/2\,\ln \left( - 3+2\,\sqrt {3} \right) \sqrt {3},\\1/6\,\pi \,\sqrt {3}-3/2\,\ln \left( 3 \right) ,\\-3\,\ln \left( 2 \right) +\pi /2,\\-3/2\,\ln \left( 3 \right) -2\,\ln \left( 2 \right) +1/2\,\pi \,\sqrt {3},\\-3/2 \,\ln \left( 2-\sqrt {3} \right) +1/2\,\sqrt {3}\ln \left( 2-\sqrt { 3} \right) -9/4\,\ln \left( 3 \right) +3/4\,\sqrt {3}\ln \left( 3 \right) -3\,\ln \left( 2 \right) -1/2\,\sqrt {2}\pi \,\cos \left( { \frac {5\,\pi }{12}} \right) \sqrt {3}+1/2\,\sqrt {2}\pi \,\cos \left( {\frac {5\,\pi }{12}} \right) -3/2\,\ln \left( -3+2\,\sqrt {3 } \right) \sqrt {3}+3/2\,\ln \left( -3+2\,\sqrt {3} \right) ,\\-3/2\, \ln \left( 2-\sqrt {3} \right) +1/2\,\sqrt {3}\ln \left( 2-\sqrt {3} \right) -9/4\,\ln \left( 3 \right) +3/4\,\sqrt {3}\ln \left( 3 \right) -3\,\ln \left( 2 \right) +1/2\,\sqrt {2}\pi \,\cos \left( { \frac {5\,\pi }{12}} \right) \sqrt {3}-1/2\,\sqrt {2}\pi \,\cos \left( {\frac {5\,\pi }{12}} \right) -3/2\,\ln \left( -3+2\,\sqrt {3 } \right) \sqrt {3}+3/2\,\ln \left( -3+2\,\sqrt {3} \right) ,\\-1/2\, \sqrt {3}\ln \left( 2-\sqrt {3} \right) -3/2\,\ln \left( 2-\sqrt {3} \right) -3/4\,\sqrt {3}\ln \left( 3 \right) -9/4\,\ln \left( 3 \right) -3\,\ln \left( 2 \right) +1/2\,\sqrt {2}\pi \,\cos \left( \pi /12 \right) +1/2\,\sqrt {2}\pi \,\cos \left( \pi /12 \right) \sqrt {3}+3/2\,\ln \left( -3+2\,\sqrt {3} \right) +3/2\,\ln \left( - 3+2\,\sqrt {3} \right) \sqrt {3} $$

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