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Is there some methods to find out if the sum of an infinite series is rational or not if we have no closed form expression for the sum?
For instance:

$$\begin{align*} &\sum_{n=1}^{\infty} \frac{n^2}{n!+1}\\ &\sum_{n=1}^{\infty} \frac{1}{n^{7/2} p_n}\\ &\sum_{n=1}^{\infty} \frac{n}{F_n}\end{align*}$$

where $p_n$ is the $n$th prime number and $F_n$ is the $n$th Fibonacci number.

And if a sum of a series have no provable closed form expression, can it still be rational?

And is there a series which is known to be rational, but not which rational?

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I don't think there's even a known closed form for 1. It seems likely that 3. is irrational based on this, but I can't think of a proof. – J. M. Sep 22 '11 at 0:49
I doubt we know any general method. But I guess a rule of thumb is that the faster the denominators grow, the less likely it is to be rational or algebraic (I base this guess on Hurwitz's and Liouville's theorems). – Joel Cohen Sep 22 '11 at 1:17
@JoelCohen: Not necessarily. A "random" convergent series almost certainly has a transcendental sum, whether the denominators grow quickly or slowly. – Robert Israel Sep 22 '11 at 4:14
For an example of a series with a rational sum and fast-growing denominators: take any positive integer $a$, and for $n \ge 1$ let $x_n = \dfrac{(n+1)! - n!}{(a+(n+1)!)(a+n!)}$. Note that this is $\dfrac{a-1+(n+1)!}{a+(n+1)!} - \dfrac{a-1+n!}{a+n!}$. Then $\sum_{n=1}^\infty x_n = \frac{1}{a+1}$. The denominators grow like $n (n!)^2$. – Robert Israel Sep 22 '11 at 4:39
@RobertIsrael: You're right, nice counter example. – Joel Cohen Sep 22 '11 at 10:34
up vote 10 down vote accepted

There are some useful results obtained in these papers (and the references therein):

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+1, very nice results! – lhf Sep 22 '11 at 10:44

It's unlikely that there are general methods. Witness the irrationality of $\zeta(2)$, which has a closed form $\pi^2/6$ found by Euler in 1735 (see Basel problem), but which was proved irrational by Hermite in 1873 only. Witness also $\zeta(3)$, whose irrationality was proved only in 1978 by Apéry, but for which no closed form is known.

I guess the closest thing to a general method is Dirichlet's irrationality criterion and its generalizations such as Hurwitz's theorem. But even these are hard to apply in any given particular case.

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Witness also $e + \pi = \sum_{n=0}^\infty \left(\frac{1}{n!} + \frac{4 (-1)^n}{2n+1}\right)$, whose irrationality is still unproven (although almost certainly it is irrational). – Robert Israel Sep 22 '11 at 1:10

No, there are no such methods (if you're talking about a proof) except for some rather special cases. Even with a closed-form expression, it is quite rare to be able to prove that something is irrational. On the other hand, given a good numerical approximation to a number you can use continued fractions to see if this number appears to be a rational with small numerator and denominators. For example, $\sum\limits_{n=1}^\infty \frac{n^2}{n!+1} \approx 4.0271515294669515849240298741047887364140370824913$ which has the continued fraction representation $\begin{split}4; & 36, 1, 4, 1, 8, 2, 3, 1, 1, 1, 2, 1, 1, 4, 1, 1, 1, 1, 2, 2, 6, 2, 33, 2, 1, 1, 1, 4, 18, 4, 1, 2, 6, 8, 3,\\& 1, 6, 1, 3, 1, 4, 4, 1, 9, 3, 8, 1, 2, 35 \ldots\end{split}$. This shows no signs of terminating, so the number is likely irrational.

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It does not seem to be periodic either and so it's probably not a quadratic irrational. – lhf Sep 22 '11 at 1:05
An interesting and possibly relevant note: $\sum\limits_{n=0}^\infty \frac{n^2}{n!} = 2e$. – Lee Sleek Jul 4 '13 at 4:17

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