# Determine if sum of series is rational or not

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?

• 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. is a poor mathematician 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

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.
• 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.
• An interesting and possibly relevant note: $\sum\limits_{n=0}^\infty \frac{n^2}{n!} = 2e$. – Lee Sleek Jul 4 '13 at 4:17