I was playing around with infinite series, specifically those that converged to rational numbers, and noticed (at least for the ones that I tried) that when I divided them by a factorial, they would become irrational.
My conjecture is if $\sum_{n=j}^{\infty} a_n$ converges to a rational number, then $\sum_{n=j}^{\infty} \frac{a_n}{n!}$ always converges to an irrational number, with $j$ being a non-negative integer and $a_n$ being some sequence. I do know that it always converges, due to the comparison test, as the series with $a_n$ converges. However, I don't know if my conjecture is true or not.
Some series that I have tested are (I actually couldn't think of many rational-converging series):
- $\sum_{n=0}^{\infty} \frac{1}{x^n} = \frac{x}{x-1}$ (for $|x| > 1$)
- $\sum_{n=1}^{\infty} \frac{1}{(\pi n)^2} = \frac{1}{6}$
- $\sum_{n=0}^{\infty} \frac{(-1)^n}{x^n} = \frac{x}{x+1}$ (for $|x| > 1$)
- $\sum_{n=0}^{\infty} \frac{1}{e \cdot n!} = 1$
When dividing them by $n!$, they all become irrational:
- $\sum_{n=0}^{\infty} \frac{1}{x^n n!} = \sqrt[x]{e}$
- $\sum_{n=1}^{\infty} \frac{1}{(\pi n)^2} \approx 0.11616$ (some hypergeometric monstrosity)
- $\sum_{n=0}^{\infty} \frac{(-1)^n}{x^n} = \frac{1}{\sqrt[x]{e}}$
- $\sum_{n=0}^{\infty} \frac{1}{e \cdot (n!)^2} \approx 0.83861$ (something to do with the modified Bessel function)
I have absolutely no idea on how to prove this, and I haven't found out very many series to test it on yet. How would I go about proving it, or do you know of some counterexamples?
One thing to note: series like $\sum_{n=0}^{\infty} \frac{1}{e \cdot n!}$ and $\sum_{n=0}^{\infty} \frac{1}{2^n n! \sqrt{e}}$ are not valid counterexamples, because without the $n!$, they either don't converge or converge to an irrational number.
Another note: an example where $a_n = 0$ for all $n \geq$ some large $N$ is a trivial solution, so perhaps we shouldn't consider it. Perhaps $a_n$ could be restricted where there cannot exist some $N \geq j$ where $a_n = 0$ for $n \geq N$.