LCM of time periods of a Fourier series

I was thinking something like this - in a Fourier series the period of the $n$th term where $p$ is the period of the function being considered would be at $x = \dfrac{2p}{n}$ . If we were to take the lcms of all these time periods, we'd get to the point where the function would repeat again. Which would be $p$

So Shouldn't the limit as $n$ goes to infinity infinity of $\dfrac{(2p)^n}{n!}$ equal $p$?

• What does this have to do with LCM? Fourier series? What is $p$? A lot of context would help. – Thomas Andrews May 30 '17 at 18:40
• @ThomasAndrews, I made a mistake there, I edited the question, hope it makes sense now – Vrisk May 30 '17 at 18:57
• Certainly the terms in a Fourier series are periodic functions with a smallest common period. In some cases the highest-frequency term is $0$ so the smallest common period, rather than being $p,$ is $p/2$ or $p/3,$ etc. However, it is unclear to me how $(2p)^n/(n!)$ is involved. Perhaps you can explain that within the question. At any rate, $\displaystyle \lim_{n\to\infty} \frac {(2p)^n}{n!} =0. \qquad$ – Michael Hardy May 30 '17 at 19:35
• @MichaelHardy, if none of the terms went to zero the periods of the respective terms from n = 1 would be $\frac{2p}{1}, \frac{2p}{2}, \frac{2p}{3}..$ All the way till infinity. I just multipled them together, you know, liked you'd do for LCMs. – Vrisk May 30 '17 at 19:48
• There's a difference between the product of several numbers and the LCM of those numbers. – Greg Martin May 30 '17 at 20:46

The LCM of $24$ and $36$ is $72,$ but $24\times 36 = 864.$ Simply multiplying doesn't generally find the LCM. Especially when you have a pattern such as that in $2p,\ 2p/2,\ 2p/3,\ 2p/4,\ 2p/5,\ \ldots$ In that case, $2p$ is the LCM.