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I'm not a mathematician nor do I otherwise use it seriously, but every once in a while I just play around with a formula that I generate (usually to look complex or designed to get really big [like a factorial function]), but this time, I think I found something interesting.

I generated the following formula: $$f(x)=\sum_{n=1}^{x}(\prod_{m=1}^{n}(\frac{1}{m}))$$ I plugged into Desmos (originally, I was testing if they had fixed their summation and product functions on their graphing calculator) and I realized that the limit was some constant.

I decided to explore this myself and, to speed up the process, I plugged in $10$ for $x$ and I got something close to the constant: $$\sum_{n=1}^{10}(\prod_{m=1}^{n}(\frac{1}{m}))\approx1.7182\ldots$$ If we plug in 100, instead, we get a similar value: $$\sum_{n=1}^{100}(\prod_{m=1}^{n}(\frac{1}{m}))\approx1.7182\ldots$$ Seeing this, I was curious about mathematical constants. I know that $\varphi=\frac{1+\sqrt{5}}{2}\approx1.6\ldots$ and $\sqrt{3}\approx1.732\ldots$, so this constant must be represented as the following: $$\varphi\leq c \leq\sqrt{3}$$ Is there a constant here, or did I just discover something (probably useless) here? If there is a constant here, what is it used for? If not, then are there potential uses?

Rushabh Mehta
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Curtis Sheppard
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I assume that by $\sum_{n = 1}^{x}$, you actually mean to sum until $\lfloor x\rfloor$, the greatest integer $\le x$. Note that $$\prod_{m = 1}^{n} \frac{1}{m} = \frac{1}{n!}.$$ And thus, in the limit $x \to \infty$, you're just calculating the sum $$\frac{1}{1!} + \frac{1}{2!} + \cdots,$$ which is (famously) equal to $\color{blue}{e} - 1$.

Aryaman Maithani
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    Generally, I think it's more aesthetically pleasing to write $$\sum\limits_{n=0}^\infty\frac1{n!}=e$$That being said, this is an obvious duplicate, so I don't think it should be answered. – Rushabh Mehta May 25 '21 at 20:41
  • Thank you! As I said, I'm not a mathematician and I mostly just toy with it when I'm bored. I suppose that $e-1$ makes sense given that I though $e$ first, but then realised that it started with a 1. Probably should have looked at it a bit more XD – Curtis Sheppard May 25 '21 at 20:52