# Why does limit: $\lim_{x\to\infty}\prod\limits^{\lfloor x\rfloor}_{n=1}\frac{\lceil x\rceil}{\lfloor x\rfloor}=e$?

This is for all $$x\in\mathbb{R}$$ and $$x\notin\mathbb{Z}$$ because it equals $$1$$ for all positive integers.

I was just messing around with floor and ceiling functions in Desmos when I came upon this. I have not been able to prove this and it is at the moment only an observation. To me, this makes absolutely no sense and I have no clue how to begin solving this. When typing it into WolframAlpha, it just gave me an answer of $$1$$ because it did not exclude integers. It also gave me a partial product formula of $$\left(\frac{\lceil x\rceil}{\lfloor x\rfloor}\right)^n$$ but this seems like it would not work for $$n=\infty$$. This is what it looks like in Desmos: Is anyone able to explain this to me? How do you even prove this? Thanks for the help in advance!

• @aschepler The OP's title doesn't match their question Jun 9, 2023 at 18:04
• OP's attached picture clearly says what the author means, just the title is not well written. @MartinR Jun 9, 2023 at 18:09
• Thanks for fixing the question. I am not quite sure why mine was incorrect, but I appreciate it nonetheless. Jun 9, 2023 at 18:10
• It's clear to me what OP means and what MathFail said. OP just has confused their Desmos graph with their title Jun 9, 2023 at 18:10
• @FShrike Thanks. Would you mind explaining what was different between the original question and what I meant. I thought because I used the same equation that I used in the Desmos, it would be the same. Sorry if this doesn't make sense, I am just quite confused. Jun 9, 2023 at 18:13

Let $$k=\lfloor x\rfloor$$

$$\prod\limits^{\lfloor x\rfloor}_{n=1}\frac{\lceil x\rceil}{\lfloor x\rfloor}=\prod\limits^{k}_{n=1}\frac{k+1}{k}=\left( 1+\frac1k\right)^k$$

therefore,

$$\lim_{x\to\infty\land x\notin\mathbb N}\prod\limits^{\lfloor x\rfloor}_{n=1}\frac{\lceil x\rceil}{\lfloor x\rfloor}=\lim_{k\to\infty}\left( 1+\frac1k\right)^k=e$$

• What is $n$? This makes as little sense to me as the OP did. Jun 9, 2023 at 18:04
• @TedShifrin the multiplicand just doesn't depend on $n$ so we get a product of constants Jun 9, 2023 at 18:05
• Note OP is not dependent on $n$, that's why it gets $e$ Jun 9, 2023 at 18:05
• Note that when $x$ is an integer, this evaluates to $1$, not $e$, so the limit isn’t actually defined.
– Eric
Jun 9, 2023 at 21:17
• Read the first sentence of OP, the author said that. @Eric Jun 9, 2023 at 22:18