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Why is the upper bound of Euler's number $=3$?

I found that the upper bound of Euler's number can be evaluated as $1 +$ the lower bound (in other word the integer part). I do not understand why is it though.

user729424
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Min Lee
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    Presumably you mean the constant $e = 2.718281828\ldots$ (this is not always what is meant when people say "Euler's number"). There is no the upper bound. There are many upper bounds, like $3$ or $4$ or $\pi$ or $1000$ or $2.718281829$. However, depending on how you define this number, $3$ might be a comparatively easy upper bound to prove. Is that what you're asking? In that case, we need to know how you define this number. – Arthur May 01 '20 at 20:57
  • I think so? It was one of my questions that I looked it up on the wolframalpha but have a no idea how to prove it or it is saying that the lowest integer that must be larger than the Euler's constant. If that's the case then it makes sense though – Min Lee May 01 '20 at 20:58
  • Ummm.... an upper bound on Euler's number is also $47$. Why do you focus on $3$? And what is the underlying point of this question? Are you just flailing here? Why not ask the same question about $\pi$ and $4$, say? Or any of a dozen other mathematical constants?? Euler's constant has some value. What are you really asking? – David G. Stork May 01 '20 at 21:02
  • I assume the question was asking any integer that is the upper bound of e. However, the question also referring that the upper bound of E can evaluate by using the comparison test between "sum 1/n!" and "sum 1/(n(n-1))".... Now I'm really confused – Min Lee May 01 '20 at 21:04
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    Are you referring to the ceiling function? Note that the floor and ceiling of a noninteger number always differ by $1$. – Jean-Claude Arbaut May 01 '20 at 21:05
  • I don't think so? My professor haven't used that term throughout the semester. – Min Lee May 01 '20 at 21:06
  • Do you want to prove this? $$
    \lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n < 3 $$     – Shiv Tavker May 01 '20 at 21:08
  • I deleted the tag [tag:Eulers-constant]; I think that's a different number from the one you're asking about – J. W. Tanner May 01 '20 at 21:09
  • It's easy to prove $e<3$ by proving instead that $\exp(-1)>1/3$ (alternating series: it's immediate that $\exp(-1)>1-1+1/2-1/6=1/3$). – Jean-Claude Arbaut May 01 '20 at 21:09
  • ahhh I think I got what you meant by that thank you so much! – Min Lee May 01 '20 at 21:17
  • An answer I gave here provides an argument that $e$ must be between $2$ and $3$. To make this rigorous would require some effort, but if you're looking for intuition, it may help. – Will Orrick May 02 '20 at 03:00

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