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Consider the function $f(n) = \Big( \dfrac{1}{n} \Big)^n$.

By setting $f'(n) = 0$, we find that the maximum of $f(n)$ occurs at $n = \dfrac{1}{e}$. Going through the calculations, there doesn't seem to be any "reason" for the appearance of $e$.

However, we do know that the definition of $e$ is given by $$ e:= \lim_{n \to \infty} \Big( 1 + \dfrac{1}{n} \Big)^n. $$

Is there any chance the similar forms of the definition of $e$ and $f(n)$ provide an "explanation" for $n = \dfrac{1}{e}$ maximizing $f(n)$? Is there any way we could just look at $f(n)$ and intuitively "see" that it has to have its maximum at $n = \dfrac{1}{e}$? Or is this just a mathematical "coincidence?"

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  • $\begingroup$ Try expanding $$\left(1+ \frac{1}{n} \right)^n = \sum_{k=0}^n \binom{n}{k} \frac{1}{n^k} = \dots.$$ $\endgroup$ – Dzoooks Apr 5 at 1:46
  • $\begingroup$ the intuition is that 2.7 is approximately e so your intuition tells you it must be e when you try to maximize it naively with a discretization. $\endgroup$ – Jorge Fernández Hidalgo Apr 5 at 1:48
  • $\begingroup$ There is some nice treatment of $e$ as the natural exponential base in "What is Mathematics" by Courant and Robbins. $\endgroup$ – George Dewhirst Apr 5 at 3:29
  • $\begingroup$ To me it's not even intuitive that the maximum of $(1/n)^n$ should occur at nonzero $n$. After all, as $n\to0$ the quantity being exponentiated goes to infinity... $\endgroup$ – Rahul Apr 5 at 4:02

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