# Intuitive reason that $(1/n)^n$ is maximized for $n = 1/e$

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?"

• Try expanding $$\left(1+ \frac{1}{n} \right)^n = \sum_{k=0}^n \binom{n}{k} \frac{1}{n^k} = \dots.$$ – Dzoooks Apr 5 at 1:46
• 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. – Jorge Fernández Hidalgo Apr 5 at 1:48
• There is some nice treatment of $e$ as the natural exponential base in "What is Mathematics" by Courant and Robbins. – George Dewhirst Apr 5 at 3:29
• 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... – Rahul Apr 5 at 4:02