### Series for $$e$$

One way of defining the number $$e$$ is

$\lim_{n \to \infty} \Big(1+\dfrac{1}{n}\Big)^n.$

The binomial theorem gives

$\Big(1+\dfrac{1}{n}\Big)^n = \dbinom{n}{0}+ \dbinom{n}{1}\dfrac{1}{n} + \dbinom{n}{2}\Big(\dfrac{1}{n}\Big)^2 + \dots + \dbinom{n}{r}\Big(\dfrac{1}{n}\Big)^r + \dots + \dbinom{n}{n-1}\Big(\dfrac{1}{n}\Big)^{n-1} + \dbinom{n}{n}\Big(\dfrac{1}{n}\Big)^n.$

Using the result that

$\lim_{n \to \infty} \dbinom{n}{k}\Big(\dfrac{1}{n}\Big)^k = \dfrac{1}{k!},$

we obtain the result

$e = \lim_{n \to \infty} \Big(1+\dfrac{1}{n}\Big)^n = \dfrac{1}{0!} + \dfrac{1}{1!}+ \dfrac{1}{2!}+\dfrac{1}{3!} + \dotsb.$

Formally, this argument involves some difficult limiting processes.

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