# Proving the limit $\lim_{n\rightarrow\infty}(1+\frac{1}{n})^n=e$ [duplicate]

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I want to prove that $$\lim_{n\rightarrow\infty}(1+\frac{1}{n})^n=e$$

There is a solution of the sum provided in my text book. There the expansion of $$(1+\frac{1}{n})^n$$ is like below:

$$(1+\frac{1}{n})^n= 1+{n\choose 1}\frac{1}{n}+{n\choose 2}\frac{1}{n^2}+{n\choose 3}\frac{1}{n^3}+.....$$

But how can they write it as an infinite expansion? Does not the expansion end in the term $$\frac{1}{n^n}$$? But then we cannot prove that its limit is $$e$$, since $$e$$ has an infinite expansion.

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• What is your definition of the number $e$? Because in some books $e$ is just defined as this limit. (and I prefer this definition by the way) – Mark May 25 at 21:26
Note that $$\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n=\exp\lim_{n\to\infty}n\ln\left(1+\frac{1}{n}\right),$$so the problem is equivalent to proving $$\lim_{x\to0^+}\frac{\ln\left(1+x\right)}{x} = 1,$$which you can do e.g. by L'Hôpital's rule. But it all depends on which definition of $$e$$, $$\exp x$$ or $$\ln x$$ you start from. As for your textbook's approach, use the fact that $$k>n\implies\binom{n}{k}=0$$ to write your limit as$$\lim_{n\to\infty}\sum_{k=0}^\infty\frac{\binom{n}{k}}{n^k}=\sum_{k\ge 0}\lim_{n\to\infty}\frac{\binom{n}{k}}{n^k}=\sum_{k\ge 0}\frac{1}{k!},$$which is one definition of $$e$$.