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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.

Please help me clear this doubt. Thanks in advance.

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marked as duplicate by RRL real-analysis May 25 at 22:54

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    $\begingroup$ 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) $\endgroup$ – Mark May 25 at 21:26
  • $\begingroup$ A limit of finite expansions could be an infinite expansion $\endgroup$ – J. W. Tanner May 25 at 21:43
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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 $e$.

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