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

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.

## marked as duplicate by RRL real-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 25 at 22:54
• 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 $$e$$.