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Why is $\lim\limits_{n\to\infty}(1+\frac1n)^n=e$?

I think it involves $\sum\limits_{n=0}^\infty\frac1{k!}=e$ but not sure how to get from one to the other.

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    $\begingroup$ At first we need how you did define $e$, because often $e$ is defined as this limit $\endgroup$ Aug 25, 2013 at 18:18
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    $\begingroup$ well before the taylor expansion I knew e as ln(1), the limit is the third definition of e that I encountered. EDIT: which is wrong, oops $\endgroup$
    – Leo
    Aug 25, 2013 at 18:19
  • $\begingroup$ $e$ is not $\ln(1)$. $\ln(1) = 0$. You probably meant $e$ is the unique real number such that $\ln(e) = 1$. But for that to be meaningful it is useful to know how you define the natural log. $\endgroup$ Aug 25, 2013 at 18:20
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    $\begingroup$ @Leo Search the site. There is also this question with several answers, and most probably many more. $\endgroup$ Aug 25, 2013 at 18:28
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    $\begingroup$ Euler original definition of ${\rm e}^{x}$ was to look for a function ${\rm f}\left(x\right)$ which satisfies ${\rm f}'\left(x\right) = {\rm f}\left(x\right), \forall\ x$. Then, ${\rm f}^{\left(n\right)}\left(0\right) = {\rm f}\left(0\right)$. Then, you get ${\rm f}\left(x\right) = {\rm f}\left(0\right)\sum_{n = 1}^{\infty}x^{n}/n!$. $\endgroup$ Sep 5, 2013 at 21:37

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Have you tried expanding by the binomial theorem? ;-) $$(1/n+1)^n=\sum_{k=0}^n\binom{n}k\frac1{n^k}=\sum_{k=0}^n\frac{n!}{k!(n-k)!}\frac1{n^k}$$then as $n\to\infty$ we find $n!/(n^k (n-k)!)\to1$ hence we have:$$\lim_{n\to\infty}\sum_{k=0}^n\frac{n!}{k!(n-k)!\cdot n^k}=\sum_{k=0}^\infty\frac1{k!}\equiv e$$

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  • $\begingroup$ Why does the limit is $1$ ? $\endgroup$
    – Belgi
    Aug 25, 2013 at 19:35
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    $\begingroup$ $\displaystyle\frac{n!}{(n-k)!n^k}=\frac{n(n-1)\cdots (n-k+1)}{n^k}\to 1$.$ $\endgroup$
    – Grobber
    Aug 25, 2013 at 19:54
  • $\begingroup$ oops -- did not see that earlier. @Belgi $n!/(n-k)!=n(n-1)\cdots(n-k+1)$ and so $n!/((n-k)!n^k)=n/n\cdot(n-1)/n\cdots (n-k+1)/n$ $\endgroup$
    – obataku
    Aug 25, 2013 at 19:56
  • $\begingroup$ observe $(n-r)/n=1-r/n$ hence as $n\to\infty$ each of our product terms tends to $1$ $\endgroup$
    – obataku
    Aug 25, 2013 at 19:57
  • $\begingroup$ @Grobber - thanks! $\endgroup$
    – Belgi
    Aug 25, 2013 at 20:01

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