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$$\lim_{n \to\infty }\sum_{k=0}^n {1 \over k!}=e$$ Which is the proof for the limit listed above?I know its a famous limit,but i cant figure out how to solve it(I'm a 12th grader).

Razvan Paraschiv
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    Not really an answer but: some people define $e$ as that series, in which case all you must do is show the series converges. Otherwise you have another definition of $e$, perhaps as $e=\lim_{n\to \infty}(1+1/n)^n$, in which case you have to show both limits converge and are actually equal. – Nap D. Lover Mar 29 '16 at 17:41
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    More concisely: what is your definition of $e$? – Clarinetist Mar 29 '16 at 17:42
  • the euler constant – Razvan Paraschiv Mar 29 '16 at 17:45
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    Duplicate of http://math.stackexchange.com/q/1054175/215011; see also http://math.stackexchange.com/questions/54448/combinatorial-proof and http://math.stackexchange.com/questions/69806/prove-the-definitions-of-e-to-be-equivalent – grand_chat Mar 29 '16 at 17:47
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    Okay, this might be hard to understand since you (quite likely) haven't been through university-level mathematics yet, and I could probably write a novel about this. When you ask for a "proof," you have to start from something and prove theorems from that something. "The Euler constant" is not a mathematical definition. For example, you could define $e$ as the number $e$ such that $\ln(e) = 1$ (but then I'd have to ask, how do you define $\ln$?). Or you could define $e$ as that limit that @LoveTooNap29 stated. – Clarinetist Mar 29 '16 at 17:49
  • The point is, you need to give us a mathematical definition to work from before we can even do anything. – Clarinetist Mar 29 '16 at 17:49
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    Hi Razvan, Following this web site's custom, I posted (right now) a complete answer to your question IN THE OLD THREAD, referenced at the top of this page. Please take a look there and let me know if you need more help. –  Mar 30 '16 at 01:59

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Use the classical definition of e, $e = \lim_{n \to \infty} (1 + \frac{1}{n})^n $ and the binomial theorem. \begin{align} e &= \lim_{n \to \infty} (1 + \frac{1}{n})^n\\ &= \lim_{n \to \infty} 1 + n \frac{1}{n} + \frac{n(n-1)}{2!} \frac{1}{n^2} + \frac{n(n-1)(n-2)}{3!}\frac{1}{n^3} + \frac{n(n-1)(n-2)(n-3)}{4!}\frac{1}{n^4} + \cdots\\ \end{align} Using the linearity of the limit and the fact that \begin{align} \lim_{n \to \infty} \frac{n(n-1)(n-2)\cdots (n - k + 1)}{k! n^k} &= \lim_{n \to \infty}\frac{n}{n}\cdot \frac{n - 1}{n} \cdot \frac{n - 2}{n} \cdots \frac{n - k + 1}{n} \frac{1}{k!}\\ &= \frac{1}{k!} \end{align} Using this result with each term of the series yields \begin{align} e &= \lim_{n \to \infty} 1 + n \frac{1}{n} + \frac{n(n-1)}{2!} \frac{1}{n^2} + \frac{n(n-1)(n-2)}{3!}\frac{1}{n^3} + \frac{n(n-1)(n-2)(n-3)}{4!}\frac{1}{n^4} + \cdots\\ &= 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \cdots\\ &= \sum^\infty_{k = 0} \frac{1}{k!}\\ &= \lim_{n \to \infty} \sum^n_{k = 0} \frac{1}{k!} \end{align}

Jeevan Devaranjan
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    This is an excellent 18th century "proof" and should be quite interesting for a 12th grader. When you escape (as you must) from the 12th grade you will question whether these operations can be justified. – B. S. Thomson Mar 29 '16 at 18:12