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$\def\e{{\rm e}}$ I recently had the task to explain the proof that $\e$ is irrational as a presentation to my classmates. To prepare this presentation, the teacher gave me a script with a proof that uses an estimation of the series $b_n = \sum_{k=0}^n1/k!\ $ to show that there are no $p,q\in\mathbb N,\ $ such that $\e = p/q.$

Because we defined $\e$ as the limit of $a_n = \big(1+\frac1n\!\big)^n\ $ I had to include the rather long proof that $\lim\limits_{n\to\infty}a_n-b_n=0,\ $ rendering the whole proof quite long. Is there a “shorter” proof, given that $\e$ is defined as the limit of $a_n$?

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    Perhaps you should shorten your proof of $\lim a_n = \lim b_n$. Denote $E = \lim b_n$. From the simple $a_n \leq E$ follows $e \leq E$. On the other hand, it is easy to see that for each $n$ we have $e \geq b_n$, and so $e \geq E$.2011-09-21
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    (Tangentially related to the question, but I am not sure if this will be ultimately useful to the OP.) Can anyone suggest the asymptotics of the error $e - (1+\frac{1}{n})^n$?2011-09-21
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    @Yuval That is actually how we did this. The problem is, that the students in my class don't like proofs containing too many “It is easy to see that...”s. The proof using this strategie filled one A4 page in the script and took 20 minutes for me to explain. (Although it is quite easy if you think about it).2011-09-21
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    @anon Fixed the first one. The second proof is not difficult, but takes too much time. That's all.2011-09-21
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    @Srivatsan, $e/(2n)+o(1/n)$. Why do you ask?2011-09-21
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    @SrivatsanNarayanan Thank you! By the way, is it clear, which proof I refer to?2011-09-21
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    @FUZxxl Do you show $(1+1/n)^n$ has a limit without using $\exp$ and $\ln$?2011-09-21
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    @francis-jamet We showed that the limit exists just as Srivatsan described.2011-09-21
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    @francis (Oops, I removed the comment.) To show existence of limit, it is enough to show the sequence is [monotonically increasing and bounded](http://planetmath.org/encyclopedia/ConvergenceOfTheSequence11nn.html).2011-09-21
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    Not what you want but see also [How Euler Did It](http://www.maa.org/editorial/euler/How%20Euler%20Did%20It%2028%20e%20is%20irrational.pdf).2011-09-21
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    @lhf Thank you for the interesting link.2011-09-21
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    @Srivatsan Narayan: $(1+1/n)^n = e (1-1/(2n) + O(1/n^2))$.2011-09-21
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    @SrivatsanNarayanan Thanks for the link.2011-09-22

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(Too long for a comment.) This is probably the way you did it, but it doesn’t seem very long to me; where did it bog down?

From the binomial theorem we have

$$a_n = \left(1 + \frac1n \right)^n = \sum\limits_{k=0}^n \binom{n}{k}n^{-k} = \sum\limits_{k=0}^n \frac{n^{\underline k}}{k! n^k} \le \sum\limits_{k=0}^n \frac{1}{k!} = b_n.$$

For the other direction let $m>1$ be an integer; then

$$a_{mn} = \sum\limits_{k=0}^{mn}\frac{(mn)^{\underline{k}}}{k!(mn)^k} \ge \sum\limits_{k=0}^n\frac{(mn)^{\underline{k}}}{k!(mn)^k} \ge \left(\frac{(mn)^{\underline n}}{(mn)^n} \right) b_n \ge \left(\frac{m-1}{m} \right)^n b_n,$$

and $\displaystyle \lim\limits_{m\to\infty} \left(\frac{m-1}{m}\right)^n = 1$, so $b_n \le e$.

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    The last step in the second line is not completely clear to me. Could you please elaborate why this holds?2011-09-22
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    @FUZxxl: $\frac{(mn)^{\underline n}}{(mn)^n}$ is the product of $n$ factors; the smallest is $(mn-n+1)/(mn)$, so they’re all $> (m-1)/m$.2011-09-22
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    Thanks for the explanation.2011-09-22