Easy proof, that $\rm e\notin \mathbb Q$

Question

$\def\e{{\rm e}}$ 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 $a_n$?

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 $e \geq E$. — Yuval Filmus, Sep 21 '11 at 20:27
(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 $e - (1+\frac{1}{n})^n$? — Srivatsan, Sep 21 '11 at 20:30
@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). — FUZxxl, Sep 21 '11 at 20:32
@anon Fixed the first one. The second proof is not difficult, but takes too much time. That's all. — FUZxxl, Sep 21 '11 at 20:34
@SrivatsanNarayanan Thank you! By the way, is it clear, which proof I refer to? — FUZxxl, Sep 21 '11 at 20:38
@FUZxxl Do you show $(1+1/n)^n$ has a limit without using $\exp$ and $\ln$? — francis-jamet, Sep 21 '11 at 20:44
@francis-jamet We showed that the limit exists just as Srivatsan described. — FUZxxl, Sep 21 '11 at 20:49
@francis (Oops, I removed the comment.) To show existence of limit, it is enough to show the sequence is monotonically increasing and bounded. — Srivatsan, Sep 21 '11 at 20:50

score 4 · Accepted Answer · answered Sep 21 '11 at 21:34

4

(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 $b_n \le e$.

answered Sep 21 '11 at 21:34

Brian M. Scott

616,228

The last step in the second line is not completely clear to me. Could you please elaborate why this holds? – FUZxxl Sep 22 '11 at 08:20
@FUZxxl: $\frac{(mn)^{\underline n}}{(mn)^n}$ is the product of $n$ factors; the smallest is $(mn-n+1)/(mn)$, so they’re all $> (m-1)/m$. – Brian M. Scott Sep 22 '11 at 10:22
Thanks for the explanation. – FUZxxl Sep 22 '11 at 10:42

Easy proof, that $\rm e\notin \mathbb Q$

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