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One of my colleagues challenged me (and his students) with the following:

"Assume you don't know that $\lim_{n\to +\infty}(1+\frac{1}{n})^n=e$.

Prove the sequence $u_n=(1+\frac{1}{n})^n$ converges to a real non rational value."

I know how to do part of this. Actually, the proof of $u_n$ convergence is quite familiar (using the monotonic and bounded real sequences theorem). However, I can't prove the non rationality of the limit.

Considering that $\lim_{n\to +\infty}(1+\frac{1}{n})^n=\ell$, I tried to assume that $\ell$ can be written by $\frac{p}{q}, p,q\in \mathbb N$ with $p$ and $q$ being co-prime, but I reached a point I can't proceed.

Here's my calculations: Let $u_n:\mathbb N\to \mathbb R, n\mapsto (1+\frac{1}{n})^n$.

STEP 1: Prove $u_n$ is monotonic

If $u_n=(1+\frac{1}{n})^n$ then $u_{n+1}=(1+\frac{1}{n-1})^{n+1}$.

So $\frac{u_{n+1}}{u_n}=\frac{\left(1+\frac{1}{n+1}\right)^{n+1}}{\left(1+\frac{1}{n}\right)^n}=\left(1-\frac{1}{(n+1)^2}\right)^n\left(1+\frac{1}{n+1}\right)$.

By Bernoulli's inequality, as $-\frac{1}{(n+1)^2}>-1$, $\left(1-\frac{1}{(n+1)^2}\right)^n\geq 1-\frac{n}{(n+1)^2}$.

So $\left(1-\frac{1}{(n+1)^2}\right)^n\left(1+\frac{1}{n+1}\right)\geq \left(1-\frac{n}{(n+1)^2}\right) \left(1+\frac{1}{n+1}\right)=1+\frac{1}{(n+1)^3}$.

As $1+\frac{1}{(n+1)^3}>1$, $\frac{u_{n+1}}{u_n}>1$, so $u_n$ is monotonically increasing. QED

STEP 2: Prove $u_n$ is bounded

As $u_n$ is monotonically increasing, by Step 1, $u_1=2$ is a lower bound for the set of $u_n$ terms.

On the other side, as $n(n-1)...(n-k+1)\leq n^3$ and $\frac{1}{k\,!}\leq \frac{1}{2^{k-1}}$, for all $k\in \mathbb N$, then

as $u_n = \sum_{k=0}^n \frac{\binom{n}{k}}{n^k}=1+\sum_{k=1}^n \frac{n(n-1)...(n-k+1)}{k\,!\,n^k}$, $u_n\leq 1+\sum_{k=1}^n \frac{1}{2^{k-1}}=3-(\frac{1}{2})^{n-1}<3$.

So $3$ is an upper bound for the set of $u_n$ terms.

Once bounded from below and bounded from above, the sequence $u_n$ is bounded. QED

STEP 3: Prove $u_n$ is convergent

Once monotonic (Step 1) and bounded (Step 2), by the monotonic and bounded real sequences theorem, the sequence $u_n$ is convergent, i.e. $\exists\,\ell\in\mathbb R:\lim_{n\to +\infty}(1+\frac{1}{n})^n=\ell$. By Step 1 and Step 2, we also conclude that $2<\ell<3$. QED

STEP 4: Prove $\ell$ can't be rational

As we saw in Step 2, $u_n =\sum_{k=0}^n \frac{n(n-1)...(n-k+1)}{k\,!\,n^k}$.

Then $\ell=\lim_{n\to +\infty}\sum_{k=0}^n \frac{n(n-1)...(n-k+1)}{k\,!\,n^k}$.

As $\lim_{n\to +\infty}\sum_{k=0}^n \frac{n(n-1)...(n-k+1)}{n^k}=1$, then $\ell=\sum_{n=0}^\infty \frac{1}{n\,!}$ (*).

Let's now suppose $\sum_{n=0}^\infty \frac{1}{n\,!}$ can be written by $\frac{p}{q}, p,q\in \mathbb N$ with $p$ and $q$ being co-prime... (and I can't proceed from here).

(*) Please note that I obviously know that $e=\sum_{n=0}^\infty \frac{1}{n\,!}$. However, the challenge asks to assume we don't know that $\lim_{n\to +\infty}(1+\frac{1}{n})^n=e$. Also, I just wrote all the steps, because it could be useful to complete the proof.

I did some research and one suggestion I saw somewhere is to assume, for some $x\in\mathbb R$, let $x=q\,!\left(\frac{p}{q}-\sum_{n=0}^q \frac{1}{n\,!}\right)$. Then $x=p(q-1)\,!-\sum_{n=0}^q q(q-1)...(q-n+1)$.

What's the next step? Is this the right approach?

Thanks

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    $\begingroup$ This goes back to Fourier. $\endgroup$ – metamorphy Aug 30 at 9:57
  • $\begingroup$ @metamorphy, what do you mean? $\endgroup$ – Pspl Aug 30 at 10:04
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    $\begingroup$ I mean, this is his way of proving the irrationality of $\sum_{n\geq 0}1/n!$. See the link. The proof proceeds by showing that your $x$ is integer (as you can see yourself already) and, on the other hand, that $0<x<1$. $\endgroup$ – metamorphy Aug 30 at 10:22
  • $\begingroup$ @metamorphy, I got it! So simple! Thanks... $\endgroup$ – Pspl Aug 30 at 13:21

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