# Proof that $(1+\frac{1}{n})^n$ can't converge to a rational number

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

• This goes back to Fourier. – metamorphy Aug 30 at 9:57
• @metamorphy, what do you mean? – Pspl Aug 30 at 10:04
• 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 $0<x<1$. – metamorphy Aug 30 at 10:22
• @metamorphy, I got it! So simple! Thanks... – Pspl Aug 30 at 13:21