Example of a concrete irrationality test 
Can you give some example of an irrational number that can be proved to be irrational with this theorem?

Theorem.
Given $a\in \mathbb{R}$, if there exists a sequence of integers $u_n,v_n \rightarrow \infty$ such that:


*

*$a$ is not equal to any $u_n/v_n$.

*The quotien sequences aproximates $a$ in this sense:
$$\lim_{n\rightarrow \infty}v_na-u_n=0 $$
then the number $a$ is irrational.
The proof is in:
https://math.stackexchange.com/q/898420
 A: First, there are some nice examples like 
 $$ e=\sum_{n\ge0}\frac1{n!} $$ or Liouville-like numbers, mentioned in the answers by Wilem2, that can be easily proved to be irrational using the theorem, but for which typically there are simpler irrationality proofs: For $e$, we quickly get that $$0<n!\bigl(e-\sum_{k\le n}\frac1{k!}\bigr)<1$$ for all large enough $n$, so $e$ cannot be rational, because otherwise for $n$ large enough the number in between 0 and 1 would be an integer. For Liouville-like numbers, the many 0s in between consecutive 1s readily imply the number does not have a periodic decimal expansion, so it should be irrational.

Now, one can prove that given any irrational $r\in\mathbb R$ there is a sequence as in the theorem converging to $r$. This is not quite obvious. Instead, I'll refer you to the theory of continued fractions. The two points to make are that if $|r-p/q|<1/q^2$ then $|qr-p|<1/q$ (so, if $q\to\infty$, then $|qr-p|\to0$) and that for any two consecutive convergents to $r$, in fact one satisfies $0<r-p/q<1/q^2$. The convergents to $r$ are the rational numbers that are obtained by iterating the following recursive procedure:
Start with $r$ irrational, let $r_0=\lfloor r\rfloor$. Note that $0<r-r_0<1$, so it is $1/x$ for some $x>1$, and we can repeat the same procedure with $x$ instead of $r$ and to obtain $r_1\in\mathbb N$ and $y>1$ such that $x=r_1+1/y$. For instance, if $r=\sqrt2$, you get $r=1+(\sqrt2-1)=1+\frac1x$, where $x=\frac1{\sqrt2-1}=\sqrt2+1=2+\frac1x$, so $$ \sqrt2=1+\frac1{2+\frac1{2+\frac1\ddots}}, $$
and the convergents to $\sqrt2$ are the rationals that appear along the way, by stopping the procedure after finitely many times, that is, the sequence
$$ 1, 1+\frac12=\frac32,1+\frac1{2+\frac12}=\frac75,1+\frac1{2+\frac1{2+\frac12}}=\frac{17}{12},\dots$$
In general, if $p_0/q_0,p_1/q_1,\dots$ are the convergents to $r$, we have that $(p_n,q_n)=1$ for all $n$ and $$ p_0/q_0<p_2/q_2<\dots<r<\dots<p_3/q_3<p_1/q_1, $$ so $p_0/q_0,p_2/q_2,\dots$ provides a sequence as desired (one can easily check just from the inequalities that $q_n\to\infty$).
That said, it is perhaps a bit unsatisfactory to use continued fractions to illustrate the theorem, because of course if $r$ has an infinite sequence of convergents, then it is irrational (by the Euclidean algorithm!), so we already know $r$ is irrational before we even exhibit the relevant sequence. That said, in certain cases, there are nice recursive relations that allow us to find $p_{n+2},q_{n+2}$ in terms of $p_n,p_{n+1},q_n,q_{n+1}$, so we can easily exhibit sequences as in the theorem, thus witnessing the irrationality of many $r$. (Of course, for some $r$, there is no nice recursive way of obtaining such sequences unless we have access to $r$ to begin with.)
Indeed, for instance for $\sqrt2$, we have $p_0=1=q_0$, $p_1=3,q_1=2$ and, in general $p_{n+2}=2p_{n+1}+p_n$ and $q_{n+2}=2q_{n+1}+q_n$. 
In general, for a given $r$, rather than always using $2$, we use $r_{n+2}$, where the integers $r_0,r_1,r_2,\dots$ are the integer parts obtained through the procedure as described above. For many irrationals $r$ there does not seem to be a nice formula for these numbers $r_n$. But there are nice formulas for $e$ and for any quadratic irrational.
(It is actually an interesting open problem whether there are nice patterns for, say $\root3\of2$ or $\pi$.)
A: Let's proof that $e=\sum_{n=0}^{\infty}\frac{1}{n!}$ is irrational with this method:
Define $v_n=n!$ and $u_n=n!\left( 1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{n!} \right) $
It's direct to see that $u_n$ is an integer, because expanding the product, each term is an integer:
$$u_n=\left( n!+\frac{n!}{1!}+\frac{n!}{2!}+\frac{n!}{3!}+...+\frac{n!}{n!} \right)$$
And also, $u_n$ tends to infinity cause each term is positive and the leading term is $n!$
Let's proof the satements of the theorem:
Clearly, $e\neq\frac{u_n}{v_n}$ because $\left( 1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{n!}+... \right) \neq\left( 1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{n!} \right) $
And for the second statement:
$$x_n=v_ne-u_n=n!\left( \sum_{i=0}^{\infty}{\frac{1}{i!}} \right) -n!\left( \sum_{i=0}^n{\frac{1}{i!}} \right) =n!\left( \sum_{i=n+1}^{\infty}{\frac{1}{i!}} \right) =\sum_{i=n+1}^{\infty}{\frac{n!}{i!}}>0
$$
Now, each term of the last serie is limited by 
$$\frac{n!}{i!}=\frac{1}{(n+1)(n+2)\cdot...\cdot(n+(i-n))}<\frac{1}{(n+1)^{i-n}}$$
and so $x_n=\sum_{i=n+1}^{\infty}{\frac{n!}{i!}}\leq \sum_{i=n+1}^{\infty}{\frac{1}{\left( n+1 \right) ^{i-n}}}=\sum_{k=1}^{\infty}{\frac{1}{\left( n+1 \right) ^k}}=\frac{1}{n+1}\left( \frac{1}{1-\frac{1}{n+1}} \right) =\frac{1}{n}\rightarrow 0
$
So we can apply the theorem to conclude that $e$ is irrational.
A: With the idea of @AndrésE.Caicedo, I will try to proof the irrationality of the Liouville Constant:
$L=\sum_{n=1}^{\infty}10^{-n!}=0.11000100000000000000000100000000000000000...$
With $v_n=10^{n!}$ and $u_n=10^{n!}\sum_{i=1}^{n}10^{-i!}$.
Clearly, $v_n$ is an integer that tends to infinity. 
For $u_n$, note that $u_n=\sum_{i=1}^{n}10^{n!-i!}$ and the power $n!-i!$ is an integers satisfying $n!-i!\geq 0$ for $i\leq n$, so each term of the serie is an integer and also, $u_n$ tends to infinity caused by the leading term $10^{n!-1}\rightarrow \infty$.
To check the hypothesis:


*

*$L \neq\frac{u_n}{v_n}$ because $L$ have more terms in the serie: $L=\sum_{i=1}^{\infty}10^{-i!}\neq \sum_{i=1}^{n}10^{-i!}=\frac{u_n}{v_n}$

*$x_n=v_nL-u_n=...=\sum_{i=n+1}^{\infty}10^{n!-i!}$. It can't be difficult to proof that this last serie is
$$\sum_{i=n+1}^{\infty}10^{n!-i!} <2\cdot 10^{n!-(n+1)!}=2\cdot 10^{-n\cdot n!}\rightarrow 0$$
But because I don't have the argument, here is another:


For $i>n$ one have $n!-i!\leq (i-1)!-i!\,=\, -(i-1)\cdot(i-1)! \leq -(i-1)$
So $n!-i! \leq -(i-1)$ and hence:
$$x_n=\sum_{i=n+1}^{\infty}10^{n!-i!}\leq \sum_{i=n+1}^{\infty}10^{-(i-1)}=\frac{1}{10^n}\sum_{i=0}^{\infty}10^{-i}=\frac{1}{10^n}\frac{1}{1-\frac{1}{10}}=\frac{1}{10^n}\frac{10}{9}\longrightarrow_{n \rightarrow\infty} 0$$
By the theorem, L is irrational.
