I am guessing that this is the exact same solution the OP (TheBigOne) got, but I am putting my answer here, so that the thread is answered. Even though this proof requires some non-elementary knowledge (Bertrand's Postulate), it is still a proof. (Although I disagree that Bertrand's Postulate is not elementary. The proof I know is quite elementary, and as i707107 said, Bertrand's Postulate is very widely known.) The solution also utilizes the fact that there are infinitely many prime natural numbers congruent to $2$ modulo $3$. (This fact has an elementary proof.)
First, let $L_k:=\text{lcm}(1,2,\ldots,k)$ for $k=1,2,3,\ldots$. Note that $L_1=1$ and $L_k\geq k(k-1)$ for $k\geq 2$. That is, for each integer $N>0$, $$\sum_{k=1}^N\,\frac{1}{L_k}\leq 1+\sum_{k=2}^N\,\frac{1}{k(k-1)}\leq 1+\sum_{k=2}^\infty\,\frac{1}{k(k-1)}=2\,.$$
This means $S:=\sum\limits_{k=1}^\infty\,\dfrac{1}{L_k}$ exists and is a positive real number less than $2$. (WolframAlpha states that $S\approx 1.77111$.)
We argue by contradiction. Suppose contrary that $S=\dfrac{a}{b}$ for some relatively prime positive integers $a$ and $b$. Let $p_1,p_2,p_3,\ldots$ be the increasing sequence of all prime natural numbers greater than $b$. Using Bertrand's Postulate, $$p_r<p_{r+1}<2\,p_r$$ for all $r=1,2,3,\ldots$. Thus, $$p_{r+1}-p_r\leq p_r-1$$ for each positive integer $r$. Note that, for infinitely many positive integers $r$, it holds that $p_r\equiv 2\pmod{3}$. Therefore, the equality $p_{r+1}-p_r=p_r-1$ does not happen (or else, $p_{r+1}\equiv 0\pmod{3}$). Hence, $p_{r+1}-p_r<p_r-1$ for infinitely many such $r$.
Now,
$$\begin{align}
L_{p_1-1}\,\left(S-\sum_{k=1}^{p_1-1}\,\frac{1}{L_k}\right)&\leq \sum_{r=1}^\infty\,\sum_{k=p_r}^{p_{r+1}-1}\,\frac{L_{p_1-1}}{L_k}\leq \sum_{r=1}^\infty\,\sum_{k=p_r}^{p_{r+1}-1}\,\frac{1}{p_1p_2\cdots p_r}
\\
&=\sum_{r=1}^\infty\,\frac{p_{r+1}-p_r}{p_1p_2\cdots p_r}<\sum_{r=1}^\infty\,\frac{p_{r}-1}{p_1p_2\ldots p_r}
\\
&=\left(1-\frac{1}{p_1}\right)+\left(\frac{1}{p_1}-\frac{1}{p_1p_2}\right)+\left(\frac{1}{p_1p_2}-\frac{1}{p_1p_2p_3}\right)+\ldots
\\
&=1\,.
\end{align}$$
This is a contradiction, as $b\mid L_{p-1}$ and $S>\sum\limits_{k=1}^{p_1-1}\,\frac{1}{L_k}$, which means $L_{p-1}\,\left(S-\sum\limits_{k=1}^{p_1-1}\,\frac{1}{L_k}\right)$ is a positive integer. Therefore, $S$ cannot be a rational number.
Best Answer
I'm not sure if it's what you're looking for, but the proof of Bertrand's postulate is somewhat related to the proof of Chebyshev's theorem, which says that $\pi(x) = \Theta(x/\log x)$; both proofs rely on careful estimation of the central binomial coefficients $\binom{2n}{n}$.
To prove Chebyshev's theorem, one can start by finding the estimate
$$ \frac{2^{2n}}{2n+1} < \binom{2n}{n} < 2^{2n}, \tag{1} $$
then using it to conclude that $\vartheta(x)/x < 4\log 2$ and hence
$$ \limsup_{x \to \infty} \frac{\pi(x)}{x/\log x} = \limsup_{x \to \infty} \frac{\vartheta(x)}{x} \leq 4 \log 2, $$
and $\psi(x)/x > (x-2)\log 2/x - \log(x+1)/x$ and hence
$$ \liminf_{x \to \infty} \frac{\pi(x)}{x/\log x} = \liminf_{x \to \infty} \frac{\psi(x)}{x} \geq \log 2, $$
where $\vartheta$ and $\psi$ are Chebyshev's functions.
To prove Bertrand's postulate, one can improve the estimate in $(1)$ to
$$ \frac{2^{2n}}{2\sqrt{n}} < \binom{2n}{n} < \frac{2^{2n}}{\sqrt{2 n}}, \tag{2} $$
which can be used to show that
$$ \vartheta(2n) - \vartheta(n) > \left(\frac{2n}{3}-1\right)\log 2-\left(\frac{\sqrt{2n}+1}{2}\right)\log 2 > 0$$
for $n \geq 2^6$. This is exactly Bertrand's postulate for $n \geq 2^6$. The remaining cases are checked by hand.
My reference for these proofs is Chandrasekharan's Introduction to Analytic Number Theory.