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Show that the infinite sum $S$ defined by -$$S=sum_{i=1}^infty frac{1}{operatorname{lcm}(1,2,...,i)}$$ is an irrational number.

I found this question while reading 'Mathematical Gems' by Ross Honsberger. After pondering over it for nearly an hour, I was able to prove it by using Bertrand's postulate which states that there is a prime between n and 2n for every natural number n>1.

This question was solved by Lajos Pósa when he was just 12 years old. Is there any elementary proof that does not use Bertrand's postulate or any complicated theorem?

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 $L_kgeq k(k-1)$ for $kgeq 2$. 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 $S:=sumlimits_{k=1}^infty,dfrac{1}{L_k}$ exists and is a positive real number less than $2$. (WolframAlpha states that $Sapprox 1.77111$.)

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, $$p_{r+1}-p_rleq p_r-1$$ for each positive integer $r$. Note that, for infinitely many positive integers $r$, it holds that $p_requiv 2pmod{3}$. Therefore, the equality $p_{r+1}-p_r=p_r-1$ does not happen (or else, $p_{r+1}equiv 0pmod{3}$). Hence, $p_{r+1}-p_r<p_r-1$ for infinitely many such $r$.

Now,
$$begin{align}
L_{p_1-1},left(S-sum_{k=1}^{p_1-1},frac{1}{L_k}right)&leq sum_{r=1}^infty,sum_{k=p_r}^{p_{r+1}-1},frac{L_{p_1-1}}{L_k}leq sum_{r=1}^infty,sum_{k=p_r}^{p_{r+1}-1},frac{1}{p_1p_2cdots p_r}
\
&=sum_{r=1}^infty,frac{p_{r+1}-p_r}{p_1p_2cdots p_r}<sum_{r=1}^infty,frac{p_{r}-1}{p_1p_2ldots p_r}
\
&=left(1-frac{1}{p_1}right)+left(frac{1}{p_1}-frac{1}{p_1p_2}right)+left(frac{1}{p_1p_2}-frac{1}{p_1p_2p_3}right)+ldots
\
&=1,.
end{align}$$
This is a contradiction, as $bmid L_{p-1}$ and $S>sumlimits_{k=1}^{p_1-1},frac{1}{L_k}$, which means $L_{p-1},left(S-sumlimits_{k=1}^{p_1-1},frac{1}{L_k}right)$ is a positive integer. Therefore, $S$ cannot be a rational number.