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?
Best Answer
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.