Elementary proof that the limit of $\sum_{i=1}^{\infty} \frac{1}{\operatorname{lcm}(1,2,...,i)}$ is irrational

Question

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?

Answer 1

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.

Answer 2

I know this question is 3 years old. But I have a nice proof that does not require Bertrand's postulate. (One can prove the irrationality of e using this method).

Consider n!S=$\frac{n!}{2}$+$\frac{n!}{6}$+$\frac{n!}{12}$+... . Also note that lcm(1,...n) $\leq$ n!, and gcd(a,a+2)=1, for all a in Z. Splitting this sum as

n!S=($\frac{n!}{2}$+$\frac{n!}{6}$+$\frac{n!}{12}$+...+$\frac{n!}{lcm(1,,n)}$)+$\frac{n!}{lcm(1,...,n+1)}$+$\frac{n!}{lcm(1,...,n+2)}$+... , we notice that $\frac{n!}{lcm(1,...,n+1)}$ might be an integer, but $\frac{n!}{lcm(1,...,n+2)}$ never is an integer, since gcd(a,a+2)=1. Thus we have everything after $\frac{n!}{lcm(1,...,n+1)}$ may not be an integer, since it is all <1.

EDIT: I MADE A MISTAKE! Assume S=$\frac{a}{b}$ and then multiply by b!, not any n!. So replace n with b, and we are done. I also assumed that gcd(a,a+2)=1. This is only true for a is odd. If a is even, gcd(a,a+1)=1. So it depends on the parity of a. So this argument still works.