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?

Sungjin Kim
  • 17,768
  • 3
  • 25
  • 69
  • 421
  • 4
  • 6
  • 6
    In these days, Bertrand's postulate or Prime number theorem are not considered as complicated anymore. It is not surprising that 12 year old person knows them. But, certainly, any proof avoiding those will be interesting. – Sungjin Kim Jan 31 '18 at 19:20
  • 1
    [This](https://artofproblemsolving.com/community/c6h511659p2874098) seems relevant. – Paul LeVan Jul 29 '18 at 18:31

2 Answers2


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.

  • 48,433
  • 4
  • 51
  • 124

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.