Prove that the number $\newcommand{\lcm}{{\rm lcm}}$
$$\mathcal{I}=\sum_{n=1}^{\infty} \frac{1}{\lcm (1, 2, \dots, n)}$$
is irrational.
Solution
Assume it's rational, say $\frac{p}{q}$. Define $p_k$ and $q_k$ such that $\frac{p_k}{q_k} = \sum \limits_{n=1}^k \dfrac{1}{\lcm(1, 2, \ldots, n)}$. We then have the lower bound
$$\frac{p}{q} - \frac{p_k}{q_k} \geq \frac{1}{qq_k} \geq \frac{1}{q \lcm(1, 2, \ldots k)}$$
We also have
$$\frac{p}{q} - \frac{p_k}{q_k} = \sum_{n=k+1}^{\infty} \frac{1}{\lcm(1, 2, \ldots, n)} = \left(\frac{1}{\lcm(1, 2, \ldots k)}\right)\left(\sum_{n=k+1}^{\infty} \frac{\lcm(1, 2, \ldots k)}{\lcm(1, 2, \ldots n)}\right)$$
where the last sum can be made arbitrarily small by setting $k+1$ to be a large prime. In particular, it can be made smaller than $\frac{1}{q}$, which induces a contradiction with our lower bound.
The exercise can also be found in AoPS.com
$$\mathcal{I}=\sum_{n=1}^{\infty} \frac{1}{\lcm (1, 2, \dots, n)}$$
is irrational.
Solution
Assume it's rational, say $\frac{p}{q}$. Define $p_k$ and $q_k$ such that $\frac{p_k}{q_k} = \sum \limits_{n=1}^k \dfrac{1}{\lcm(1, 2, \ldots, n)}$. We then have the lower bound
$$\frac{p}{q} - \frac{p_k}{q_k} \geq \frac{1}{qq_k} \geq \frac{1}{q \lcm(1, 2, \ldots k)}$$
We also have
$$\frac{p}{q} - \frac{p_k}{q_k} = \sum_{n=k+1}^{\infty} \frac{1}{\lcm(1, 2, \ldots, n)} = \left(\frac{1}{\lcm(1, 2, \ldots k)}\right)\left(\sum_{n=k+1}^{\infty} \frac{\lcm(1, 2, \ldots k)}{\lcm(1, 2, \ldots n)}\right)$$
where the last sum can be made arbitrarily small by setting $k+1$ to be a large prime. In particular, it can be made smaller than $\frac{1}{q}$, which induces a contradiction with our lower bound.
The exercise can also be found in AoPS.com
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