# Convergence of a double sum of reciprocals of lowest common multiples

Let $m,n$ be positive integers and let ${\rm lcm}(m,n)$ denote the lowest common multiple of $m$ and $n$. Now consider the double sum

$$\sum_{n=1}^{\infty}\sum_{m=1}^{\infty} \frac{1}{{\rm lcm}(m,n)}.$$

Is there a way to bound this sum from above or prove something about its convergence? If so, how can I go about this?

Hint. Note that $$\sum_{n=1}^{\infty}\sum_{m=1}^{\infty} \frac{1}{{\rm lcm}(m,n)}\geq\sum_{n=1}^{\infty}\sum_{m=n}^{n} \frac{1}{{\rm lcm}(n,m)}= \sum_{n=1}^{\infty}\frac{1}{n}.$$
Clearly, $$(m,n)\le mn$$ and hence $$\frac{1}{mn}\le \frac{1}{(m,n)}$$ But $$\sum_{m,n=1}^\infty \frac{1}{mn}=\sum_{m=1}^\infty \frac{1}{m}\sum_{n=1}^\infty \frac{1}{n}=\infty.$$ Hence $$\sum_{m,n=1}^\infty \frac{1}{(m,n)}=\infty$$
Note that $$\sum_{n=1}^{\infty}\sum_{m=1}^{\infty} \frac{1}{{\rm lcm}(m,n)}\ge \sum_{n=1}^\infty\frac1{\text{lcm}(1,n)}=\sum_{n=1}^\infty\frac1n$$ which is divergent.