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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?


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

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

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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.

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