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Is there a closed form for the double sum with least common multiple: $$\sum_{n \geq 1} \sum_{k \geq 1} \frac{1}{\text{lcm}^3 (n,k)}$$

For truncated sums, Mathematica gives:

$$\sum_{n = 1}^{2500} \sum_{k = 1}^{2500} \frac{1}{\text{lcm}^3 (n,k)}=1.707289827$$

$$\sum_{n = 1}^{5000} \sum_{k = 1}^{5000} \frac{1}{\text{lcm}^3 (n,k)}=1.707290976$$

$$\sum_{n = 1}^{10000} \sum_{k = 1}^{10000} \frac{1}{\text{lcm}^3 (n,k)}=1.707291287$$

It's very close to $1+1/ \sqrt{2}$, but not quite.

By the way, how do we prove it converges?

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    Convergence follows from $\text{lcm}(n,k)\geq \sqrt{nk}$ giving $\frac{1}{\text{lcm}^3(n,k)}\leq \frac{1}{n^{3/2}k^{3/2}}$ and the sum is bounded by $(\sum \frac{1}{n^{3/2}})^2 = \zeta^2(3/2)$. – 2017-02-19
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    there is this link http://mathoverflow.net/questions/33600/double-sum-of-reciprocal-powers-of-integer-lcms – 2017-02-19
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    @zwim, thank you! So, apparently, the closed form is $$\frac{\zeta ^3 (3)}{\zeta(6)}=1.707291401$$ – 2017-02-19
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    The text is not clear, I don't know if it is a bound or exact value. Seems last = symbol when going from gcd(u,v)=1 to u,v=1 should be $\le$ instead. typo or not typo ? – 2017-02-19
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    @zwim, note that the question you linked asks for a finite sum, which is why this answer talks about bound – 2017-02-19
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    Or maybe simpler $$\frac{945 \zeta(3)^3}{\pi^6}$$ – 2017-09-09

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For any $m\geq 1$, the number of couples $(n,k)$ such that $\text{lcm}(n,k)=m$ can be easily understood. Assuming that the factorization of $m$ is given by $p_1^{\alpha_1}\cdots p_k^{a_k}$, there are $(2\alpha_1+1)\cdots (2\alpha_k+1)$ such couples. If we denote with $g(u)$ the multiplicative function whose value at $p^\alpha$ is $2\alpha+1$, the original series equal the Dirichlet series: $$ \sum_{m\geq 1}\frac{g(m)}{m^3}=\prod_{p\in\mathcal{P}}\left(1+\frac{g(p)}{p^3}+\frac{g(p^2)}{p^6}+\frac{g(p^3)}{p^9}+\ldots\right)=\prod_{p\in\mathcal{P}}\left(1-\frac{1}{p^6}\right)\left(1-\frac{1}{p^3}\right)^{-3} $$ that can be easily represented in closed form, $\color{red}{\large\frac{\zeta(3)^3}{\zeta(6)}}$, through Euler's product.

In a similar way:

$$\forall s>1,\qquad \sum_{m,n\geq 1}\frac{1}{\text{lcm}(m,n)^s} = \sum_{m\geq 1}\frac{g(m)}{m^s}=\color{red}{\frac{\zeta(s)^3}{\zeta(2s)}}.$$