TITLE: Is there a known asymptotic for $A(X):= \sum_{1 \leq i,j \leq X} \frac{1}{\mathrm{lcm}(i,j)}$?
QUESTION [11 upvotes]: My guess is that there exists a constant $C$ such that $A(X) \sim C (\log X)^2$.

REPLY [10 votes]: If we are talking about "elementary", then just multiply the original sum by the sum of inverse squares and note that if we have three numbers $a=a'd, b=b'd, n^2$ where $(a',b')=1$, $d,n$ are arbitrary, then $A=a'n, B=b'n, d$ are arbitrary $3$ integers with the product $LCM(a,b)n^2$. The original triple sum has the bounds $a'd, b'd\le X$, $n$ formally unrestricted, but restricting it to $[1,N]$ for fixed $N$ changes the triple sum $1+O(1/N)$ times. But then we can squeeze the bounds on $A,B,d$ between $Ad\le X, Bd\le X$ (treating the implicit $n=(A,B)$ as unrestricted) to get the lower bound and $Ad\le NX, Bd\le NX$ (assuming $n\le N$ now) to get the upper bound, so we get an answer for the triple sum of $\frac 1{ABd}$ between roughly speaking $\frac13\log^3 X$ and $\frac13\log^3(XN)$ with additive error $O(\log^2(XN))$ (no number theory in this sum!), which have the same asymptotic up to $1+O(\frac{\log N}{\log X})$. Taking $N$ about $\sqrt{\log X}$, we get the total multiplicative error $1+O(\frac{1}{\sqrt{\log X}})$, which is, of course, suboptimal but who cares. :-)