In a different context, I encountered the divergent series $1+2+2+3+3+3+\cdots$, and I was wondering about its summation. Putting $f(x)=1+2x+2x^2+3x^3+3x^4+3x^5+\cdots$, we have $$(1-x)f(x)=1+x+x^3+x^6+\cdots=\frac12 x^{-1/8}\theta_2(0, \sqrt{x}).$$ This doesn't seem to help a lot, but made me wonder: Is there such a thing as "theta regularisation"?
And how to go about the original series?
A similar question concerns the "summing" of
$1+$
$3+2+$
$6+5+4+$
$10+\dots$