Dear Abraham, is there any question you want to formulate in regards to your findings? I'd say a number of clearly stated questions would help enormously.

Could you please give an indication as for where the first (product) property comes from?

Thanks. As I corrected myself above, the metric in the case of finite-volume hyperbolic manifolds is not a product near the boundary, but there is a vanishing correction term. However, this is the presence of cusps that's real concern. Also, I'm not aiming at computing the invariant or getting a very good formula for it: I need to know that some non-trivial invariant of the cusped boundary manifold has to be integral. Just to show that being "bounding" is, in principle, a non-trivial property.

@IgorBelegradek: Thanks, Igor! Just in case, my question is related to this paper (arxiv.org/pdf/math/0007197.pdf) of Long and Reid about geometric cobordisms of hyperbolic 3-manifolds.

@DannyRuberman: Thanks, Danny. Ok, I see: it might be hard, and I'm no real differential geometer. However, this question appears extremely interesting in the context of geometric cobordisms of hyperbolic 3-manifolds. If you're interested, I can write more by e-mail.

Knowing this will make sure that some sequences represented by coefficients of certain divergent series do not have "simple" recursive formulas behind them - that's the motivation. However, when I'm confronted with a case when $y(z)$ satisfies some equation $y^{(n)}(z)=Q(y^{(n−1)},y^{(n−2)},…,y′,z)$, with $n\geq 2$, the above reasoning would not work. That's why I was asking for clarification and references to make sure I'm getting things right. For now I've learnt a few things from your answers, and my problem has got a clearer shape. Thank you again!

The general line of thought (again, can be blurry): if $y(z)$ satisfies a linear diff equation (with coefficient polynomials in $z$), then we can substitute in it all the derivatives of $y(z)$ with polynomials in $\xi = y(z)$ and get a polynomial equation $P(z,y(z))=0$. Then the implicit function theorem says that $y(z)$ must be analytic, though it cannot be, since it's represented by a divergent everywhere (except $z=0$) series.

Sorry for being unclear. Indeed, you construed my question correctly. Basically, I think that one can show that if a divergent formal power series satisfies some differential equation $y^{\prime} = Q(y(z))$, with $Q\in\mathbb{C}[z][\xi]$ of degree $\geq 2$ in $\xi$, then $y$ cannot satisfy a simpler, linear differential equation. May be some other conditions are necessary.

Would the situation change if I ask for $f(z)$ in $P(z, f(z))=0$ to be a formal power series first $f(z) \in \mathbb{C}[[z]]$, and then assert it should be analytic (may be the only one such solution is identically $0$)? I'd believe that if $P$ involves also derivatives of $f$ then it's a wrong statement (Riccati's equation is of this type and it has formal series solutions that are divergent except $z=0$). However I wonder what is known in general about this type of equations (hopefully I don't ask anything silly again).