Can we derive results from infinite sequences of integration by parts?

I think a good example is the erfc function expanded around $x = \infty$: \begin{align} \text{erfc}(x) &:= \frac1{\sqrt\pi}\int_x^\infty e^{-y^2} \, dy \\ &= \frac1{\sqrt\pi}\int_x^\infty \frac 1y y e^{-y^2} \, dy \\ &= \frac1{\sqrt\pi} \frac 1{2x} e^{-x^2} - \frac1{\sqrt\pi}\int_x^\infty \frac 1{2y^2} e^{-y^2} \, dy \\ &= \frac1{\sqrt\pi} \frac 1{2x} e^{-x^2} - \frac1{\sqrt\pi}\int_x^\infty \frac 1{2y^3} y e^{-y^2} \, dy \\ &= \frac1{\sqrt\pi} \frac 1{2x} e^{-x^2} - \frac1{\sqrt\pi} \frac 1{4x^3} e^{-x^2} + \frac1{\sqrt\pi}\int_x^\infty \frac 3{4y^4} e^{-y^2} \, dy \\ & = \cdots \end{align} Note that the last term gives you a good estimate of the remainder. For example $$ \frac1{\sqrt\pi}\int_x^\infty \frac 3{4y^4} e^{-y^2} \, dy \le \frac1{\sqrt\pi} \frac 3{4x^4} \int_x^\infty e^{-y^2} \, dy < \frac1{\sqrt \pi} \frac 3{8x^5} e^{-x^2} .$$ (I may have got some details wrong, but I think you can see the idea.)