how did they find A_n, from a(x), /sci/?

>>8568983
>a_0 + (a_1 - a_0)x + (a_2 - a_1)x^2 + ...
>=
>1 + x + 3x^2 + ...

Oh, gee, I wonder.

Not sure what tgey did with A as the partial fractions bit shpuld be -1/2 , but they ignore this? Will have a go at solving soon

[math] \dfrac {1} {1-c} = \sum_{n=0}^\infty c^n [/math]

>>8568983
a_n is the coefficient of x^n in a(x)

If a(x) is a power series generating function, and can be written in terms of elementary functions, then to find each a_i, you have to find the i'th derivative of a(x) divided by the i'th factorial