product of the power series expansion of geometric and exponential function

Question

I know the power series expansion around the point $0$ for $\frac{1}{1-t}$ and $\exp(t)$ for $|t| < 1$. Now let $u \in \mathbb{R}$ and define

$$ f(t) := \frac{1}{1-t} \exp\left\{ \frac{-ut}{1-t} \right\}. $$ What is the power series expansion of $f$ around $0$?

Answer 1

I don't think there's a closed-form expression for the coefficients, but they satisfy the recurrence $$ (n+1) a_n+(u-3-2n)a_{n+1}+(n+2) a_{n+2} = 0 $$ with $a_0 = 1$ and $a_1 = 1-u$. This comes from the differential equation $$ (t-1)^2 y' + (u+t-1) y = 0 $$ satisfied by $f$.