How to equate the coefficients of $t^n/n!$ in the following equation?

Question

How to equate the coefficients of $t^n/n!$ in the below equation?

$$\sum_{n=0}^\infty F(n,r) \frac{t^n}{n!}=\frac{t^{r-1}}{r!}+\sum_{n=r+1}^\infty (1-2^{n-r})\frac{G(n,r)}{r} \frac{t^n}{n!}.$$

Answer 1

First, notice the right-hand side has fewer terms, missing terms of the form $a_n t^n$ when $n < r-1$ or $n=r$. But this is the same as adding terms of the form $0 \cdot t^n$. So given the equation holds for all real $t$ in some interval,

$$ \begin{align*} F(n,r) &= 0 \qquad \mathrm{if\ } n < r-1 \\ F(r,r) &= 0 \end{align*} $$

For the $t^{r-1}$ term, we get the term of the $F$ sum when $n=r-1$, and the $G$ sum does not contribute, but the lone addend before it does.

$$F(r-1, r) \frac{1}{(r-1)!} = \frac{1}{r!}$$ $$F(r-1, r) = \frac{1}{r}$$

Finally, equating coefficients of $t^n$ when $n \geq r+1$,

$$F(n,r) \frac{1}{n!} = (1 - 2^{n-r})\frac{G(n,r)}{r} \frac{1}{n!}$$ $$F(n,r) = (1 - 2^{n-r}) \frac{G(n,r)}{r} \qquad \mathrm{if\ } n \geq r+1$$