If we have these generating functions:
$$G(t)=\frac{t}{e^t-1}$$
and
$$F(t)=\frac{te^{tx}}{e^t-1}$$
We also know that the infinite sum of geometric sequence for $|r|>1$. That is,
$$\sum_{k=0}^\infty a r^k=\frac{a}{r-1},$$
where $a$=first term $a_1$, and $r$=common ratio.
Then, I thought that $G(t)$ is a geometric sequence with $a=t$ and $r=e^t$,
and $F(t)$ is a geometric sequence with $a=te^{tx}$ and $r=e^t$ .
Am I right?
Thanks for your any kind help.
If $|z|<1$ then $$ \sum_{k=0}^\infty z^k = \frac1{1-z}. $$ So if $e^t<1$, that is, $t<0$, then $$ \sum_{k=0}^\infty e^{tk} = \frac1{1-e^t}, $$ and hence $$\sum_{k=0}^\infty te^{tk} = \frac t{e^t-1} = G(t),$$
$$ \sum_{k=0}^\infty te^{tx}e^{tk} = \frac {te^{tx}}{e^t-1} = F(t). $$