Show that the series $\sum_{l \ge k}\binom{l} {k} z^{l-k}$ has radius of convergence equal $1$. And find a closed form for its sum.
My attempt:
$$\binom{l} {k} = \frac{l!}{k!(l-k)!}$$
Let $a_l = \frac{l!}{k!(l-k)!}$ And so lim as $l$ tends to infinity of $\frac{a_{l+1}}{a_l} = 1$.
And hence $R=1$
And what about the other part of the question? I don't know how to solve it and I didn't understand it. Any help please?