$$
\begin{align}
\sum_{n=0}^\infty a_n(1-z^n)
&=(1-z)\sum_{n=0}^\infty\sum_{k=0}^{n-1}a_nz^k\\
&=(1-z)\sum_{k=0}^\infty\sum_{n=k+1}^\infty a_nz^k\\
&=(1-z)\sum_{k=0}^\infty t_{k+1}z^k\\
&=(1-z)\sum_{k=0}^{N-1}t_{k+1}z^k+(1-z)\sum_{k=N}^\infty t_{k+1}z^k\\
\left|\,\sum_{n=0}^\infty a_n(1-z^n)\,\right|
&\le|1-z|\sum_{k=1}^N|t_k|+\frac{|1-z|}{1-|z|}\sup_{k\gt N}|t_k|
\end{align}
$$

Suppose that f is defined on a set $\Omega\subset C$ that contains an open set U. Suppose that f is complex differentiable at some u in U, can it be said that f' is complex differentiable at u too?
I know that the conclusion I stated above holds if f is complex differentiable on an open set containing closure of a disk.
(i.e., the conlusion holds 'locally')

I'm struggling with this proof right now (it's from my Real Analysis course)

Let $S$ be an ordered set. Let $B \subset S$ be bounded (above and below). Let $A \subset B$ be a nonempty subset. Suppose all the infimums and supremums exist. Show that $$inf(B) \leq inf(A) \leq sup(A) \leq sup(B_)$$

I have an idea of why this works. I drew a picture to illustrate the situation.