Conway (Complex Analysis) Proposition 3.2.5, question on part of proof.

Question

The proposition:

Prop 3.2.5. Let $f(z) := \sum_{n=0}^{\infty} a_{n}(z-a)^{n}$ have radius of conv. $R > 0$. Then:

(a) $\forall k\geq 1$, $\sum_{n=k}^{\infty} n(n-1)\cdots(n-k+1)a_{n}(z-a)^{n-k}$ has radius of conv. $R$.

(b) $f$ is infinitely differentiable on $B(a, R)$, and....[more stuff].

Question: Assuming (a) is proved, I have a question about (*) below:

(Proof of part (b)). For $|z| < R$ put $g(z) := \sum na_{n}z^{n-1}$, $s_{n}(z)$ the $n$-th partial sum, and $R_{n}(z) := \sum_{k\geq n+1} a_{k}z^{k}$. Fix $w \in B(0, R)$ and $|w| < r < R$. We wish to show $f'(w)$ exists and equals $g(w)$. Let $\delta > 0$ such that $\overline{B}(w, \delta) \subset B(0, r)$. Let $z \in B(w, \delta)$. Then $$\frac{f(z) - f(w)}{z-w} - g(w) = \Big[ \frac{s_{n}(z) - s_{n}(w)}{z-w} - s_{n}'(w)\Big] + (s_{n}'(w) - g(w)) + \Big[ \frac{R_{n}(z) - R_{n}(w)}{z-w}\Big].$$ Now, $$\frac{R_{n}(z) - R_{n}(w)}{z-w} = \sum_{k=n+1}^{\infty} a_{k}\Big( \frac{z^{k} - w^{k}}{z-w}\Big).$$ But $$\frac{|z^{k}-w^{k}|}{|z-w|} = |z^{k-1} + z^{k-2}w + \cdots + w^{k-1}| \leq kr^{k-1}.$$ Hence, $$\Big|\frac{R_{n}(z)-R_{n}(w)}{z-w}\Big| \leq \sum_{k=n+1}^{\infty} |a_{k}|kr^{k-1}.$$ Since $r < R$, we have $\sum_{k=1}^{\infty} |a_{k}|kr^{k-1}$ converges. So, for any $\epsilon > 0$ there is $N_{1}$ s.t. for any $n \geq N_{1}$, $$\Big|\frac{R_{n}(z)-R_{n}(w)}{z-w}\Big| < \epsilon/3 \hspace{5ex} (*)$$ ......

How do we know this term will go to zero, as indicated by $\epsilon/3$ ?

Answer 1

Should've noticed that $\sum_{k=N+1}^{\infty} [blah] = s - s_{N} < \epsilon$.

Answer 2

It looks like you are confusing sequences and series. If $s_n\to s $, then $s-s_n\to0$; that's the tail that Conway (and almost any other proof with series) is using.