Derivative of Complex Power Series/Proof 2

Theorem
Let $\xi \in \C$ be a complex number.

Let $\langle a_n\rangle$ be a sequence in $\C$.

Let $\displaystyle f \left({z}\right) = \sum_{n=0}^\infty a_n \left({z - \xi}\right)^n$ be a power series in a complex variable $z \in \C$ about $\xi$.

Let $R$ be the radius of convergence of the series defining $f \left({z}\right)$.

Let $\left \vert{z - \xi}\right \vert < R$.

Then:
 * $\displaystyle f' \left({z}\right) = \sum_{n=1}^\infty n a_n \left({z - \xi}\right)^{n-1}$

Lemma

 * $\displaystyle \lim_{n\to\infty} \left[ \frac {n\left(n-1\right)}{2} \right]^{1/n} = 1$

Proof of Lemma
Choose any $\alpha>1$. It follows from the ratio test that:


 * $\displaystyle \lim_{n\to\infty} \frac{1}{\alpha^n} \frac {n\left(n-1\right)}{2}=0$

Therefore, for all sufficiently large $n$,


 * $\displaystyle \frac {n\left(n-1\right)}{2} \le \alpha^n$

and so:


 * $\displaystyle \left[ \frac {n\left(n-1\right)}{2} \right]^{1/n} \le \alpha$

It follows that:


 * $\displaystyle \lim_{n\to\infty} \left[ \frac {n\left(n-1\right)}{2} \right]^{1/n} \le \alpha$

Since $\alpha>1$ was arbitrary, we can conclude that:


 * $\displaystyle \lim_{n\to\infty} \left[ \frac {n\left(n-1\right)}{2} \right]^{1/n} \le 1$

It is clear that the following holds for sufficiently large $n$:


 * $\displaystyle \left[ \frac {n\left(n-1\right)}{2} \right]^{1/n} \ge 1^{1/n} = 1$

Therefore:


 * $\displaystyle \lim_{n\to\infty} \left[ \frac {n\left(n-1\right)}{2} \right]^{1/n} = 1$

Proof
Define:
 * $\displaystyle g\left(z\right)=\sum_{n=1}^\infty na_n\left(z-\xi\right)^{n-1}$

Fix an $\epsilon>0$ satisfying $\epsilon<R-\left\vert z-\xi\right\vert$.

Let:
 * $\displaystyle M=\sum_{n=2}^\infty\frac{n\left(n-1\right)}{2}\left\vert a_n\right\vert\left(R-\epsilon\right)^{n-2}$

We use the root test to prove convergence of this series:

The last equality follows from the lemma and:


 * $\displaystyle \limsup_{n\to\infty} \left \vert a_n \right \vert^{1/n}=\frac{1}{R}$

Suppose that $\left\vert h\right\vert\le R-\epsilon-\left\vert z-\xi\right\vert$.

It follows by the triangle inequality that $\left\vert z-\xi+h\right\vert\le\left\vert z-\xi\right\vert+\left\vert h\right\vert\le R-\epsilon$.

By the triangle inequality, Difference of Two Powers, and Closed Form for Triangular Numbers, the following holds:

Letting $h\to 0$ gives $f'(z)=g(z)$, as desired.

Remark
The proof for real power series is identical.