User:Anghel/Sandbox

Theorem
Let $\xi \in \C$.

For all $z \in \C$, define the power series $\displaystyle S \left({z}\right) = \sum_{n \mathop = 0}^\infty a_n \left({z - \xi}\right)^n$ and $\displaystyle S' \left({z}\right) = \sum_{n \mathop = 1}^\infty n a_n \left({z - \xi}\right)^{n-1}$.

Let $R$ be the radius of convergence of $S \left({z}\right)$, and let $R'$ be the radius of convergence of $S' \left({z}\right)$.

Then $R =R'$.

Proof
Suppose that $z \in \C$ with $\left\vert{z - \xi}\right\vert < R'$.

Then $S' \left({z}\right)$ converges absolutely by definition of radius of convergence, so:

From the $n$th Root Test, it follows that $S \left({z}\right)$ converges absolutely.

Hence, $R \ge R'$.

Suppose that $z \in \C$ with $\left\vert{z - \xi}\right\vert < R$.

Find $z_o \in \C$ such that $\left\vert{z - \xi}\right\vert < \left\vert{z_0 - \xi}\right\vert < R$, so $S \left({z_0}\right)$ converges absolutely.

From the $n$th Root Test, it follows that $\left\vert{a_n \left({z_0 - \xi}\right)^n }\right\vert^{1/n} < 1$ for all $n \ge N$ for some $N \in \N$.

Then:

From the $n$th Root Test, it follows that $S' \left({z}\right)$ converges absolutely.

Hence, $R' \ge R$, so $R' = R$.