Theorem: Power Series is Differentiable on Interval of Convergence

Let $\xi \in \R$ be a real number.
Let $\ds \map f x = \sum_{n \mathop = 0}^\infty a_n \paren {x - \xi}^n$ be a (real) power series about $\xi$.
Let $\map f x$ have an interval of convergence $I$.
Then $\map f x$ is continuous on $I$, and differentiable on $I$ except possibly at its endpoints.
Also:
:$\ds \map {D_x} {\map f x} = \sum_{n \mathop = 1}^\infty n a_n \paren {x - \xi}^{n - 1}$

Proof:
Let the radius of convergence of $\map f x$ be $R$.
Suppose $x \in I$ such that $x$ is not an endpoint of $I$.
Then there exists $x_0 \in I$ such that $x$ lies between $x_0$ and $\xi$.
Thus:
:$\size {x - \xi} < \size {x_0 - \xi} < R$
Consider the series:
:$\ds \sum_{n \mathop = 2}^\infty \size {\frac {n \paren {n - 1} } 2 a_n X_0^{n - 2} }$ where $X_0 = x_0 - \xi$
From Radius of Convergence from Limit of Sequence, we have:
:$\ds \frac 1 R = \limsup_{n \mathop \to \infty} \size {a_n}^{1/n} = \limsup_{n \mathop \to \infty} \paren {\size {\frac {n \paren {n - 1} } 2} a_n}^{1/n}$
Thus we may deduce that:
:$\ds \sum_{n \mathop = 2}^\infty \size {\frac {n \paren {n - 1} } 2 a_n X_0^{n - 2} }$
converges.
Put $\delta = \size {x - x_0}$.
For values of $y$ such that $0 < \size {x - y} < \delta$, we consider:
:$\ds \Delta = \frac {\map f y - \map f x} {y - x} - \sum_{n \mathop = 1}^\infty n a_n \paren {x - \xi}^{n - 1} = \sum_{n \mathop = 1}^\infty a_n \paren {\frac {Y^n - X^n} {Y - X} - n X^{n - 1} }$
where $X = x - \xi, Y = y - \xi$.
We want to show that $\Delta \to 0$ as $Y \to X$.
From Difference of Two Powers, we have:
{{begin-eqn}}
{{eqn | l = \frac {Y^n - X^n} {Y - X} - n X^{n-1}
      | r = \paren {Y^{n - 1} + Y^{n - 2} X + \cdots + Y X^{n - 2} + X^{n - 1} } - n X^{n - 1}
      | c = 
}}
{{eqn | r = \paren {Y^{n - 1} - X^{n - 1} } + X \paren {Y^{n - 2} - X^{n - 2} } + \cdots + x^{n - 2} \paren {Y - X}
      | c = 
}}
{{end-eqn}}
It follows that from all of these terms we can extract $\left({Y - X}\right)$ as a factor.
So the {{RHS}} reduces to a product of $\paren {Y - X}$ with the sum of $n \paren {n - 1} / 2$ terms of the form $X^r Y^s$ where $r + s = n - 2$.
The number $n \paren {n - 1} / 2$ arises from Closed Form for Triangular Numbers:
:$1 + 2 + \cdots + \paren {n - 1} = \dfrac {n \paren {n - 1} } 2$
Since:
:$\size X = \size {x - \xi} < \size {x_0 - \xi} = \size {X_0}$
:$\size Y = \size {y - \xi} < \size {x - \xi} + \delta = \size {X_0}$
we have:
:$\size {\dfrac {Y^n - X^n} {Y - X} - n X^{n - 1} } < n \paren {n - 1} / 2 \size {X_0}^{n - 2} \size {Y - X}$
So:
:$\ds \size \Delta \le \size {Y - X} \sum_{n \mathop = 2}^\infty \frac {n \paren {n - 1} } 2 \size {a_n X_0^{n - 2} } \to 0$ as $Y \to X$
So $f$ is differentiable at all $x \in I$ which is not an endpoint, and that:
:$\ds \map {D_x} {\map f x} = \sum_{n \mathop = 1}^\infty n a_n \paren {x - \xi}^{n - 1}$
Now we need to investigate the question of left hand continuity and right hand continuity at the endpoints of $I$.
Abel's Theorem tells us that if $\ds \sum_{k \mathop = 0}^\infty a_k$ is convergent, then:
:$\ds \lim_{x \mathop \to 1^-} \paren {\sum_{k \mathop = 0}^\infty a_k x^k} = \sum_{k \mathop = 0}^\infty a_k$
{{proof wanted}}