Power Series is Differentiable on Interval of Convergence

Theorem
Let $\xi \in \R$ be a real number.

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

Let $f \left({x}\right)$ have an interval of convergence $I$.

Then $f \left({x}\right)$ is continuous on $I$, and differentiable on $I$ except possibly at its endpoints.

Also:
 * $\displaystyle D_x \left({f \left({x}\right)}\right) = \sum_{n=1}^\infty n a_n \left({x - \xi}\right)^{n-1}$

Proof
Let the radius of convergence of $f \left({x}\right)$ 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:
 * $\left|{x - \xi}\right| < \left|{x_0 - \xi}\right| < R$

Consider the series:
 * $\displaystyle \sum_{n=2}^\infty \left|{\frac {n \left({n-1}\right)} 2 a_n X_0^{n-2}}\right|$ where $X_0 = x_0 - \xi$

From Radius of Convergence from Limit of Sequence‎, we have:
 * $\displaystyle \frac 1 R = \limsup_{n \to \infty} \left|{a_n}\right|^{1/n} = \limsup_{n \to \infty} \left({\left|{\frac {n \left({n-1}\right)} 2}\right| a_n}\right)^{1/n}$

Thus we may deduce that:
 * $\displaystyle \sum_{n=2}^\infty \left|{\frac {n \left({n-1}\right)} 2 a_n X_0^{n-2}}\right|$

converges.

Put $\delta = \left|{x - x_0}\right|$.

For values of $y$ such that $0 < \left|{x - y}\right| < \delta$, we consider:
 * $\displaystyle \Delta = \frac {f \left({y}\right) - f \left({x}\right)} {y - x} - \sum_{n=1}^\infty n a_n \left({x - \xi}\right)^{n-1} = \sum_{n=1}^\infty a_n \left({\frac {Y^n - X^n} {Y - X} - n X^{n-1}}\right)$

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:

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 $\left({Y - X}\right)$ with the sum of $n \left({n-1}\right) / 2$ terms of the form $X^r Y^s$ where $r + s = n - 2$.

The number $n \left({n-1}\right) / 2$ arises from the result $1 + 2 + \cdots + \left({n-1}\right) = \dfrac {n \left({n-1}\right)} 2$.

Since:
 * $\left|{X}\right| = \left|{x - \xi}\right| < \left|{x_0 - \xi}\right| = \left|{X_0}\right|$
 * $\left|{Y}\right| = \left|{y - \xi}\right| < \left|{x - \xi}\right| + \delta = \left|{X_0}\right|$

we have:
 * $\left|{\dfrac {Y^n - X^n} {Y - X} - n X^{n-1}}\right| < n \left({n-1}\right) / 2 \left|{X_0}\right|^{n-2} \left|{Y - X}\right|$

So:
 * $\displaystyle \left|{\Delta}\right| \le \left|{Y - X}\right| \sum_{n=2}^\infty \frac {n \left({n-1}\right)} 2 \left|{a_n X_0^{n-2}}\right| \to 0$ as $Y \to X$

So $f$ is differentiable at all $x \in I$ which is not an endpoint, and that $\displaystyle D_x \left({f \left({x}\right)}\right) = \sum_{n=1}^\infty n a_n \left({x - \xi}\right)^{n-1}$.

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 $\displaystyle \sum_{k=0}^\infty a_k$ is convergent, then $\displaystyle \lim_{x \to 1^-} \left({\sum_{k=0}^\infty a_k x^k}\right) = \sum_{k=0}^\infty a_k$.