Power Series is Differentiable on Interval of Convergence

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

Let $$f \left({x}\right) = \sum_{n=0}^\infty a_n \left({x - \xi}\right)^n$$ 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 at its endpoints.

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

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 $$x$$ and $$\xi$$.

Thus $$\left|{x - \xi}\right| < \left|{x_0 - \xi}\right| < R$$.

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

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

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

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

For values of $$y$$ such that $$0 < \left|{x - y}\right| < \delta$$, we consider:

$$\Delta = \frac {f \left({y}\right) - f \left({x}\right)} {y - x} - \sum_{n=1}^\infty n a_n \left({x - \xi}\right)^{n-1} = \sum_{n=1}^\infty a_n \left({\frac {Y^n - X^n} {Y - X} - n X^{n-1}}\right)$$

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 $$1 + 2 + \cdots + \left({n-1}\right) = \frac {n \left({n-1}\right)} 2$$.

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 $$\left|{\frac {Y^n - X^n} {Y - X} - n X^{n-1}}\right| < n \left({n-1}\right) / 2 \left|{X_0}\right|^{n-2} \left|{Y - X}\right|$$.

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

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

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