Convergence of Taylor Series of Function Analytic on Disk

Theorem
Let $F$ be a complex function.

Let $x_0$ be a point in $\R$.

Let $R$ be an extended real number greater than zero.

Let $F$ be analytic at every point $z \in \C$ satisfying $\left\lvert{z - x_0}\right\rvert < R$.

Let $f = F {\restriction_{\R}}$ be a real function.

Then:
 * the Taylor series of $f$ about $x_0$converges to $f$ at every point $x \in \R$ satisfying $\left\lvert{x - x_0}\right\rvert < R$

Corollary: Taylor Series reaches closest Singularity
Let the singularities of a function be the points at which the function is not analytic.

Let $F$ be analytic everywhere except at a finite number of singularities.

Let $R \in \R_{>0}$ be the distance from $x_0$ to the closest singularity of $F$.

Then:
 * the Taylor series of $f$ about $x_0$ converges to $f$ at every point $x \in \R$ satisfying $\left\lvert{x - x_0}\right\rvert < R$

Corollary: Taylor Series of Analytic Function has infinite Radius of Convergence
Let $F$ be analytic everywhere.

Then:
 * the Taylor series of $f$ about $x_0$ converges to $f$ at every point in $\R$

Proof
Let $r$ be a real number satisfying:
 * $0 < r < R$

Let $x$ be a real number satisfying:
 * $\left\lvert{x - x_0}\right\rvert < r$

$f$ has a Taylor series expansion about $x_0$ with radius of convergence greater than zero as $f$ is analytic at $x_0$.

The Taylor's formula with remainder for $f$ about $x_0$ is:
 * $f \left({x}\right) = \displaystyle \sum_{i \mathop = 0}^n \frac {\left({x - x_0}\right)^i} {i!} f^{\left({i}\right)} \left({x_0}\right) + R_n \left({x}\right)$

where
 * $R_n \left({x}\right) = \dfrac 1 {n!} \displaystyle \int_{x_0}^x \left({x - t}\right)^n f^{\left({n \mathop + 1}\right)} \left({t}\right) \mathrm d t$

Our first aim is to prove:
 * $\displaystyle \lim_{n \mathop \to \infty} R_n \left({x}\right) = 0$

For the case $x = x_0$, the interval of integration in the expression for $R_n \left({x}\right)$ has zero length.

Therefore, $R_n \left({x}\right) = 0$.

Accordingly, $\displaystyle \lim_{n \mathop \to \infty} R_n \left({x}\right) = 0$ is true for this case.

Now we consider the case $x \ne x_0$.

We have:
 * $0 < r - \left\lvert{x - x_0}\right\rvert$ as $\left\lvert{x - x_0}\right\rvert < r$

Observe that:
 * $\left\lvert{x - x_0}\right\rvert \ge \left\lvert{t - x_0}\right\rvert$

Therefore:
 * $r - \left\lvert{x - x_0}\right\rvert \le r - \left\lvert{t - x_0}\right\rvert$
 * $0 < r - \left\lvert{x - x_0}\right\rvert \le r - \left\lvert{t - x_0}\right\rvert$
 * $0 < \left\lvert{r - \left\lvert{x - x_0}\right\rvert}\right\rvert\le \left\lvert{r - \left\lvert{t - x_0}\right\rvert}\right\rvert$

We have:

Let $y \in \R$ be equal to $x_0 + r$ if $x > x_0$ and $x_0 - r$ if $x < x_0$.

Note that $y > x$ if $x > x_0$ and $y < x$ if $x < x_0$.

The general situation is:
 * $x_0 \le t \le x < y$ if $x > x_0$
 * $y < x \le t \le x_0$ if $x < x_0$

Let us study $\left\lvert{x - t}\right\rvert$ in the expression above for the bound for $\left\lvert{R_n \left({x}\right)}\right\rvert$:

Also, we have:

We combine these two results to get:

We use this result in the expression for the bound for $\left\lvert{R_n \left({x}\right)}\right\rvert$:

We have:
 * $\displaystyle \frac r {\left\lvert{x - x_0}\right\rvert} > 1$ as $\left\lvert{x - x_0}\right\rvert < r$ and $x \ne x_0$

Therefore:
 * $\displaystyle \lim_{n \mathop \to \infty} \frac n {\left({\frac r {\left\lvert{x - x_0}\right\rvert} }\right)^n} = 0$ by the lemma

Letting $n$ approach $\infty$ in the expression for the bound for $\left\lvert{R_n \left({x}\right)}\right\rvert$, we get:

So:

Accordingly, $\displaystyle \lim_{n \mathop \to \infty} R_n \left({x}\right) = 0$ is true for the case $x \ne x_0$.

Thus, $\displaystyle \lim_{n \mathop \to \infty} R_n \left({x}\right) = 0$ holds for every $x$ satisfying $\left\lvert{x - x_0}\right\rvert < r$ where $r < R$.

Since we can choose $r$ as close to $R$ as we like, we conclude that $\displaystyle \lim_{n \mathop \to \infty} R_n \left({x}\right) = 0$ holds for every $x$ that satisfies $\left\lvert{x - x_0}\right\rvert < R$.

Therefore, the Taylor series expansion of $f \left({x}\right)$ about $x_0$ converges to $f \left({x}\right)$ for every $x$ that satisfies $\left\lvert{x - x_0}\right\rvert < R$.

Lemma
Let $y > 1$.

Then:
 * $\displaystyle \lim_{n \mathop \to \infty} \frac n {y^n} = 0$

Proof
Note that $\ln y > 0$ as $y > 1$.