I'm solving old exam problems in real analysis. Thus, only such methods may be used. I've been trying to solve the problem below and have encountered some issues.
Let $f\in C^\infty(\mathbb{R})$ in a neighborhood of the point $x_0$. Assume that there exists positive numbers $\delta$ and $M$ such that for any $x\in(x_0-\delta,x_0+\delta)$ one has the estimate $$\left|\frac{d^kf(x)}{dx^k}\right|\leq M\frac{k!}{\delta^k}.$$ Show that under these assumptions $$f(x)=\sum_{k=0}^\infty\frac{1}{k!}\frac{d^kf(x_0)}{dx^k}(x-x_0)^k.$$ Note that this means that the estimate above implies that $f(x)$ is analytic at $x_0$.
My first thought is to try to find something that looks what we want by dividing the estimate by $k!/\delta^k$ on both sides. Equivalently, multiply with $\delta^k/k!$ on both sides. Here $$\left|\frac{d^kf(x)}{dx^k}\right|\leq M\frac{k!}{\delta^k}\Leftrightarrow\frac{\delta^k}{k!}\left|\frac{d^kf(x)}{dx^k}\right|\leq M\frac{k!}{\delta^k}\frac{\delta^k}{k!}=M$$ for all $x\in(x_0-\delta,x_0+\delta)$. That $M$ is a fixed number means that we have a bound of the left-hand side of the equation. That $x\in[x_0-\delta,x_0+\delta]$ means that $|x_0-x|\leq\delta$. Thus $(x_0-x)^k\leq\delta^k$. Hence $$\frac{1}{k!}\left|\frac{d^kf(x)}{dx^k}\right|(x_0-x)^k\leq\frac{1}{k!}\left|\frac{d^kf(x)}{dx^k}\right|\delta^k\leq M.$$ This means that the summand is bounded.
I don't really know how to progress from this step, or if anything that I've deduced is of any value. I guess that showing pointwise or uniform convergence might give us what we want, e.g. showing that this converges to a function, but how do we know that is the function that we wanted to begin with? I feel like I'm missing something obvious here.
If you take $x\in (x_0 - \rho, x_0+\rho)$ with $\rho < \delta$ then the terms of the Taylor series will be bounded by $M\frac{\rho^k}{\delta^k}$ and you can apply the Weierstrass M-test. Hence the series converges absolutely and uniformly on $(x_0 - \rho, x_0+\rho)$ and $f$ is analytic at $x_0$.
EDIT: This argument only shows that the Taylor series defines an analytic function $g$ around $x_0$. To see that $g=f$ one may fix an arbitrary point $x\in (x_0 - \rho, x_0+\rho)$ and use the Lagrange remaider for the Taylor expansion of $f$ up to order $n$ and then use the estimates to show that $|f(x)-g(x)| =0$ letting $n\to \infty$.
$$|f(x)-g(x)| = \left|\frac{f^{(n+1)}(x^\ast)}{(n+1)!}(x-x_0)^{n+1}-\sum_{k=n+1}^\infty\frac{f^{(k)}(x_0)}{k!}(x-x_0)^k\right|\leq \\ \sum_{k=n+1}^\infty\frac{|f^{(k)}(x_0)|}{k!}|x-x_0|^k+\frac{|f^{(n+1)}(x^\ast)|}{(n+1)!}|x-x_0|^{n+1} \leq \sum_{k=n+1}^\infty M\frac{k!}{\delta^k}\frac{|x-x_0|^k}{k!} + M\frac{(n+1)!}{\delta^{n+1}} \frac{|x-x_0|^{n+1}}{(n+1)!}=\sum_{k=n+1}^\infty M\frac{|x-x_0|^k}{\delta^k} + M\frac{|x-x_0|^{n+1}}{\delta^{n+1}}$$ Since $|x-x_0|<\delta$, $\sum_{k=n+1}^\infty M\frac{|x-x_0|^k}{\delta^k}$ is the tail of a convergent geometric series, hence it tends to $0$ as $n\to \infty$. Also $M\frac{|x-x_0|^{n+1}}{\delta^{n+1}}\to 0$ as $n\to \infty$ and we conclude that $|f(x)-g(x)| \leq 0$.