# Does this taylor series converge

We have that there exists $$b>0$$ such that $$|f^{(n)}(x)| \leq \frac{1}{b}$$ for $$x \in \mathbb{R}, n \in [0,\infty), n \in \mathbb{N}$$. If a taylor series is constructed for f centered at $$x=a$$, then does this taylor series converge?

We can construct this taylor series by using the formula: $$\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n$$. I am trying to think of a counterexample to the question. Let b=1, then the sine function satisfies all of the conditions above. However, the taylor series for a sine function does converge. If I choose any value of b, then I can scale the sine fucntion (for instance, to 0.5sinx, 0.0001sinx, and so on) and it seems like there will always be a taylor series that converges. Does this mean that this statement is true?

From the preamble we assume $$f^{(n)}(x)$$ exists and is bounded by $$1/b$$ for all $$x$$. Then we can apply Taylor's theorem with remainder. Choose $$x$$ and for any $$n$$, there exists $$\xi \in \mathbb R$$ (actually between $$a$$ and $$x$$), such that, \begin{align} \left \lvert f(x) - \sum_{r=0}^{n-1} f^{(r)}(a) \frac{(x-a)^r}{r!}\right\rvert =\left\lvert f^{(n)}(\xi) \frac{(x-a)^n}{n!} \right\rvert \leqslant \frac{1}{b}\frac{\lvert x-a\rvert^n}{n!} \end{align} and the right side converges to zero for all $$x$$ as $$n \to \infty$$. Thus the Taylor's series will always converge for all $$x$$.