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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?

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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 $x$.

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