# How to arrive at this sufficient condition for convergence of Taylor Series?

So I have been studying the convergence of Taylor series from Tom M Apostol. There is a theorem which states the sufficient condition for convergence of Taylor series. Quoting the theorem we've

Assume $$f$$ is infinitely differentiable in an open interval $$I = (a - r, a + r)$$, and assume that there is a positive constant $$A$$ such that $$|f^{(n)}(x)| \le A^n$$ for $$n = 1, 2, 3, \dots$$ and every $$x$$ in $$I$$. Then the Taylor's series generated by $$f$$ at $$a$$ converges to $$f(x)$$ for each $$x$$ in $$I$$.

My question is how to arrive at $$|f^{(n)}(x)| \le A^n$$ deductively ? I tried looking it up on the internet but haven't found anything that solves my issue.

Thanks!

• Do you mean how to deduce that such a bound would be a sufficient condition for convergence? It is a strange question asking how to deduce a hypothesis. Hypotheses are starting points. One could guess, or motivate a hypothesis. In the case of necessary and sufficient conditions one could deduce it, but this one is only a sufficient condition. – logarithm May 13 at 13:13

The assertion $$|f^n(x)| \le A^n$$ is a hypothesis in this theorem. You can't "arrive at it deductively" in general. When it happens to be true for any particular function $$f$$ then you can conclude something about the Taylor series for that function.
For example, the $$n$$th derivatives of $$\sin$$ are $$\sin$$ and $$\cos$$. Their absolute values are always at most $$1$$. This theorem then implies that the Taylor series for $$\sin$$ converges everywhere to the value of the function.