# does a Taylor series converge to f(x) if the radius of convergence is $R=\infty$

If I want to check whether a Taylor series converges to its function f(x). Is it enough to check if the radius of convergence of the Taylor series is infinite?

Or do I have to use the remainder theorem and show the remainder converges to zero?

Edit: I'm still confused about convergence of taylor series. If I have a function f(x) which is defined on $$x\in ]-R,R[$$, and I found it's taylor series T(x) with a radius of convergence R. Doesn't the taylor series converge to f(x)?

Why do I necessarily need to check that the remainder $$R_n(x)$$ converges to zero for it to be true?

In other words, what does the taylor series converge to, if it doesn't converge to f(x)? (Because I thought if a taylor series was convergent it would always be towards f(x)? But that's probably where the whole issue is?).

• $f(X)=e^{-1/x}$ for $x >0$ $f(x)=0$ for $x \leq 0$. All the terms in the Taylor series are $0$ for this function. Nov 26 '20 at 23:43
• @KaviRamaMurthy, but what if I work with regular functions like e^x, cosx, ln(x), sinx*cosx, etc. If I found the radius of convergence to be the same as the interval that the original function is defined. Would the series converge? Nov 26 '20 at 23:49
• For $\mathrm e^x, \sin x,\cos x, \sinh x,\cosh x$, it does converge. Nov 26 '20 at 23:53
• @Bernard but can I conclude that because the radius of convergence is infinite? Or that isn't enough? Nov 26 '20 at 23:58
• @sjm23 If by "normal" you mean "analytic", then yes by definition.
– user239203
Nov 26 '20 at 23:58

Example of a non-analytic infinitely differentiable function of a real variable: $$f(x)=\begin{cases} \mathrm e^{-1/x^2}&\;\text{ if }x\ne 0,\\ 0&\;\text{ if }x=0. \end{cases}$$ It is easy to show that $$f^{(n)}(0)=0$$ for all $$n$$, hence its Taylor series at $$0$$ is $$0,\:\ne f(x)$$ if $$x\ne 0$$.