# Taylor series remainder equaling zero

I asked this question a while ago and I figured out my initial problem. I basically learned that convergence of series doesn't mean that it converges to that function since there could be error

My question now is, if there is no error between a function and Taylor series, then does that mean that Taylor series is equal to that function (at least on that interval?)

And, for example, for $$e^x$$ or $$\sin(x)$$. if I prove the error is zero, then do I need to prove the radius of convergence of that series (perhaps by ratio test?)?

• I think that you are confusing Taylor series and infinite series. Mar 8, 2021 at 14:22

Look at the case of $$e^x$$. The $$n$$th Taylor polynomial is $$T_n (x) = \sum\limits_{k = 0}^n {\frac{{x^k }}{{k!}}} .$$ It is easy to show, using the Lagrange form of the remainder, that $$\left| {e^x - T_n (x)} \right| \le \frac{{x^{n + 1} }}{{(n + 1)!}}\max (1,e^x )$$ for any real $$x$$. The right-hand side converges point-wise to zero as $$n\to +\infty$$. Thus, the function sequence $$T_n(x)$$ converges to $$e^x$$ for any real $$x$$, i.e., $$\lim_{n\to +\infty}T_n(x)=e^x$$ for any real $$x$$. By the definition of an infinite series $$\mathop {\lim }\limits_{n \to + \infty } T_n (x) =: \sum\limits_{k = 0}^\infty {\frac{{x^k }}{{k!}}} .$$ Consequently, $$e^x = \sum\limits_{k = 0}^\infty {\frac{{x^k }}{{k!}}}$$ for any real $$x$$. You do not need to look at the radius of convergence, because we have just shown that this equality holds for all real $$x$$, i.e., the series on the right-hand side converges for all real $$x$$ (and its sum is $$e^x$$). Does this answer your question?