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?)?

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

1 Answer 1


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?

  • $\begingroup$ Hi, "You do not need to look at the radius of convergence, because we ahve just shown that this equality holds " that addresses my concerns. Because in some examples in the book, they look at the radius of convergence, but in others, they just look at the remainder. $\endgroup$
    – user612996
    Mar 8, 2021 at 21:05
  • $\begingroup$ @Sat Sometimes you are given a power series without its sum in closed form (it can well happen that you cannot express the sum in closed form), and you have to decide its region of convergence. Because it is a full series without a remainder, you have to look at the radius of convergence. $\endgroup$
    – Gary
    Mar 9, 2021 at 7:15

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