Does the remainder term of Taylor series necessarily tend to 0?

It is known that the $n$th remainder term of Taylor series $R_n=f^{(n)}(c)\dfrac{(b-a)^n}{n!}$, assuming $f^{(k)}(x)$ is differentiable in interval $[a,b]$ for all $k$ and $c\in[a,b]$.

But is it possible that $f^{(n)}(c)$ tends to infinity as $n$ increases such that $R_n$ doesn't converge to 0 even though $\dfrac{(b-a)^n}{n!}$ tends to 0. In such case Taylor series may not be a good approximation of the function even when $n$ is large.

The famous example is $e^{-1/x^2}$ whose Taylor series is zero.
• I understand that Taylor series is 0 here since $f^{(k)}(x)$ are 0 for all k, but does $f^{k}(c_k)$ tends to $\infty$? It seems to me $f^{(k)}(x)$ for $x\neq0$ has no chance to tend to $\infty$, and hence the $R_n$ has no chance to be equal to $e^{-1/x^2}$ (but they should be equal given that Taylor series is 0). Where has gone wrong? May 15 '17 at 15:59
• A typo above, it should be $f^{(k)}(0)$ are 0 for all k (since we are talking about Maclaurin series). Actually, it's a pretty good answer and your example is also "valid". I just have this specific follow-up question since it seems $f^{(k)}(c)$ has no way to tend to $\infty$ so as to guarantee $f(x)=R_n(x)$ in the end. Am I wrong? May 15 '17 at 17:13
In order for the Taylor series of $f(z)$ at $z=a$ to converge to $f(z)$ at some point with $|z-a| = R$, $f$ must be analytic in the disk $\{z \in \mathbb C: |z - a| < R\}$. Even if $f$ is real-analytic on the whole real line, if $f$ has a pole or branch point $p$ in the complex plane the series will diverge for $|z-a| > |z-p|$. For example, the Maclaurin series for $f(z) = 1/(1+z^2)$ diverges for $|z| \ge 1$.
• thanks, just wonder "analytic function" is a concept usually introduced in complex analysis (seems my calculus textbook doesn't mention this)? Also, just to ensure that $f^{(k)}(c_k)$ tends to $\infty$ here? May 15 '17 at 15:51