TITLE: Convergence of the taylor polynomial $P_{2n}$ to $f$ implies convergence of $P_n$?
QUESTION [6 upvotes]: Let $f:(a,b) \to \mathbb{R}$ be infinitely differentiable at $x_0 \in (a,b)$, and let $P_n$ be its taylor polynomial of order $n$ around $x_0$. Fix a point $x \in (a,b)$, and suppose that $\lim_{n \to \infty }P_{2n}(x)=f(x)$. 
Is it true that $\lim_{n \to \infty }P_n(x)=f(x)$?
More generally, is it true that if $P_{2n}(x)$ converges (not necessarily to $f(x)$) then $P_n(x)$ also converges?

REPLY [3 votes]: This conclusion is not valid.
Consider $(a,b) = (-2,2), \, x_0 = 0,\, x = 1$. There exists a $C^\infty$-function $f$ such that $f(0) = f(1) = 0$ and $f^{(k)}(0) = (-1)^kk!$ for $k > 0$. Then $P_n(x) = \sum_{j=1}^n(-x)^j$ and therefore $P_{2n}(1) = 0 = f(1)$ for all $n$. But $P_n(1)$ does not converge.