# Composition of series and Taylor expansion

Let's say $$p_n (z)$$ is the Taylor-expansion of a function $$a(z)$$ up to the $$n$$-th order. (Consider $$a:\mathbb{R} \rightarrow \mathbb{R}$$). Now I have a series $$x_n\rightarrow x$$ for $$n \rightarrow\infty.$$ I know that $$p_m \rightarrow a$$ and $$x_n \rightarrow x$$ but is it true that $$p_n (x_n) \rightarrow a(x)$$ ?

On the one hand this seems totally clear to me but I can't come up with a proper argument...

• When you write $p_m\to a$, you mean pointwise convergence ? – Gabriel Romon Jul 18 at 12:43
• You are using the symbol $a$ with two distinct meanings. – José Carlos Santos Jul 18 at 12:43
• It is true if $x_n(0)=0$ (supposing you have a Maclaurin expansion). – Bernard Jul 18 at 12:45
• Sorry, yes, I mean pointwise – Alvo Jul 18 at 13:00

If the sequence $$(p_n)_{n\in\mathbb N}$$ converges uniformly to $$a$$ in an interval $$[\alpha,\beta]$$, if $$(\forall n\in\mathbb N):x_n\in[\alpha,\beta]$$, and if $$\lim_{n\to\infty}x_n=x_0$$, then, yes,$$\lim_{n\to\infty}p_n(x_n)=f(x_0).$$
• Uniform convergence happens for example when each derivative $f^{(n)}$ is bounded by some $M_n$ and $M_n=o((n+1)!)$. – Gabriel Romon Jul 18 at 12:57