Taylor polynomial: the higher the degree, the better the approximation? 
Let $f$ be an infinite times differentiable function. Is it true that:
  the higher the degree $n$ of the Taylor polynomial $T_{n,f,x_0}$ of $f$ around $x_0$, the better the approximation?

Some thoughts. Given $n$, polynomial $T_{n,f,x_0}$ is the best approximation of $f$ near $x_0$ that fulfils the requirement of equal derivatives with $f$ at $x_0$. So regarding polynomials of degree  at most $n$, $T_{n,f,x_0}$ is the winner. 
On the other hand, although one would like $T_{n,f,x_0}$ to fit better function $f$, as $n$ grows larger, it seems to me that there is no reason for this to happen. Of course, one should define what "fit better" means. In our case, it would be something like:
$$\sup_{x\in I}|T_{n+1,f,x_0}(x)-f(x)|\leqslant \sup_{x\in I}|T_{n,f,x_0}(x)-f(x)|$$
in a neighborhood $I$ of $x_0$.
Of course, I must admit that the cases I see graphically most of the times, fulfill the last requirement, by fitting better and better the graph of $f.$ 
Thank in advance for the help.
 A: I will assume without loss of generality that $x_0 = 0$ and we denote by $I$ an open neighborhood of $0$.
In general it is false that the approximation of the $n^{\text{th}}$ Taylor polynomial of a smooth function is getting better as $n$ grows.
In proposition 1, we see that there exists a function such that on every neighborhood of $0$ the supremum of the Taylor remainder goes to infinity as $n \to +\infty$.
However, for a smooth function you can always find a neighborhood of $0$ on which the $N$ first supremun of the Taylor remainder decrease but the width of your neighborhood depend on $N$, see proposition 2.

Proposition 1. There exists a function $f \in \mathcal{C}^\infty(I)$ such that for all $\varepsilon > 0$ and $[-\varepsilon,\varepsilon] \subset I$, we have
  $$
\lim_{n \to +\infty} \sup_{x\in [-\varepsilon,\varepsilon]} |T_n(x) - f(x)| = +\infty
$$
  where $T_n(x) = \sum_{k=0}^n \frac{f^{(k)}(0)}{k!}\, x^k$ is the $n^{\text{th}}$ order Taylor polynomial.



*

*First somewhat of a heuristic for the function $f$, for $x$ small enough, we have roughly
$$
|T_n(x) - f(x)| \approx \frac{f^{(n+1)}(0)}{(n+1)!} |x|^{n+1} \quad\text{and}\quad
|T_{n+1}(x) - f(x)| \approx \frac{f^{(n+2)}(0)}{(n+2)!} |x|^{n+2}
$$
so the inequality $|T_{n+1}(x) - f(x)| \le |T_n(x) - f(x)|$ become
$$
\frac{f^{(n+2)}(0)}{(n+2)!} |x|^{n+2} \le \frac{f^{(n+1)}(0)}{(n+1)!} |x|^{n+1} \quad\implies\quad |x| \le \frac{f^{(n+1)}(0)}{f^{(n+2)}(0)} (n+2).
$$
If $\frac{f^{(n+1)}(0)}{f^{(n+2)}(0)} (n+2)$ goes to $0$ then you would not be able to find a neighborhood of $0$ to have your inequality for all $n$.
To construct a counterexample, we will choose a function such that $$\lim_{n \to +\infty} \frac{f^{(n+1)}(0)}{f^{(n+2)}(0)} (n+2) = 0.$$.

*Proof of proposition 1. We choose a function $f \in \mathcal{C}^\infty(\mathbb{R})$ such that $f^{(n)}(0) = (n!)^2$, such a function exist by the Borel's lemma or by this post for an explicit construction for this particular example.
We have that, for all $n \ge 0$, $$T_n(x) = \sum_{k = 0}^n k!\, x^k$$ and the series $\sum_{k\ge 0} k!\, x^k$ has a zero radius of convergence because, for $x \neq 0$, $\lim_{k \to +\infty} |k!\, x^k| = +\infty$.
That mean that for $\varepsilon > 0$, there exists $a \in (0,\varepsilon)$ such that the sequence $(|T_n(a)|)_{n\ge 0}$ diverge to infinity.
By the triangle inequality, we have
$$
\sup_{x\in [-\varepsilon,\varepsilon]} |T_n(x) - f(x)| \ge |T_n(a) - f(a)| \ge |T_n(a)| - |f(a)|
$$
then we obtain
$$
\lim_{n \to +\infty} \sup_{x\in [-\varepsilon,\varepsilon]} |T_n(x) - f(x)| = +\infty.
$$
This prove that your statement is false for general smooth function.

*Why do you graphically observe that it seem to work?
The answers of @zhw. showed the following statement is true:

Proposition 2. Let $f \in \mathcal{C}^\infty(I)$, for a fixed integer $N$ there exists $\varepsilon > 0$ such that $[-\varepsilon,\varepsilon] \subset I$ and for all $n \le N$, we have
  $$
\sup_{x \in [-\varepsilon,\varepsilon]} |T_{n+1}(x) - f(x)| \le \sup_{x \in [-\varepsilon,\varepsilon]} |T_n(x) - f(x)|.
$$

A: WLOG $x_0=0.$ Claim: Given $n,$ there exists $r>0$ such that
$$|f(x)-T_{n+1}(x)| \le |f(x)-T_{n}(x)|,\,|x|<r.$$
To prove this, note first that if $f^{(n+1)}(0)=0,$ then $T_{n+1}=T_n,$ and there is nothing to prove. So assume $f^{(n+1)}(0)\ne 0.$ By the Langrange form of the remainder, we have
$$f(x)-T_{n}(x) = \frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1},$$
where $c$ is between $0$ and $x.$
Because $f^{(n+1)}(0)\ne 0,$ there is $s>0$ and a positive constant $A$ such that $|f(x)-T_{n}(x)| \ge A|x|^{n+1}$ for $|x|<s.$ But using Lagrange again, we have the standard estimate
$$|f(x)-T_{n+1}(x)|=O(x^{n+2})$$
as $x\to 0.$ Since $O(x^{n+2})$ is bounded above by $A|x|^{n+1}$ for small $x,$ we have the desired result.
We can say more: either $f^{(n+1)}(0)= 0,$ in which case $T_{n+1}=T_n,$ or $f^{(n+1)}(0)\ne 0,$ in which case there will exist $r>0$ such that
$$|f(x)-T_{n+1}(x)| < |f(x)-T_{n}(x)|,\,0<|x|<r.$$
I've taken a few liberties in getting the main points across. Ask if you have questions.
A: I thought I would contribute another answer. This is on the problem of showing that even for an analytic function, there may be no $r>0$ such that $|f-T_{n}|\le|f-T_{n-1}|$ holds in $(-r,r)$ for all $n.$
Let $E=\{2,4,6,\dots\}.$ For $n\in E,$ define the polynmials
$$p_n(x) = \frac{x^n}{n^n}-\frac{x^{n+1}}{n^{n-1}}.$$
Now set $f(x)= \sum_{n\in E} p_n(x).$ This $f$ equals a power series convergent everywhere in $\mathbb R.$
Claim: For $n\in E,$ $|f(1/n)-T_{n-1}(1/n)| < |f(1/n)-T_{n}(1/n)|.$
Corollary: If $r>0,$ then $|f-T_{n}|\le|f-T_{n-1}|$ fails somewhere in $(-r,r)$ for all even $n>1/r.$
Lemma: For $n\in E,$
$$\tag 1\sum_{k=1}^{\infty}|p_{n+2k}(1/n)|<\frac{2}{n^{2n}}\frac{1}{n^4}\frac{1}{1-1/n^4}$$
Proof: For each $k$ we have
$$|p_{n+2k}(1/n)| \le \frac{(1/n)^{n+2k}}{(n+2k)^{n+2k}} + \frac{(1/n)^{n+2k+1}}{(n+2k)^{n+2k-1}} $$ $$< \frac{1}{n^{n+2k}}\frac{1}{n^{n+2k}} + \frac{1}{n^{n+2k+1}}\frac{1}{n^{n+2k-1}} = \frac{2}{n^{2n+4k}}.$$
Now sum over $k$ and recognize the geometric series to get the right side of $(1).$
Proof of claim: Note that
$$f(1/n)-T_{n-1}(1/n) = p_n(1/n) +p_{n+2}(1/n) + \cdots.$$
Now $p_n(1/n)=0.$ (That's why I chose the power series the way I did!) From the Lemma we then see
$$\tag 2 |f(1/n)-T_{n-1}(1/n)| < \frac{2}{n^{2n}}\frac{1}{n^4}\frac{1}{1-1/n^4} < \frac{2}{n^{2n}}\frac{1}{16}\frac{16}{15} = \frac{2}{15}\frac{1}{n^{2n}}.$$
On the other hand,
$$f(1/n)-T_{n}(1/n) = -\frac{(1/n)^{n+1}}{n^{n-1}} +p_{n+2}(1/n) + \cdots.$$
Thus, using the Lemma again,
$$ |f(1/n)-T_{n}(1/n)| \ge \left|\frac{(1/n)^{n+1}}{n^{n-1}}\right| -|p_{n+2}(1/n)| - |p_{n+4}(1/n)| - \cdots $$ $$\ge \frac{1}{n^{2n}} -\frac{2}{n^{2n}}\frac{1}{n^4}\frac{1}{1-1/n^4}$$ $$\tag 3\ge\frac{1}{n^{2n}} -\frac{2}{15}\frac{1}{n^{2n}}=\frac{13}{15}\frac{1}{n^{2n}}.$$
Thus $(2)<(3)$ and the claim is proved.
