What does it mean that a taylor series generated for a function f(x) doesnt converge to f(x)? If a some function f(x) is continous and has derivatives of all orders on some interval I, and assuming that f(x) can be expressed as a power series on I. And now you generate a taylor series for f(x), this series is may or may not converge on that interval I, to the function f(x).
My question is how can a taylor series generated for a function not converge to that function? That just doesnt seem intuitiv to me. And if it doesnt converge to that function, does it mean it still approximates the function, you can still choose n as high as you want and get a very good approximation of the function f(x), the series just never stops?
 A: Infinitely differentiable functions (of one real variable) are not real analytic (i.e. represented by a power series) for free. You have been exposed to a counter-example in the comments:
$$
f(x)=
\cases{e^{-1/x^2}, &$x \neq 0$ \cr
0, &$x=0$.\cr
}
$$
But real analytic functions can be rather different than holomorphic functions. For example, the function
$$
f(x)=\frac{1}{1+x^2}
$$
is real analytic at any point, but its power series
$$
\sum_{n=0}^\infty (-1)^n x^{2n}
$$
converges only in the interval $-1\leq x \leq 1$. You can read more here.
A: If $f$ can be expressed as power series on $I$ (what you assume) then the Taylor series will converge to your function $f$ on $I$ (assuming you expand about the middle of $I$ if $I$ is finite). If $I$ is infinite you can chose every point of $I$ and it will work. Even if $I$ is finite it might be possible to expand about other points then the middle and still get a converging Taylor series on $I$. Also this series converges to your function $f$.
A: You do not seem to know Taylor's Theorem with remainder. If the function has many derivatives, we may write approximations with error terms that can sometimes be estimated;
$$  f(x) = f(a) + f'(a)(x-a) + f''(c)\frac{(x-a)^2}{2},   $$
where $c$ is some real number that lies between $x$ and $a$ ( and so depends on both). The error term, with $c,$ is usually called the remainder. there are other ways to write it: https://en.wikipedia.org/wiki/Taylor%27s_theorem#Explicit_formulae_for_the_remainder
We get better accuracy near $a$ by using more derivatives,
$$  f(x) = f(a) + f'(a)(x-a) + f''(a)\frac{(x-a)^2}{2} + f'''(c)\frac{(x-a)^3}{6}.   $$
Here we have returned to $f''(a),$ instead the unknown $c$ between $x$ and $a$ now shows up in $f'''(c).$ 
A: It may be useful to look at an example of a $C^\infty$ function whose Maclaurin series has radius of convergence $0$.
Consider 
$$ f(x) = \sum_{n=0}^\infty 2^{-n} \cos(n^2 x)$$
Note that this series (which is not a Taylor series) converges for all real $x$.  The derivatives are
$$ f^{(k)}(x) = \sum_{n=0}^\infty 2^{-n} n^{2k} g_k(n^2 x)$$
where $g_k(t) = \dfrac{d^k}{dt^k}(\cos(t))$ is either $\pm \cos(t)$ or $\pm \sin(t)$ depending on $k$.
This also converges for all real $x$; the function is $C^\infty$.
At $x=0$ the odd-numbered derivatives are $0$, because the function is even, and the even-numbered derivatives are
$$ f^{(2j)}(0) = (-1)^j \sum_{n=0}^\infty 2^{-n} n^{4j} $$
The absolute value of the coefficient of $x^{2j}$ in the Maclaurin series
of $f(x)$ is
$$ \dfrac{\left|f^{(2j)}(0)\right|}{(2j)!} = \sum_{n=0}^\infty \dfrac{2^{-n} n^{4j}}{(2j)!} \ge \dfrac{2^{-2j} (2j)^{4j}}{(2j)!} \sim \dfrac{j^{2j-1/2} e^{2j}}{2 \sqrt{\pi}}$$
(where I used Stirling's approximation for the factorial in the last step)
and the Root Test shows that the radius of convergence is $0$.
