If I have a function $f(x)$ that is continuous and differentiable$(-\infty,\infty)$, its Taylor series$$\sum_{n=0}^\infty\frac{f^n(a)}{n!}\left(x-a\right)^n$$ should always converge as regardless of what value $x-a$ may take, $n!,\,n\to\infty$ takes priority, right?

So I can generalize and say all functions can be expressed as a converging series?

Question

If I have a function $f(x)$ that is continuous and differentiable$(-\infty,\infty)$, its Taylor series$$\sum_{n=0}^\infty\frac{f^n(a)}{n!}\left(x-a\right)^n$$ should always converge as regardless of what value $x-a$ may take, $n!,\,n\to\infty$ takes priority, right?

So I can generalize and say all functions can be expressed as a converging series?

blockcolder · Answer

Not really. The Taylor series of (1+x)^k doesn't converge outside (-1,1).

blockcolder · Answer

Also, here's a function to consider:
$$\large f(x)=
\begin{cases}
e^{-1/{x^2}} &; x\neq0\
0 &;x=0
\end{cases}$$