Okay We know that
$$if \ |f^{(n+1)}≤M|\ then;\\R_n ≤ M\frac{(x-c)^{n+1}}{(n+1)!}$$
Now we say that the series converges if $\lim R_n -> 0$
But isn't $(n+1)!>(x-c)^{n+1}$ after $\exists N$ such that $n>N$. I mean shouldn't $\lim \frac{(x-c)^{n+1}}{(n+1)!}=0$ as $n-> \infty$
Therefore $$0≤|R_n| ≤ |M\frac{(x-c)^{n+1}}{(n+1)!}|$$ $R_n -> 0$ by sandwich.
But how come some functions have remainder not going to zero?
Simple. They are not analytic functions, and as a consequence, the remainder does not go to zero.
The classic example is usually given by
$$f(x)=\begin{cases}e^{-1/x^2}&;x\ne0\\0&;x=0\end{cases}$$
whereupon the Taylor expansion at $x=0$ might give you a good chuckle. Similar examples of such cases may be found in the given link above.