Taylor Polynomial converges to the original function? If $$P_n(x)=x-\frac{x^2}{2}+\frac{x^3}{3}-..+\frac{x^{2n+1}}{2n+1}$$ (It's taylor series of $\ln(1+x)$ near x=0. Then can I say that:
$\lim_{n\to\infty}{P_n(x)}=\ln(1+x)$, please explain why or why not.
 A: Hint: There is specifically no need to consider the final exponent as odd $(2n + 1)$. But assuming that it is so, consider the following expansion which holds $$\frac{1}{1 + t} = 1 - t + t^{2} - \cdots + (-1)^{2n}t^{2n} + (-1)^{2n + 1}\frac{t^{2n + 1}}{1 + t}$$ for all $t \neq -1$. Integrate this in interval $[0, x]$ and try to approximate the integral $$I_{n} = \int_{0}^{x}\frac{t^{2n + 1}}{1 + t}\,dt$$ It will be found that $I_{n} \to 0$ as $n \to \infty$ if and only if $-1 < x \leq 1$.
A: You cannot say this. For instance, $P_n(3)$ doesn't even converge as a series, because the $n$th term goes to infinity. 
You can say that for some small interval (possibly a single point!) around the center point of the series, it converges to $f$. You might guess that if the series converges, then it converges to the original function from which you built the Taylor series, but even that is not true. There's a function $h$ whose value and all its derivatives at the origin are zero, so the Taylor series is the zero polynomial, which converges everywhere, but the function itself is not zero. It's
$$
h(x) = \begin{cases}  0 & x = 0 \\
e^{-\frac{1}{x^2}} & x \ne 0
\end{cases}.
$$
Determining the interval of convergence, and whether the resulting function equals the original, is part of what you should learn as you study Taylor series. 
