What does it really mean for the power series of a function to converge? So I was watching a Khan Academy video about power series https://www.khanacademy.org/math/integral-calculus/sequences_series_approx_calc/power-series-algebra/v/rep-function-with-geometric-series;
It had the function $h(x)=\frac 1 {3+x^2}=\frac 1 {3(1+ \frac {x^2} 3)}=\frac {\frac 1 3} {1-(\frac {-x^2} 3)}$.
We can turn this into the power series $\sum_{n=0}^{\infty} \frac 1 3 ({\frac {-x^2} 3})^n$, which we know converges when $\lvert {\frac {-x^2} 3} \rvert <1$. After some algebraic manipulations we end up with $-\sqrt{3}<x<\sqrt{3}$. So when x is in that interval, the series which represents the function $h(x) = \frac 1 {3+x^2}$ converges. So far so good...
But here's where I get confused: the function $y = \frac 1 {3+x^2}$ is defined everywhere. If the series above is supposed to be representing this function, why does it only converge on the small interval $(-\sqrt{3}, \sqrt{3})$? In other words, what does the convergence of a power series actually mean about the function it approximates, if not where the function is defined?
 A: The convergence of the power series of a functions converging in some given domain means that within that domain the function and the series are identical as functions, i.e.: for each value within the convergence domain, substituting into the power series not only gives us a convergent series but also that it converges to the value of the function in that point...but only within the convergence domain! 
Thus, for example:
$$\frac14=h(1)=\frac1{3+1^2}=\frac13\sum_{n=0}^\infty(-1)^n\frac{1^{2n}}{3^n}=\frac13\sum_{n=0}^\infty\frac{(-1)^n}{3^n}$$
Observe that the resulting series in this case is an easy-to-calculate-otherwise geometric series:
$$\sum_{n=0}^\infty\frac{(-1)^n}{3^n}=\frac1{1+\frac13}=\frac34$$
A: well the power series will be valid in the complex plane as well. It will generally go as far as it can in $C$ without hitting a singularity. In this case, $\sqrt{3} i , -\sqrt{3} i$ are singularities so the power series has to break down as $|z|$ approaches $\sqrt{3}.$
A: As you said, in the real domain, the function $h(x)=\frac 1 {3+x^2}$ is defined everywhere. The problem arises when you build the series expansion which is an approximation of the function around the point at which it has been built and the series converges only inside its radius of convergence.
For illustration purposes, plot on the same graph $$h(x)=\frac 1 {3+x^2}$$ $$g(x)= \frac{1}{3}-\frac{x^2}{9}+\frac{x^4}{27}-\frac{x^6}{81}+\frac{x^8}{243}-\frac{x^{10}}
   {729}$$ for $-2 \leq x \leq 2$. You should notice that they are very close to eachother for $|x| \leq 1.5$ but they become very different for larger values of $|x|$. Increasing the number of terms used in the series does not qualitatively change the problem.
