What does the range of convergence in Maclaurin series mean? I tried to calculate the following Maclaurin series:
$$
f(x) = \sqrt{1+x^2} = 1+\frac{x}{2}+\sum_{n=2}^{\infty}(-1)^{n+1}\frac{x^{2n}(1)(3)...(2n-3)}{2^n n!}
$$
With Ratio Test, I found the range of convergence is $x\in(-1,1)$:
$$
\lim_{n\to\infty} |\frac{x^{2(n+1)}(1)(3)...(2n-3)(2n-1)}{2^{n+1} (n+1)!} \frac{2^{n}n!}{
x^{2n}(1)(3)...(2n-3)}| = \lim_{n\to\infty}|\frac{x^2 (2n-1)}{2(n+1)}| = x^2(<1)
$$ 
To me, it means that The Maclaurin series only converge to a solution (and thus only valid) when $x\in(-1,1)$. However, since the $x$ in the original function $f(x) = \sqrt{1+x^2}$ can take on any value ($x\in\mathbb{R}$), the Maclauin series cannot give us a solution when $x \leq -1$ or $x \geq 1$. This is the case even though the original function $\sqrt{1+x^2}$ can take on these values. Is my interpretation correct?   
Edit: Let's limit our discussion to $x \in \mathbb{R}$. Thanks!
 A: The analysis is correct.  The function $\sqrt{1+x^2}$ has the series representation
$$\sqrt{1+x^2}=\sum_{n=0}^\infty\binom{1/2}{n}x^{2k} \tag 1$$
that is valid for $|x|\le 1$ where 
$$\binom{1/2}{n}=\frac{(-1)^{n-1}(2n)!}{(n!)^2\,4^n\,(2n-1)}$$
The series in $(1)$ diverges for $|x|>1$.

Note that we can develop quite easily a series representation that is valid for $|x|\ge 1$ by writing $\sqrt{1+x^2}=|x|\sqrt{1+x^{-2}}$.  Then, we have
$$\sqrt{1+x^2}=|x|\sum_{n=0}^\infty\binom{1/2}{n}x^{-2k} \tag 2$$
which is valid for $|x|\ge 1$.  The series in $(2)$ diverges for $|x|<1$.

NOTE: Convergence at the end points
One way to show that the series in $(1)$ and $(2)$ converge at $|x|=1$ is to use Stirling's Formula
$$n!=\sqrt{2\pi n}\left(\frac ne\right)^n\left(1+O\left(\frac1n\right)\right)$$
Then, we see that 
$$\begin{align}
\binom{1/2}{n}&\sim (-1)^{n-1}\frac{\sqrt{2\pi (2n)}\left(\frac {2n}{e}\right)^{2n}}{(2\pi n)\left(\frac {n}{e}\right)^{2n}4^n(2n-1)}\\\\
&=\frac{(-1)^{n-1}}{\sqrt{\pi n}(2n-1)}
\end{align}$$
And the series $\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{\sqrt{\pi n}(2n-1)}$ converges absolutely. 
