Walter Van Assche gives a modern account of Pade Approximation, a variant of these were used in the proof that e is transcendental.

In the case that $f(z)$ is the Cauchy transform of a compactly supported measure $\mu(x)$,

\[ f(z) = \int \frac{1}{z-x} d\mu(x) \]

Then $P_n(x)$ is an orthogonal polynomial with respect to $\mu(x)$ while

\[ Q_{n-1}(z) = \int_a^b \frac{P_n(z) - P_n(x)}{z-x} \, \mu(x) \]

Then we approximate $f(z)$ as a rational function

\[ P_n(z) - f(z) Q_{n-1}(z) = \int_a^b \frac{P_n(x)}{z-x}\, d\mu(x) \]

Then there exists a function $r(z)$ such that the pointwise convergence of the Pade approximants is exponential as we move along the diagonal.

\[ \lim_{n \to \infty} \left| f(z) - \frac{Q_{n-1}(z)}{ P_n(z)}\right|^{1/n} = \frac{1}{r^2} \]

**ORIGINAL ANSWER**
The coefficients of the series $a_1, a_2, a_3, \dots$ are an infinite amount of data. The radius of convergence is a property of this sequence $1/R = \limsup |a_n|^{1/n}$ using the root test.

A paper by Hubert S. Wall, dating back to 1929, relates Pade approximants the Stieltjes moment problem and continued fractions.

**EDIT**: My guess is Pade approximant is *not* always better than Taylor series.

As a counter example let's find $[0/1]_{z+1}$:

\[ z + 1 \approx \frac{1}{1-z} \mod z^2 \]

Taylor series is exact and the 1-1 approximant diverges.

In your example, $\sqrt{\frac{1+\frac{z}{2} }{1+2z}} \to \frac{1}{2}$ for large values and likely the 1,1 approximant does the same, making it a good global fit. The 0,2 or 2,0 approximants will have different global behavior. I was unable to find any precise comparison, overall.

Continued fractions are known to be "best approximations" in a certain sense.

A best rational approximation to a
real number x is a rational number
d/n, d > 0, that is closer to x than
any approximation with a smaller
denominator.

Wikipedia's example is to approximate

\[ 0.84375 = \cfrac{1}{1 + \cfrac{1}{5 + \cfrac{1}{2 + \cfrac{1}{2}}}} = [0;1,5,2,2]\]

We can stop in the middle of this process to get the closes fraction given an upper bound on the denominator.

\[ 0.84375 \approx 1,\frac{5}{6}, \frac{ 11}{13 } , \frac{27}{32}\]

The farey fractions up to denominator 6 are

$$0,\frac{1}{6}, \frac{1}{5},\frac{1}{4},\frac{1}{3},\frac{2}{5},\frac{1}{2},\frac{3}{5},\frac{2}{3},\frac{3}{4},\frac{4}{5},\mathbf{\frac{27}{32}},\frac{5}{6} 1$$

Pade approximants are best approximations of functions and can be calculated used a kind of continued fraction.

You get an approximation $\frac{p(x)}{q(x)}$ for given degrees $m = \deg p, n = \deg q$. You are trying to find a polynomial greatest common divisor between your Taylor series and a monomial,

\[ \gcd(T_{m+n}(x), x^{m+n+1} ) \]

You can do this Euclid algorithm doing polynomial long division and taking the remainder at each step:

\[ \frac{p(x)}{q(x)} \equiv T_{m+n}(x) \mod x^{m+n+1} \]

Here is the table for $e^z$ from Wikipedia: (also Pade approximant to exponential function)

\[ \begin{array}{c||c|c|c} & 0 & 1 & 2 \\\\ \hline \hline
0 & \frac{1}{1} & \frac{1}{1-z} & \frac{1 }{1 - z + \frac{1}{2}z^2 } \\\\ \hline
1 & \frac{1+z}{1} & \frac{1+ \frac{1}{2} z}{ 1- \frac{1}{2} z} & \frac{1 + \frac{1}{3}z }{ 1 - \frac{2}{3}z + \frac{1}{6}z^2 } \\\\ \hline
2 & \frac{1 - z + \frac{1}{2}z^2 }{1 } & \frac{ 1 - \frac{2}{3}z + \frac{1}{6}z^2 }{1 + \frac{1}{3}z } & \frac{ 1+ \frac{1}{2} z + \frac{1}{12} z^2}{ 1- \frac{1}{2} z + \frac{1}{12} z^2 } \end{array}
\]

Let's try to work out the steps for m=2, n=2 (not in Wikipedia). This involves the GCD of the 4th Taylor polynomial $1 + z + \frac{1}{2}z^2 + \frac{1}{6}z^3 + \frac{1}{24}z^4$ and $z^{5}$.

\[ \frac{1}{24} z^5 = (z-4)\left(1 + z + \frac{1}{2}z^2 + \frac{1}{6}z^3 + \frac{1}{24}z^4\right) + \left(4 + 3z + z^2 + \frac{1}{6}z^3 \right) \]

\[
1 + z + \frac{1}{2}z^2 + \frac{1}{6}z^3 + \frac{1}{24}z^4 =
\left( \frac{1}{4}z - \frac{1}{2} \right)\left( 4 + 3z + z^2 + \frac{1}{6}z^3 \right)
+ \left( 3 + \frac{3}{2}z + \frac{1}{4}z^2 \right) \]

\[
4 + 3z + z^2 + \frac{1}{6}z^3 =
\frac{2}{3}z \left( 3 + \frac{3}{2}z + \frac{1}{4}z^2 \right) + (z+4) \]

\[
3 + \frac{3}{2}z + \frac{1}{4}z^2 = \left( \frac{1}{4}z + \frac{1}{2}\right)(z+4) + 1\]

Using the 1st two long divisions, we get the (2,2) Pade approximant.

\[ 1 + z + \frac{1}{2}z^2 + \frac{1}{6}z^3 + \frac{1}{24}z^4 \approx
\frac{ 3 + \frac{3}{2}z + \frac{1}{4}z^2 }{(z-4) \left( \frac{1}{4}z - \frac{1}{2} \right)+1 } = \frac{ 1+ \frac{1}{2} z + \frac{1}{12} z^2}{ 1- \frac{1}{2} z + \frac{1}{12} z^2 } \]

Alternatively compare coefficients of your rational approximation and polynomial
$a_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4 \approx \frac{p_0 + p_1 x + p_2 x^2}{q_0 + q_1 x + q_2 x^2 } $
Then you can solve the system of equations:
\begin{eqnarray*}
a_0 &=& p_0 \\\\
a_1 + a_0 q_1 &=& p_1 \\\\
a_2 + a_1 q_1 + a_0 q_2 &=& p_2 \\\\
a_3 + a_2 q_1 + a_1 q_0 &=& 0 \\\\
a_4 + a_3 q_1 + a_2 q_0 &=& 0
\end{eqnarray*}

Cramer rule gives you the correct fraction at the end:

\[ \frac{ \left|\begin{array}{ccc} a_1 & a_2 & a_3 \\\\ a_2 & a_3 & a_4 \\\\
a_0 x^2 & a_0 x + a_1 x^2 & a_0 + a_1 x + a_2 x^2 \end{array} \right|} { \left|\begin{array}{ccc}
a_1 & a_2 & a_3 \\\\ a_2 & a_3 & a_4 \\\\
x^2 & x & 1 \end{array} \right|}
= \frac{ \left|\begin{array}{ccc} 1 & 1/2 & 1/6 \\\\ 1/2 & 1/6 & 1/24 \\\\
x^2 & x + x^2 & 1 + x + \frac{1}{2} x^2 \end{array} \right|} { \left|\begin{array}{ccc}
1 & 1/2 & 1/6 \\\\ 1/2 & 1/6 & 1/24 \\\\
x^2 & x & 1 \end{array} \right|}\]