Quick question on convergence 
Shouldn't the radius of convergence be defined as:
$$\frac{1}{R} = \lim_{n\rightarrow \infty} \left(a_n \right)^{\frac{1}{n}}$$
Not sure what they are doing above..
 A: The formula for the radius of convergence of a power series $\sum\limits_{n=0}^\infty a_n z^n$ that always works is the Cauchy-Hadamard formula
$$\frac{1}{R} = \limsup_{n\to\infty} \lvert a_n\rvert^{1/n},\tag{CH}$$
with the natural interpretation if the limes superior is $0$ or $+\infty$. It is not always the best method to determine the radius of convergence, however, so having other means to determine it at one's disposal is a good thing.
A frequently used formula, if all coefficients (from some point on) are nonzero, and if the limit in question exists, is the ratio formula (an easy variation of the familiar ratio test for absolute convergence of any series)
$$\frac{1}{R} = \lim_{n\to\infty} \left\lvert \frac{a_{n+1}}{a_n}\right\rvert.\tag{RF}$$
The chain of inequalities
$$\liminf_{n\to\infty} \left\lvert \frac{a_{n+1}}{a_n}\right\rvert \leqslant \liminf_{n\to\infty} \lvert a_n\rvert^{1/n} \leqslant \limsup_{n\to\infty} \lvert a_n\rvert^{1/n} \leqslant \limsup_{n\to\infty} \left\lvert \frac{a_{n+1}}{a_n}\right\rvert\tag{1}$$
shows that if the limit in $(\mathrm{RF})$ exists, then $\lvert a_n\rvert^{1/n}$ also converges, and to the same limit, whence the formula for the radius follows. But all inequalities in $(1)$ can be strict, so the Cauchy-Hadamard formula is strictly stronger than the ratio formula. If the limit of the ratios does not exist, then the upper respectively lower limits yield lower respectively upper bounds for the radius of convergence, which in some cases may be enough for the task at hand.
An obvious shortcoming of the ratio formula is that - in the form above, at least - it cannot be used for power series with infinitely many zero coefficients. We can in some cases amend that, and generalise the ratio formula to cover more cases, by splitting the power series into several equally-spaced parts,
$$\sum_{n=0}^\infty a_n z^n = f(z) = \sum_{r=0}^{k-1} f_r(z),$$
where
$$f_r(z) = \sum_{m=0}^\infty a_{m\cdot k+r} z^{m\cdot k + r}$$
for some $k > 1$. The case in hand is $k = 2$. The radius of convergence of the series of $f$ is then the minimum of the radii of convergence of the $f_r$. A part $f_r$ with all but finitely many coefficients vanishing has infinite radius of convergence, and can thus be ignored in the determination of the overall radius of convergence. If only finitely many $a_{m\cdot k + r}$ are zero and the limit of the ratios $\left\lvert \frac{a_{(m+1)k+r}}{a_{mk+r}}\right\rvert$ exists, then we have
$$\frac{1}{R_r^k} = \lim_{m\to\infty} \left\lvert \frac{a_{(m+1)k+r}}{a_{mk+r}}\right\rvert \tag{RF'}$$
as a generalisation of $(\mathrm{RF})$. If a sequence $\bigl(a_{mk+r}\bigr)_{m\in\mathbb{N}}$ of coefficients of an $f_r$ has infinitely many zero terms and infinitely many nonzero terms, it may be possible to split that up analogously to find the radius of convergence.
That splitting up of the power series only works, however, if the coefficients follow a regular pattern. It is frequently of use when dealing with odd or even functions, but rarely otherwise.
If the function represented by the power series is known, usually the most efficient way to determine the radius of convergence of the power series is to use the fact that it is the distance between the centre of the power series and the nearest singularity [Note: this does not hold for real-analytic functions defined on (intervals in) $\mathbb{R}$, one must consider the holomorphic extension of such functions defined in an open subset of $\mathbb{C}$].
