Radius of convergence and ratio test My book says that given a power series $\sum_{n = 1}^\infty c_nz^n$ where the $c_n$ are complex the radius of convergence of the series is $\dfrac{1}{L}$ where $L = \lim \sup \sqrt[n]{|c_n|}$.  So the radius of convergence is defined using the root test.  Since we can also apply the ratio test, is it fair to say, that the radius of convergence is $\dfrac{1}{L}$ where $L = \lim \sup \bigg|\dfrac{c_{n+1}}{c_n}\bigg|$?
 A: Consider the series 
$$1+2z+3^2z^2+2^3z^3+3^4z^4+2^5z^5+3^6z^6+\cdots.$$
The required limsup of the $n$-th roots is $3$, and the radius of convergence is $\frac{1}{3}$. 
Now look at the ratios $\dfrac{c_{n}}{c_{n+1}}$. You will notice they are very badly behaved, and tell us essentially nothing about the radius of convergence. 
A: The ratio test is strictly weaker than the root test in the sense that if the ratio test gives an answer, then so does the root test and they are the same.  However, as the other answers show, there are many series for which it gives no answer.  Even though you are using $\limsup$ rather than $\lim$, the limit ratio may fail to be defined if the series contains infinitely many zeros, which make the corresponding ratios undefined, as in julien's answer.
In general, the ratio test is easier for common series because they involve algebraic operations for which computing the $n$th roots is difficult but for which the ratios can be computed pretty easily.  But for theoretical purposes it is inferior.
A: The second formula does not hold.  The ratio test only gives us bounds on the radius of convergence and only if applicable. To begin with, it requires the $c_n$'s be eventually nonzero. For instance
$$
\frac{1}{1-z^2}=\sum_{n\geq 0}z^{2n}.
$$
has radius of convergence $1$. But $c_{2n}=1$ and $c_{2n+1}=0$. So the following is not defined:
$$
\not\exists\; \limsup \frac{|c_{n+1}|}{|c_{n}|}.
$$
Now if the $c_n$'s are eventually nonzero, note that
$$
\limsup \frac{|c_{n+1}z^{n+1}|}{|c_nz^n|}=|z|\limsup \frac{|c_{n+1}|}{|c_{n}|}.
$$
and
$$
\liminf \frac{|c_{n+1}z^{n+1}|}{|c_nz^n|}=|z|\liminf \frac{|c_{n+1}|}{|c_{n}|}.
$$
A series $\sum a_n$ with $a_n\geq 0$ converges whenever $\limsup a_{n+1}/a_n<1$ and diverges whenever $\liminf a_{n+1}/a_n>1$. Also, note that
$$
\limsup a_n=\frac{1}{\liminf \frac{1}{a_n}}.
$$
Hence our series converges if
$$
|z|\lt \frac{1}{\limsup\frac{|c_{n+1}|}{|c_n|}}=\liminf\frac{|c_n|}{|c_{n+1}|}
$$
and diverges if
$$
|z|\gt \frac{1}{\liminf\frac{|c_{n+1}|}{|c_n|}}=\limsup\frac{|c_n|}{|c_{n+1}|}
$$
It follows that the radius of convergence $R$ has
$$
\liminf \frac{|c_{n}|}{|c_{n+1}|}\leq R\leq \limsup \frac{|c_{n}|}{|c_{n+1}|}.
$$
For an example where this is strict, take $c_{2n}=2^n\cdot 3^n=6^n$ and $c_{2n+1}=2^n\cdot 3^{n+1}=3\cdot 6^n$. Then
$$
\frac{1}{R}=\limsup \sqrt[n]{|c_n|}=\sqrt{6}<3=\limsup \frac{|c_{n+1}|}{|c_n|}.
$$
Note that if $\lim \frac{|c_{n+1}|}{|c_n|}$ exists, then upper and lower limits coincide and we get a convenient formula
$$
R=\lim \frac{|c_{n}|}{|c_{n+1}|}.
$$
