Ratio test vs Cauchy-Hadamard I'm sorry if this had been asked before, but there are many other questions that make it difficult to find an answer to my own.
I know that the radius of convergence of a power series can be determined by the Cauchy-Hardamard theorem, but what is the relationship between the value given therein and using the ratio test find the supremumn of all $x$ (i.e. $\sum a_n x^n$ ) such that the ratio test proves convergence. Many sources prescribe using the ratio test to find the radius of convergence. I suspect this (ratio) method be weaker?
What is the relationship between these methods?
 A: The ratio test only considers the ratio of adjacent terms and so is useless (at least naively) in general when ratio could get unbounded.  For example, it would naively fail to come up with a nonzero radius for the very simple MacLaurin series of $$\cos(2z)+\sin(3z)$$, because the coefficient of $z^n$ are
$$
\begin{cases}
\pm\dfrac{2^n}{n!} & \text{ if }n\equiv 0\pmod2\\
\pm\dfrac{3^n}{n!} & \text{ if }n\equiv 1\pmod2
\end{cases}
$$
and so $\left|\dfrac{a_{2n+1}}{a_{2n}}\right|=\dfrac{3^{2n+1}}{2^{2n}(2n+1)}\to\infty$, causing the ratio test to only give meaningful result at $z=0$.  In this case it is easy to come up with an ad hoc fix of splitting the odd and even terms but what about with long string of constant coefficients and then have large jump, then stay constant for even longer, then jump at even larger ratio, etc.?
If the (limit) ratio test succeeds, then it would give the same result as Cauchy-Hadamard, for the simple reason that if $|a_{n+1}/a_n|<r$ for all $n$ (sufficiently large) then $\lim |a_n|^{1/n}$ exists and is $\leq r$.
A: The formula for the convergence radius deduced from the Cauchy-Hadamard theorem looks at every coefficient $a_n$ seperately. The formula deduced from the ratio test analyses the ratio of two consecutive coefficients. 
The ratio formula is generally easier to use, but can only be applied when the values of the $a_n$ are different from zero. For instance, to determine the radius of convergence of series like
$$\sum_{n=0}^\infty \frac{(-1)^{2n}}{(2n)!}x^{2n} \quad\text{ or }  \quad \sum_{n=1}^\infty 2^n x^{n!} $$
we have to use the Cauchy-Hadamard formula.
