find the radius of convergence? Determine the radius of convergence of the following  power series.
$$
\sum_{n=0}^{\infty}{a_n}(x-2017)^n\: \text{ with }\: a_n = \begin{cases} 1/2\:\text{ if  $n$  is even} \\ 1/3\:\text{ if  $n$ is odd} \end{cases}
\tag{1}\label{1}
$$
As far as I know.... to find the radius of convergence it must be true that 
$$
\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| < 1\tag{2}\label{2}
$$
But now I'm  stuck here  ... assuming that \eqref{2} holds, how  can I find the radius  of convergence?
 A: Frank Lu has already answered almost comprehensively in the comments to this question: however I think it is a nice idea to precise some points. The standard way of calculating the radius of convergence of a power series is perhaps by using the Cauchy-Hadamard formula ([1], remark 5.2, p. 517)
$$
R=\left(\limsup_{n\to\infty} \sqrt[n]{|a_n|}\right)^{-1}\tag{I}\label{I}
$$
Due to the properties of $\limsup$, the value of $R$ calculated by formula \eqref{I} always exists.
The formula for $R$ involving the inverse limit quotient as calculated for the ratio convergence test for the series $\sum a_n$, i.e.
$$
R=\lim_{n\to\infty}\frac{|a_n|}{|a_{n+1}|}\tag{II}\label{II}
$$
works only if the ratio ${|a_n|}/{|a_{n+1}|}$ converges ([1], theorem 5.4, p. 518), and in this case obviously gives the same value as calculated by \eqref{I}. In the question posed by the OP, limit \eqref{II} does not exists since the sequence is oscillating, therefore the correct answer should be calculated only by formula \eqref{I}, i.e.
$$
R=\left(\limsup_{n\to\infty} \sqrt[n]{|a_n|}\right)^{-1}=1
$$
[1] Emanuel Fisher (1983), "Intermediate Real Analysis", Springer Verlag.
A: The series can be decomposed in two convergent parts,
$$\frac12\sum_{n=0}^\infty(x-2017)^{2n}+\frac13\sum_{n=0}^\infty(x-2017)^{2n+1}$$
and both have radius of convergence $1$. For $x=2016$ and $x=2018$, they diverge and so does the sum (there is no cancellation).
