Determining the radius of convergence of a series that I don't know how to express in sigma Question:

Problem 1: Consider the ODE: $(1+x^3)y'' -6xy = 0$

(a) Compute the first $3$ nonzero terms of power series expansion about $x = 0$ for two linearly independent solutions.
(b) Use the ratio test to determine the radius of convergence of the series. Could your result
    have been predicted by inspection?


The solution for $y = a_0(1+x^3) + a_1(x + \frac{x^4}{2}- \frac{x^7}{14}+ \frac{x^{10}}{35}- \frac{x^{13}}{65}+\cdots)$
In order to use the ratio test $$ \lim_{n \rightarrow \infty}|\frac{a_{n+1}(x-x_0)^{n+1}}{a_n (x-x_0)^n}|$$
I need to know how to express $(x + \frac{x^4}{2}- \frac{x^7}{14}+ \frac{x^{10}}{35}- \frac{x^{13}}{65}+\cdots)$ in the sigma form. Am I right?
However, I cannot come up with a proper way to express the equation in a sigma form. Is there an alternate way to perform the ratio test? 

You should get a coefficient recursion ($a_{-1}=a_{-2}=0$)
$$
x^n: (n+2)(n+1)a_{n+2}+(n-1)(n-2)a_{n-1}-6a_{n-1}=0
\\~\\\iff (n+1)[(n+2)a_{n+2}+(n-4)a_{n-1}]=0
\\~\\\implies
a_{n+3}=-\frac{n-3}{n+3}a_n.
$$
so that $a_2=0$ and thus $a_{3k+2}=0$ for all $k$ and also $a_6=0$ and thus $a_{3k}=0$ for $k\ge 2$. 
The remaining part $\sum_ka_{3k+1}x^{3k+1}$ has radius of convergence equal $1$ by the quotient test.
From $(n+3)na_{n+3}=-n(n-3)a_n$ it follows that these products are constant with alternating sign, thus $a_{3k+1}=(-1)^{k-1}\frac{2}{(3k+1)(3k-2)}a_1$ giving the solution as
$$
y(x)=a_0(1+x^3)+2a_1\sum_{k=0}^\infty\frac{(-1)^{k-1}x^{3k+1}}{(3k+1)(3k-2)}.
$$

In the normalized form, $y''-\frac{6x}{x^3+1}y=0$ has analytical coefficients with convergence radius $1$, thus any solutions are also analytical with at least the same convergence radius, and at least one solution with exactly this radius.
