I am to use the method of Frobenius to find a series solution of the ODE
$$xy^{\prime\prime} + y = 0$$
I have derived the indicial equation $$r^2 -r = 0$$ from which we see that either $r = 0$ or $r= 1$. We then propose the Frobenius series solution $$y(x) = \sum_{k=0}^{\infty}a_kx^{k+r}$$ and upon differentiation, we have $$y^{\prime\prime}(x) = \sum_{k=0}^{\infty}(k+r)(k+r-1)a_kx^{k+r-2}$$ and so substituting this into our ODE gives $$\sum_{k=0}^{\infty}(k+r)(k+r-1)a_kx^{k+r-1}+\sum_{k=0}^{\infty}a_kx^{k+r}=0$$ which can also be written as $$r(r-1)a_0x^{r-1} + \sum_{k=1}^{\infty}(k+r)(k+r-1)a_kx^{k+r-1}+\sum_{k=1}^{\infty}a_{k-1}x^{k+r-1}=0$$ Thus $a_0$ can be chosen arbitrarily so long as it satisfies certain conditions and, in general, for $n\geqslant 1$ we have the recurrence relation
$$(n+r)(n+r-1)a_n + a_{n-1}=0$$
The question is, how do I progress from here to obtain a series solution? The recurrence relation seems quite complicated so I am not sure that it is correct. I substituted the values of $r$ into this relation and I was a little discouraged by the number of factorials in there.