How to find radius of convergence with power series from differential equations So I have a question that says find the radius of convergence after I have found the power series solution of a given differential equation. I know to find the radius of convergence you take
$$ p=\lim_{n \rightarrow \infty} \left\lvert\frac{C_n}{C_{n+1}}\right\rvert $$
but, I don't understand where $C_n$ and $C_{n+1}$ come from. Could someone please explain how I would find $C_{n}$ and $C_{n+1}$?
 A: An elementary (and not particularly smart) approach could be:
Let's say the problem is $$\begin{cases}y'(x)=f(x,y(x))\\ y(0)=y_0\end{cases}$$
Since $C_n=\dfrac{y^{(n)}(0)}{n!}$, you "only" need to compute $y^{(n)}(0)$.
You know that $y'(0)=f(0,y_0)$.
Deriving in $x$ the first equation you get $$y''(x)=\frac{\partial f}{\partial x}(x,y(x))+y'(x)\frac{\partial f}{\partial y}(x,y(x))$$
Whence $y''(0)=\dfrac{\partial f}{\partial x}(0,y_0)+y'(0)\dfrac{\partial f}{\partial y}(0,y_0)$. Notice that you have already calculated $y'(0)$ the step before.
Keep deriving
$$y^{(3)}(x)=\\=
\dfrac{\partial^2 f}{\partial x^2}(x,y(x))+2y'(x)\dfrac{\partial^2 f}{\partial x\partial y}(x,y(x))+y''(x)\dfrac{\partial f}{\partial y}(x,y(x))+(y'(x))^2\dfrac{\partial^2 f}{\partial y^2}(x,y(x))
$$
Again, you can evaluate everything in $x=0,\ y=y_0$ and get $y^{(3)}(0)$.
The formulas rapidly worsen the more you derive, but perhaps the specific instance of the problem simplifies the calculations.
A: $p(x) = \sum c_n x^n$
The ratio test: a series converges (power-series of otherwise) the ratio of all successive elements greater than some N are less than 1. 
That is, if there exists an $N$ such that $n>N \implies |\frac{c_{n+1} x^{n+1}}{c_{n} x^{n}}|<1$ then the series converges.
$|\frac{c_{n+1} x^{n+1}}{c_{n} x^{n}}| = |\frac{c_{n+1}}{c_{n}}||x|$
$\lim_\limits{n\to\infty}$$|\frac{c_{n}}{c_{n+1}}| = r$ and $\frac {|x|}r < 1$
G. Sassatelli tells me that I might have misundrstood your problem.
How do you find the power series....
You say Suppose: $y = \sum_\limits{n=0}^{\infty} c_n x^n$
$y' = \sum_\limits{n=0}^{\infty} n c_n x^{n-1}$
And if necessary
$y'' = \sum_\limits{n=0}^{\infty} (n)(n-1) c_{n} n x^{n-2}$
Now you plug these into your original equation.
Set $c_0 = y(0)$
find and equation such that $c_{n+1} = f(c_n)$
And then try to find a general formula that for $c_n$
A simple example
$y' = y, y(0)= 1\\
y' = \sum_\limits{n=0}^{\infty} n c_n x^{n-1}\\
y' = \sum_\limits{n=1}^{\infty} (n+1)c_{n+1} x^{n}\\
\sum_\limits{n=1}^{\infty} (n+1)c_{n+1} x^{n} = \sum_\limits{n=0}^{\infty} c_{n} x^{n}\\
(n+1) c_{n+1} = c_n\\
c_n = \frac{c_0}{n!}\\
y = \sum_\limits{n=0}^{\infty} \frac{x^{n}}{n!}$
Now, in this case the radius of convergence is infinite, but that is not always true.
