<s>[INST] Generating function for Pell numbers Problem: The Pell numbers $p_n$ are defined by the recurrence relation \begin{align*} p_{n+1} = 2p_n + p_{n-1} \end{align*} for $n \geq 1$. The initial conditions are $p_0 = 0$ and $p_1 = 1$. 
a) Determine the generating function \begin{align*} P(x) = \sum_{n=0}^{\infty} p_n x_n \end{align*} for the Pell numbers. What is the radius of convergence?
b) Determine (on the basis of the found generating function) an explicit formula for $p_n$.
Solution: Together with the initial conditions we have \begin{align*} P(x) &= 0 + x + \sum_{n=2}^{\infty} p_n x_n \\ &= x + \sum_{n=1}^{\infty} p_{n+1} x^{n+1} \\ &= x + \sum_{n=1}^{\infty} (2p_n + p_{n-1}) x^{n+1} \\ &= x + \sum_{n=1}^{\infty} 2p_n x^{n+1} + \sum_{n=1}^{\infty} p_{n-1} x^{n+1} \\ &= x + 2x \sum_{n=1}^{\infty} p_n x^n + x \sum_{n=1}^{\infty} p_{n-1} x^n.     \end{align*}
Now we look at each series separately to see what we've got. The first series on the left expands as $(x + p_2 x^2 + p_3 x^3 + ...)$. This is nothing but the original $P(x)$ (because we can ignore the constant term $0$ right?). So for the first series we've got $2x P(x)$. 
The second series on the right expands as $(0 + x^2 + p_2 x^3 + ...)$. We can factorize $x$ out such that we get $P(x)$ again. So everything together we have:
\begin{align*} P(x) = x + 2xP(x) + x^2 P(x), \end{align*} which gives us \begin{align*} P(x) (1-2x-x^2) = x, \end{align*} or \begin{align*} P(x) = \frac{x}{(1-2x-x^2)}. \end{align*}
But then I don't know how to determine the radius of convergence, and how to do b).
Any help would be appreciated.
 [/INST] 
The radius of convergence is the distance to the nearest singularity

For $(b)$ you can advance as (based on your calculations) using partial fraction

$$P(x)= \frac{x}{(1-2x-x^2)}= \frac{A}{x-a} + \frac{B}{x-b} $$

where $a,b$ are the roots of $1-2x-x^2$ and $A$ and $B$ need to be determined. The calculations gives: 
$$\frac{-1-\sqrt{2}}{2 \sqrt{2} (x+\sqrt{2}+1)}+\frac{1-\sqrt{2}}{2 \sqrt{2} (x-\sqrt{2}+1)}$$
Then use the geometric series expansion.
Note: 

$$ \frac{1}{a-t} = \frac{1}{a}\sum_{n=0}^{\infty} \frac{t^n}{a^n}  $$

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