Can every recurrence relation be solved?

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Motivation

A possible way to solve an ODE is to express the solution as: $y= \sum_{n=0}^\infty a_nx^n$. We substitute in the ODE and then calculate the coefficients $a_n$.

For example, $y''+y=0$ will give: $$a_{n+2}=-\frac{a_n}{(n+2)(n+1)}$$

This gives:

$$a_{2n}=\frac{(-1)^n}{(2n)!}a_0$$

$$a_{2n+1}=\frac{(-1)^n}{(2n+1)!}a_1$$

But I've read (Boyce, di Prima):

the determination of a formula for an in terms $a_0$ and $a_1$ will be fairly complicated, if not impossible.

Is there any example where we can't isolate $a_n$ (linear or non-linear)? I'm not sure if he requires the solution to be a closed-form.

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Solve them, as in finding a "nice" formula, definitely not. If you check the book "A = B" by Petkovsek, Wilf and Zeilberger you'll find ways of checking if certain sums (which can rather trivally reduced into recurrence equations) have closed form or not. Not surprisingly, many have no closed form.