I have been solving differential equations by series solution method and I wanted to ask if there is a method to check if the series we get is periodic in its variable. In particular, if we get a solution for some ODE as
$$y \left( x \right) = \sum\limits_{n = 0}^{\infty} a_n x^n$$
where $a_n$ is governed by some recurrence relation, then can we predict if the solution is
Edit:-
I have performed some try: If the function $y$ is periodic of period, say $p > 0$, then it must satisfy $y \left( x \right) = y \left( x + p \right)$. This, when substituted in series and then comparing the coefficients of equal powers of $x$, we get the following result for $n \geq 0$,
$$a_n = \sum\limits_{k = 0}^{\infty} \binom{k + n}{n} a_{k + n} p^k$$
If we add another condition of periodicity $y' \left( x \right) = y' \left( x + p \right)$, we get the same result for $n \geq 1$.
If we just cancel out the first term, we get the following result for every $n \in \mathbb{N} \cup \left\lbrace 0 \right\rbrace$
$$\sum\limits_{k = 1}^{\infty} \binom{k + n}{n} a_{k + n} p^k = 0$$
Now, from here what can we say about the coefficients $a_n$ so that the function $y$ is periodic?
Edit:-
I have been thinking and the above equation tells us that the series $\sum\limits_{k = 1}^{\infty} \binom{k + n}{n} a_{k + n} p^k$ converges. For that to happen, a necessary condition is that
$$\lim\limits_{k \rightarrow \infty} \binom{k + n}{n} a_{k + n} p^k = 0$$
for each $n \in \mathbb{N}$. So, if the recurrence relation for $a_n$ is known from the series solution method of differential equations, can we find some conditions so that the above limit holds true? In particular, I have the recurrence relation as $a_0, a_1 \in \mathbb{R}$ are arbitrary, $a_2 = 0$ and for $n \geq 1$ we get
$$a_{n + 2} = \dfrac{1}{\left( n + 2 \right) \left( n + 1 \right)} \left[ a_n - 2 \sum\limits_{k = 0}^{\left[ \frac{n}{2} \right]} \dfrac{\left( n - 2k \right) a_{n - 2k} \left( -1 \right)^k}{\left( 2k + 1 \right)!} - \sum\limits_{k = 0}^{\left[ \frac{n}{2} \right]} \dfrac{a_{n - 2k} \left( -1 \right)^k}{\left( 2k \right)!} \right]$$
Can we find the rate of convergence of this recurrence relation and then comment on the limit mentioned above?