Given the Taylor series $\sum a_k (x - x_0)^k$ of an analytic function, it is possible to determine whether the function is periodic more-or-less directly from the coefficients $a_0, a_1, \ldots$ of the series (equivalently, the derivatives $f'(x_0), f''(x_0), \ldots$ of $f$ at $x_0$)?
(For simplicity, I'm happy to restrict if necessary to the case that the series converges on all of $\mathbb{R}$. Note too that it's no real loss of generality to take $x_0 = 0$.)
(This was motivated by this related question.)
Just to summarise as an answer the line of thought in the comments: determining whether a function is periodic requires looking at an infinite number of its coefficients, since any finite Taylor series can be the initial segment of a periodic function, and indeed a periodic function with arbitrary period. So there is no decision algorithm for the question if the Taylor series is provided by an oracle giving one coefficient at a time.
On the other hand if you can evaluate $f(q)$ at every rational number, you can find all members of the discrete countable set of zeros of $f$ in $\mathbb{R}$; then testing whether $f(x-z) - f(x) \equiv 0$ at each such zero $z$ gives you an infinitary method relying only on the most obviously natural operations on the whole series $(a_n)$ and making a countable number, (as an ordinal, $2 \cdot \omega$) of decisions, to come to an answer.
Clearly this cannot count as "reasonably direct" because that would be a deeply uninteresting answer to an interesting question. So to make the question precise we would have to frame it in a way that lies between these two cases. What that should be, I have no idea.