Is there an algorithm to determine if a power series is periodic? We know that the sine function is periodic by its geometric definition. The Taylor/MacLaurin series expansion about 0 which is the basis of actual mechanisms for computing it is:
$$\sin(x) = \sum_{n=0}^\infty  \frac{(-1)^n}{(2n+1)!}x^{2n+1}$$
This series manages to be periodic with period $2\pi$ because it has an alternating sign.
Is there a way to tell if an arbitrary power series is periodic? More informally, if someone gave us the above summation for $\sin$ without telling us it was a trigonometric function, is there a procedure for discovering that it is periodic and finding the period?
Amendment: As pointed out in the answers, there is clearly no algorithm if the coefficients are allowed to be arbitrary, thus containing an unbounded amount of information. I should have asked "Under what limitations to a function defining the coefficients of a power series does there an exist algorithm for determining if the power series is periodic?"
In particular, if $f(n)$ is limited to a rational expression that would be accepted as a "closed-form" expression, as it is in the case of $\sin(x)$, does such an algorithm exist? If $f(n)$ is limited to being a simple arithmetic computation from $n$, can we determine if the function is periodic?
 A: Not really an answer so much as a comment on the other answers. Of course it's clear that there is no Turing machine that does this, since the answer is not determined  by finitely many coefficients.
But asking for a Turing machine is surely not what the question means! For example, there is a "test" to determine whether the radius of convergence is infinite; check whether $\limsup|a_n|^{1/n}=0$. That can't be done by a Turing machine but it seems clear to me that it does  count as a "test" in the sense in which the OP meant the question.
About the question:
It seems very unlikely that there does exist a test for periodicity in terms of the coefficients. Of course one cannot say for sure without a suitable definition of what sort of "test" one wants...
A: Edit: This Answer Applied to the Original Question, which talked about power series in general. When restricting to power series that can be expressed in a finite amount of information, this argument no longer holds.

No. There is no way to tell if an arbitrary power series is periodic or not. We can only do this consistently for a subset of power series. Every power series is written as:
$$\sum_{n=0}^\infty  f(n)\cdot x^n$$
The information about the power series is completely contained within the function $f(n)$. Now, since we're talking about arbitrary power series, $f$ can be any function from the function space:
$$ \mathbb{N} \mapsto \mathbb{R} $$
OK, so let's say there was a consistent and complete method we could use to tell, given this $f$ whether or not the associated power series was periodic. Then we'd be able to encode a Turing machine which, given any sequence $f(n)$ would return a "yes" or "no" answer saying if the function was periodic or not.
Now, this machine would have to terminate in a finite number of steps. This in turn means it could only read a finite sample of values from our $f(n)$ e.g. it might just read three values $\{f(1), f(2778), f(99)\}$. Now, of course, the number of values could depend on $f$, but this won't matter- we'll still get our contradiction, just wait.
So, suppose we pass to our machine the periodic function $f$. Naturally, it will return "yes", telling us that we have a periodic function. But as argued above, it will only have read a finite amount of values for $f(n)$. Then let $i$ be one of the infinite natural numbers for which $f(i)$ was not read by our algorithm. And here's the killer stroke. Define:
$$g(n)=\left\{\begin{array}{cc}
f(n) & n \neq i \\
f(n) + 1 & n=i
\end{array}\right.$$
Clearly, our machine, being deterministic, will give us the same result- because it will never read $g(i)$, so it's result will be the exact same as $f(i)$. But it's also clear that $g$ is not periodic. So our machine will return "yes" for a non-periodic function, meaning it's inconsistent, contrary to our assumption that it was consistent and complete.
In fact: This proof works even though we're giving the Turing machines the unreasonable power of being able to completely read a real number to infinite precision.
Aside: I'd say that probably what you're looking for is some subset of functions $f(n)$ that can be specified with a finite amount of information. But that hypothesis would need to be made rigorous.
A: As it turns out, this is an undecidable problem. Here is the argument:
First, let's start with the fact that it is impossible to know whether or not an arbitrary series converges or diverges in general. Recall that by definition of a periodic function $f(x+T)=f(x)$ where $T$ is the period. 
Let's take an arbitrary function $f\in C^{\omega}$. The Taylor series converges $\forall x\in \mathbb{R}$. Let $S(T)$ denote the Taylor series for $f(x+T)$ and $S$ denote the Taylor series for $f(x)$. Let $\tilde S=S(T)-S$. We need to show the series $\tilde S$ converges to $0$. This is a necessary requirement for periodicity of $f$. 
However, since we do not know a priori where the function $f$ is periodic, we can "test" it by the above argument. A priori, we do not know if the above series $\tilde S$ converges, so there is no way to tell in general. 
A: Consider the following absolutely convergent series
$$ f(x)=\sum\limits_{k=0}{a[k]{{x}^{k}}} $$
Assume that the period, $\Lambda =1$  (this can be relaxed to an arbitrary real number later). The restriction
$$ f(x)-f(x+1)=0 $$
generates the follow set of conditions on the coefficients a[k]
$$ (U(\infty )-I)\left[ a(k) \right]=[0] $$
Where U is the (infinite dimensional) upper triangular Pascal Matrix, and I is the identity matrix, and a[k] is a column vector formed from the coefficient list. This matrix equation is just another way to express the periodicity constraint (with period = 1). Note that the matrix is singular so there are an infinite set of coefficient lists that satisfy this equation.
Each row*a[k] gives a constraint that the a[k]'s must satisfy if the related series is periodic. The general constraint is
$$ \sum\limits_{k=j}{\left( \begin{matrix}
   k, j-1\end{matrix} \right)a[k]=0} $$
where the term in parenthesis is the Binomial Coefficient. For example, the first two constraint equations are
$$ \begin{matrix}
   \sum\limits_{k=1}{a[k]=0}  \\
   \sum\limits_{k=2}{k_{{}}^{{}}a[k]=0}  \\
\end{matrix} $$
In the case where the period is not equal to 1, a simple rescaling of x is implied, which can be made explicit by appropriately inserting an additional scaling matrix (diagonal in powers of $\Lambda$) into the above matrix equation.
Hence the suggested procedure (algorithm) would be to apply each constraint equation in turn until failure, or if the a[k] are sufficiently well behaved the general case (for arbitrary j) would need to be proved. 
