TITLE: Infinite sum of elements in a finite field
QUESTION [6 upvotes]: This is a bit of a curiosity that intrigues me. Let $p$ be a prime and consider the sum of reciprocals of squares divisible by $p$. This is just
$$
\dfrac{1}{p^2}\sum_{n=1}^\infty \dfrac{1}{n^2} = \dfrac{\pi^2}{6p^2}.
$$
Then the sum of the reciprocals of squares not divisible by $p$ is 
$$
S = \sum_{\substack{n=1\\ p \nmid n}}^\infty \dfrac{1}{n^2}= \sum_{n=1}^\infty\dfrac{1}{n^2} - \dfrac{\pi^2}{6p^2} = \dfrac{\pi^2}{6} \left( 1 - \dfrac{1}{p^2} \right).
$$
Now here's a bit of trouble (for me at least). We can view the terms of $S$ as elements in the field $\mathbb{F}_p$. Indeed, there's no division by the characteristic of the field, $p$. Since the field is closed under addition, the addition of each term is an element of the field. Therefore the addition of every element in the sum gives an element in the field. However $\frac{\pi^2}{6} ( 1 - \frac{1}{p^2} )$ is clearly not an element of $\mathbb{F}_p$.
I think the problem lies in that the limit of the partial sums does not exist modulo $p$, at least in this case. However, are there instances where it makes sense to add an infinite number of elements in a finite field, say perhaps when the sequence of terms has finite period? Thanks

REPLY [5 votes]: Alex S already explained the problem with defining convergence of an infinite sum. Let me supplement that with the following well known simple fact. 
Assume that $p>3$. Then the number of non-zero squares modulo $p$ is $(p-1)/2>1$. This means that the squares form a non-trivial subgroup of $\Bbb{F}_p^*$. So do their reciprocals (in fact it's the same subgroup).
In a field the sum of elements of a non-trivial finite subgroup of the multiplicative group $G$ is always zero. This is because if $a\in G, a\neq1$, then
$$
S:=\sum_{x\in G}x=\sum_{a^{-1}x\in G}x=\sum_{y\in G}ay=aS.
$$
Thus $(a-1)S=0$ implying the claim.
In your case this means that for all integers $k$ we have
$$
\sum_{n=kp+1}^{(k+1)p-1}\frac1{n^2}=0\in\Bbb{F}_p.
$$
If $p=3$ or $p=2$, we can easily check that the sum over any corresponding range is $\equiv-1\pmod p$. Going over $p$ such ranges then contributes something $\equiv0\pmod p$ in these cases.
So in all cases the sequence of partial sums is periodic. The length of the period is $4$, $9$ or $p$ depending on whether $p=2,3$ or $>3$.