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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

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  • $\begingroup$ Hmm I think that even all sequences over finite fields are ultimately periodic, so that this maybe not be enough to make sense of an infinite sum here? $\endgroup$ – user152169 Jul 11 '15 at 3:46
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    $\begingroup$ You'd need to introduce a topology making this a topological group before you can even talk about infinite sums. I can't think of any particular useful ones for $\mathbb{F}_p$ off the top of my head, though. $\endgroup$ – Avi Steiner Jul 11 '15 at 3:46
  • $\begingroup$ Interesting. Could you explain a bit more why? $\endgroup$ – user152169 Jul 11 '15 at 3:48
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    $\begingroup$ You need the topology so you can talk about convergence of sequences. You need addition to be continuous with respect to this topology so that you can talk about convergence of infinite series. $\endgroup$ – Avi Steiner Jul 11 '15 at 3:51
  • $\begingroup$ Thanks. I wish I remembered better those undergrad analysis courses.. Could you suggest something to read for this, preferably something more or less related to my question? Thanks again. $\endgroup$ – user152169 Jul 11 '15 at 3:57
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As @AviSteiner has notes, infinite sums are defined in terms of a convergence of partial sums. Discussion of convergence of sequences requires a topology. Furthermore, a convergent sequence is not guaranteed to converge to a unique point unless the topology is Hausdorff. Since I expect you would like to say that your infinite sum of points in the field converges to something unique (provided it converges at all), we should make the topology on our finite field Hausdorff. Incidentally, the only Hausdorff topology on a finite set is discrete. In the discrete topology, the only convergent sequences are constant after a while. This leads to a rather disappointing answer: in order to guarantee a unique solution to your series, you must have that all but finitely many terms in the series are the additive identity.

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  • $\begingroup$ :( too bad. Hmm what if instead of the finite field, we take its (infinite) closure $\overline{\mathbb{F}_p}$? Note that the terms of $S$ belongs to this infinite set. $\endgroup$ – user152169 Jul 11 '15 at 4:25
  • $\begingroup$ For an infinite set, there are a lot of Hausdorff topologies you can define. I do not know if there exists a natural topology on $\mathbb F_p$, but I suspect there is. You might look at topological vector spaces. $\endgroup$ – Alex S Jul 11 '15 at 4:39
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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 $>3$.

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    $\begingroup$ Nice :). I like this. $\endgroup$ – user152169 Jul 11 '15 at 5:51

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