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I have asked a while ago about the nature of the following series: $$\sum \limits _{n\geq 1}\frac{2^n\mod n}{n^2}$$ over here: Does the sum $\sum_{n \geq 1} \frac{2^n\operatorname{mod} n}{n^2}$ converge?. I am now asking if the series $$\sum \limits _{p}\frac{2^p\mod p}{p^2}$$ for p prime is also divergent. I am trying to remake in this context the famous proof of Euler that starts with the fact that: $$\sum \limits _{n\geq 1}\frac{1}{n}=\prod \limits _{p}\ (1+\frac{1}{p}+\frac{1}{p^2}+\frac{1}{p^3}+...) $$. So, I am trying to obtain: $$\sum \limits _{n\geq 1}\frac{2^n\mod n}{n^2}=\prod \limits _{p} something$$. Can it be done?

Aurelian Florea
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  • Do you have calculated the sum upto , say , $10^{6}$ and $10^7$ ? This does not yet tell us whether it will converge , but it would give a first impression what we can expect. Whether we can express it with a nice formula using the primes is another story. I cannot rule it out, but my first impression is that this does not work. – Peter Sep 20 '22 at 09:13
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    If the sum runs only over the primes , Fermat's little theorem tells us $2^p\mod p=2$ for every odd prime $p$ , hence the sum clearly converges. – Peter Sep 20 '22 at 09:21
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    @WillSherwood No, this would be twice the sum over $\frac{1}{n^2}$ , if $n$ runs over all positive integers. The sum here seems to slowly diverge for $n\to \infty$ , if it runs over all positive integers. The sum over the primes is approximately $0.4044948$ – Peter Sep 20 '22 at 09:27
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    Also, the sequence $a(n) = 2^n \pmod n$ is not multiplicative, i.e., for two coprime numbers $m$ and $n$ it is not true that $a(n)a(m) = a(nm)$. Thereby there cannot be an euler product. – LurchiDerLurch Sep 20 '22 at 09:35
  • @LurchiDerLurch Ok, it can't be an Euler product and probably it is not possible to do it unless in terms of probabilities. Take a look at the particular case here: https://math.stackexchange.com/questions/4535282/nature-of-the-prime-numbers-series-sum-limits-p-frac2pp3-mod-2p5 My argument is that for a chaotic enough behavior in the above example the the distribution of (2p+p3) mod(2∗p+5) over 0..2p+5 is constant. In this case it could happen that the series it's still divergent as you only get an p+5/2 for the denominator. Or am I wrong. This is why I've asked about the product! – Aurelian Florea Sep 23 '22 at 15:47

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As Peter wrote in the comments, $2^p \equiv 2 \pmod p$ by Fermat's little theorem, when $p>2$. For $p=2$, we have $2^2 \equiv 0 \pmod 2$. Hence, we can bound the sum with the naturals:

$$0 < \sum \limits _{p}\frac{2^p\mod p}{p^2} = \sum_p \frac 2{p^2} < \sum_{n \in \mathbb N} \frac 2 {n^2} = \pi^2/3.$$

Therefore, the sum converges.

Will Sherwood
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