Reciprocal Prime Sums

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Orkimond
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Reciprocal Prime Sums

OK, bear with me the whole way through because this gets interesting (sorry about the set up, I don't know how to use the maths thing)
I found an alternate expression for the sum of the reciprocals of the squares of the primes using a simple sieving technique:

$\sum_{n\geq 1} \frac{1}{n^2} - 1 - \left(\sum_{n\geq 1} \frac{1}{n^2} -1\right)\left(\frac{1}{2^2} + \frac{1}{3^2}\left(1-\frac{1}{2^2}\right) + \frac{1}{5^2}\left(1 - \frac{1}{3^2}\right)\left(1 - \frac{1}{2^2}\right) + \frac{1}{7^2}\left(1 - \frac{1}{5^2}\right)\left(1 - \frac{1}{3^2}\right)\left(1 - \frac{1}{2^2}\right) + \ldots\right) =$

$= \frac{\pi^2}{6} - 1 - \left(\frac{\pi^2}{6} - 1\right)\left(\frac{1}{2^2} + \frac{1}{3^2}\left(1 - \frac{1}{2^2}\right) + \frac{1}{5^2}\left(1 - \frac{1}{3^2}\right)\left(1 - \frac{1}{2^2}\right) + \frac{1}{7^2}\left(1 - \frac{1}{5^2}\right)\left(1 - \frac{1}{3^2}\right)\left(1 - \frac{1}{2^2}\right) + \ldots\right)$

Now I managed to prove that this equals pi2/20. As such I found that the sieve:

$\left(\frac{1}{2^2} + \frac{1}{3^2}\left(1 - \frac{1}{2^2}\right) + \frac{1}{5^2}\left(1 - \frac{1}{3^2}\right)\left(1 - \frac{1}{2^2}\right) + \frac{1}{7^2}\left(1 - \frac{1}{5^2}\right)\left(1 - \frac{1}{3^2}\right)\left(1 - \frac{1}{2^2}\right) + \ldots\right) = \frac{7\pi^2 - 60}{10\pi^2 - 60}$

Now, using the same sieve for the sum of the reciprocals of the primes we get:

$\sum_{n\geq 1} \frac1n - 1 - \left(\sum_{n\geq 1} \frac1n -1\right)\left(\frac12 + \frac13 \left(1 - \frac12\right) + \frac15 \left(1 - \frac13\right)\left(1 - \frac12\right) + \frac17 \left(1 - \frac15\right)\left(1 - \frac13\right)\left(1 - \frac12\right) +\ldots\right) = \sum_{p} \frac1p$

Now any mathematician worth their salt knows that both the harmonic series and the reciprocal prime series converge, but the above sieve section does not because some rearranging gives:

$1 - \frac{\sum_{p} \frac1p }{\sum_{n \geq 1} \frac1n - 1} = \frac12 + \frac13 \left(1 - \frac12\right) + \frac15 \left(1 - \frac13\right)\left(1 - \frac12\right) + \frac17 \left(1 - \frac15\right)\left(1 - \frac13\right)\left(1 - \frac12\right) + \ldots$

And:

$\sum_{p} \frac1p < \sum_{n\geq 1} \frac1n - 1$

So my question is this, what is the closed form of $\frac12 + \frac13 \left(1 - \frac12\right) + \frac15 \left(1 - \frac13\right)\left(1 - \frac12\right) + \frac17 \left(1 - \frac15\right)\left(1 - \frac13\right)\left(1 - \frac12\right) + \ldots$
Any ideas at all would be helpful, it's a hard problem and I need some inspiration.

If someone can type my equations using the maths thing I'll edit those into this post
Last edited by Orkimond on Sat Jul 25, 2009 11:41 am UTC, edited 1 time in total.

jaap
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Re: Reciprocal Prime Sums

Orkimond wrote:Now any mathematician worth their salt knows that both the harmonic series and the reciprocal prime series converge, but the above sieve section does not because some rearranging gives:

When you rearrange such a series you can get any answer for its limit. See Riemann Series Theorem.

Orkimond
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Re: Reciprocal Prime Sums

That only works if a series is conditionally convergent, both of the mentioned series are not.

jaap
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Re: Reciprocal Prime Sums

The harmonic series diverges. You are subtracting another diverging series from it. This combined series is conditionally convergent. Subtracting infinity from infinity is meaningless, and rearranging the terms in this subtraction to get another diverging series (the sum of reciprocals of primes) even more so.

antonfire
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Re: Reciprocal Prime Sums

Because this thread desperately needs a serious boost in readability (though it may be too late to save it anyway):

Orkimond wrote:OK, bear with me the whole way through because this gets interesting (sorry about the set up, I don't know how to use the maths thing)
I found an alternate expression for the sum of the reciprocals of the squares of the primes using a simple sieving technique:
$\sum_{n\geq 1} \frac{1}{n^2} - 1 - \left(\sum_{n\geq 1} \frac{1}{n^2} -1\right)\left(\frac{1}{2^2} + \frac{1}{3^2}\left(1-\frac{1}{2^2}\right) + \frac{1}{5^2}\left(1 - \frac{1}{3^2}\right)\left(1 - \frac{1}{2^2}\right) + \frac{1}{7^2}\left(1 - \frac{1}{5^2}\right)\left(1 - \frac{1}{3^2}\right)\left(1 - \frac{1}{2^2}\right) + \ldots\right) =$
$= \frac{\pi^2}{6} - 1 - \left(\frac{\pi^2}{6} - 1\right)\left(\frac{1}{2^2} + \frac{1}{3^2}\left(1 - \frac{1}{2^2}\right) + \frac{1}{5^2}\left(1 - \frac{1}{3^2}\right)\left(1 - \frac{1}{2^2}\right) + \frac{1}{7^2}\left(1 - \frac{1}{5^2}\right)\left(1 - \frac{1}{3^2}\right)\left(1 - \frac{1}{2^2}\right) + \ldots\right)$

Now I managed to prove that this equals pi2/20. As such I found that the sieve:
$\left(\frac{1}{2^2} + \frac{1}{3^2}\left(1 - \frac{1}{2^2}\right) + \frac{1}{5^2}\left(1 - \frac{1}{3^2}\right)\left(1 - \frac{1}{2^2}\right) + \frac{1}{7^2}\left(1 - \frac{1}{5^2}\right)\left(1 - \frac{1}{3^2}\right)\left(1 - \frac{1}{2^2}\right) + \ldots\right) = \frac{7\pi^2 - 60}{10\pi^2 - 60}$

Now, using the same sieve for the sum of the reciprocals of the primes we get:
$\sum_{n\geq 1} \frac1n - 1 - \left(\sum_{n\geq 1} \frac1n -1\right)\left(\frac12 + \frac13 \left(1 - \frac12\right) + \frac15 \left(1 - \frac13\right)\left(1 - \frac12\right) + \frac17 \left(1 - \frac15\right)\left(1 - \frac13\right)\left(1 - \frac12\right) +\ldots\right) = \sum_{p} \frac1p$

Now any mathematician worth their salt knows that both the harmonic series and the reciprocal prime series converge, but the above sieve section does not because some rearranging gives:
$1 - \frac{\sum_{p} \frac1p }{\sum_{n \geq 1} \frac1n - 1} = \frac12 + \frac13 \left(1 - \frac12\right) + \frac15 \left(1 - \frac13\right)\left(1 - \frac12\right) + \frac17 \left(1 - \frac15\right)\left(1 - \frac13\right)\left(1 - \frac12\right) + \ldots$

And:
$\sum_{p} \frac1p < \sum_{n\geq 1} \frac1n - 1$

So my question is this, what is the closed form of [imath]\frac12 + \frac13 \left(1 - \frac12\right) + \frac15 \left(1 - \frac13\right)\left(1 - \frac12\right) + \frac17 \left(1 - \frac15\right)\left(1 - \frac13\right)\left(1 - \frac12\right) + \ldots[/imath]?
Any ideas at all would be helpful, it's a hard problem and I need some inspiration.

If someone can type my equations using the maths thing I'll edit those into this post

Next time, do it yourself. LaTeX isn't that hard.
Jerry Bona wrote:The Axiom of Choice is obviously true; the Well Ordering Principle is obviously false; and who can tell about Zorn's Lemma?

Orkimond
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Joined: Tue Oct 14, 2008 4:04 am UTC

Re: Reciprocal Prime Sums

I'm dividing not subtracting

Also, I figured it out: $\frac12 + \frac13 \left(1 - \frac12\right) + \frac15 \left(1 - \frac13\right)\left(1 - \frac12\right) + \frac17 \left(1 - \frac15\right)\left(1 - \frac13\right)\left(1 - \frac12\right) + \ldots$ converges to 1

[imath]\sum_{p} \frac1p[/imath] diverges asymptotically to [imath]\ln{\ln{x}} + O(1)[/imath]
[imath]\sum_{n\geq 1} \frac1n[/imath] diverges asymptotically to [imath]\ln{x} + O(1)[/imath]
$1 - \frac{\sum_{p} \frac1p}{\sum_{n\geq 1} \frac1n - 1} \Rightarrow 1 - \frac{\ln{\ln{x}}}{ln(x)-1} \Rightarrow 1 - \frac{\ln{\ln{x}}}{ln(x)} => 1$
Last edited by Orkimond on Sat Jul 25, 2009 11:56 am UTC, edited 1 time in total.

Orkimond
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Joined: Tue Oct 14, 2008 4:04 am UTC

Re: Reciprocal Prime Sums

antonfire, you are a beautiful man

jaap
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Re: Reciprocal Prime Sums

Orkimond wrote:I'm dividing not subtracting

You have
$\sum_{n\geq 1} \frac1n - 1$
and subtract
$\left(\sum_{n\geq 1} \frac1n -1\right)\left(\frac12 + \frac13 \left(1 - \frac12\right) + \frac15 \left(1 - \frac13\right)\left(1 - \frac12\right) + \frac17 \left(1 - \frac15\right)\left(1 - \frac13\right)\left(1 - \frac12\right) +\ldots\right)$
to somehow get
$\sum_{p} \frac1p$

All three are divergent series. I don't believe you can simply transfer what you found in the convergent case with squares of reciprocals to this divergent case and have it make sense. It simply says infinity minus infinity is infinity.

Orkimond
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Re: Reciprocal Prime Sums

You can make it a sum for elements below M and then take a limit at the end...

EDIT: Also that isn't derived from the earlier analysis, it is a separate sieve, it just looks similar.

antonfire
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Re: Reciprocal Prime Sums

By the way,
$\left(\frac{1}{2^2} + \frac{1}{3^2}\left(1 - \frac{1}{2^2}\right) + \frac{1}{5^2}\left(1 - \frac{1}{3^2}\right)\left(1 - \frac{1}{2^2}\right) + \frac{1}{7^2}\left(1 - \frac{1}{5^2}\right)\left(1 - \frac{1}{3^2}\right)\left(1 - \frac{1}{2^2}\right) + \ldots\right) = \frac{7\pi^2 - 60}{10\pi^2 - 60}$is certainly false, as you can see by noting that the first two terms of the LHS already sum to something bigger than the RHS.

You are right that
$\left(\frac{1}{2} + \frac{1}{3}\left(1 - \frac{1}{2}\right) + \frac{1}{5}\left(1 - \frac{1}{3}\right)\left(1 - \frac{1}{2}\right) + \frac{1}{7}\left(1 - \frac{1}{5}\right)\left(1 - \frac{1}{3}\right)\left(1 - \frac{1}{2}\right) + \ldots\right) = 1$since the n'th partial sum is one minus the n'th partial product of [imath]\prod_p \left(1-\frac1p\right)[/imath], and that product goes to 0.
Jerry Bona wrote:The Axiom of Choice is obviously true; the Well Ordering Principle is obviously false; and who can tell about Zorn's Lemma?

Orkimond
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Joined: Tue Oct 14, 2008 4:04 am UTC

Re: Reciprocal Prime Sums

You're right, and I'm trying to figure out what I did... Damn, my equation for the sieve is wrong, wow, that ruins my whole post, but I think I can fix it, I'll repost once I'm done

Orkimond
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Re: Reciprocal Prime Sums

I think:
$\left(\frac{1}{2^2} + \frac{1}{3^2}\left(1 - \frac{1}{2^2}\right) + \frac{1}{5^2}\left(1 - \frac{1}{3^2}\right)\left(1 - \frac{1}{2^2}\right) + \frac{1}{7^2}\left(1 - \frac{1}{5^2}\right)\left(1 - \frac{1}{3^2}\right)\left(1 - \frac{1}{2^2}\right) + \ldots\right) = \sum_{n\geq 1} \frac{1}{n^2} - 1$

Although I may be making that up

Also, kudos on the topology signature, I lul'd

antonfire
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Re: Reciprocal Prime Sums

Like in the other sum, the n'th partial sum of that thing is one minus the n'th partial product of [imath]\prod_{p} \left(1-p^{-2}\right)[/imath]. However, we know that [imath]\prod_{p} \frac{1}{1-p^{-2}} = \sum_{n\geq 1} n^{-2} = \zeta(2) = \frac{\pi^2}{6}[/imath]. So, that sum converges to [imath]1-\frac{6}{\pi^2}[/imath].
Jerry Bona wrote:The Axiom of Choice is obviously true; the Well Ordering Principle is obviously false; and who can tell about Zorn's Lemma?

Orkimond
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Joined: Tue Oct 14, 2008 4:04 am UTC

Re: Reciprocal Prime Sums

I found that as my upper bound earlier. Apparently its also equal to the sum...

Now I'm going to explain my mistake:
$\left(\frac{1}{2^2} + \frac{1}{3^2}\left(1 - \frac{1}{2^2}\right) + \frac{1}{5^2}\left(1 - \frac{1}{3^2}\right)\left(1 - \frac{1}{2^2}\right) + \frac{1}{7^2}\left(1 - \frac{1}{5^2}\right)\left(1 - \frac{1}{3^2}\right)\left(1 - \frac{1}{2^2}\right) + \ldots\right)$
Was meant to be a sieve such that:
$\left(\sum_{n\geq 1} \frac{1}{n^2} -1\right)\left(\frac{1}{2^2} + \frac{1}{3^2}\left(1-\frac{1}{2^2}\right) + \frac{1}{5^2}\left(1 - \frac{1}{3^2}\right)\left(1 - \frac{1}{2^2}\right) + \frac{1}{7^2}\left(1 - \frac{1}{5^2}\right)\left(1 - \frac{1}{3^2}\right)\left(1 - \frac{1}{2^2}\right) + \ldots\right) = \sum_{c \epsilon C} \frac{1}{c^2}$
Where C is the set of all composite numbers
But in fact it repeats every composite number by the number of distinct prime factors that composite number has, so 1/36 appears 2ce but 1/81 appears 1ce only.

Orkimond
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Re: Reciprocal Prime Sums

so that means that:

$\left(\sum_{n\geq 1} \frac{1}{n^2} -1\right)\left(\frac{1}{2^2} + \frac{1}{3^2}\left(1-\frac{1}{2^2}\right) + \frac{1}{5^2}\left(1 - \frac{1}{3^2}\right)\left(1 - \frac{1}{2^2}\right) + \frac{1}{7^2}\left(1 - \frac{1}{5^2}\right)\left(1 - \frac{1}{3^2}\right)\left(1 - \frac{1}{2^2}\right) + \ldots\right) = \sum_{c \epsilon C} \frac{\omega(c)}{c^2}$
Where [imath]\omega(c)[/imath] is the distinct prime divisor function.
So:
$\sum_{c \epsilon C} \frac{\omega(c)}{c^2} = \left(\frac{\pi^2}{6}-1\right)\left(1-\frac{6}{\pi^2}\right) = \frac{\pi^4-12\pi^2+36}{6\pi^2} = \frac{(\pi^2-6)^2}{6\pi^2}$

antonfire
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Re: Reciprocal Prime Sums

You can check that this is wrong too, by summing the first few terms. In this case, I think you have to go up to 45 or so.

Note that, for example, all instances of 1/302 cancel out in your sum, although you would have me believe that you end up with 3 of them in the end.

What you actually get if you expand the product is this:
$\left(\frac{\pi^2}{6}-1\right)\left(1-\frac{6}{\pi^2}\right) = \left(\sum_{a \neq 1} a^{-2}\right)\left(1 - \prod_{p}\left(1-p^{-2}\right)\right) = \left(\sum_{a \neq 1} a^{-2}\right)\left(\sum_{b\neq 1} -\mu(b)b^{-2}\right)=$$=\sum_{n} \left(\sum_{ab=n;\ a, b \neq 1} -\mu(b)\right) n^{-2} = \sum_{n} \phi(n) n^{-2}$

since for n not equal to 1, [imath]\sum_{ab=n} \mu(b) = 0[/imath], we have
$\phi(n) = \begin{cases} 0 & n = 1 \\ 1+\mu(n) & n \neq 1 \end{cases}$

That is, [imath]\phi(n)[/imath] is 0 when n is one or when n is a product of an odd number of distinct primes, 2 when n is a product of a (positive) even number of distinct primes, and 1 in other cases.

Of course, you can see the same thing in a much easier way:

$\left(\frac{\pi^2}{6}-1\right)\left(1-\frac{6}{\pi^2}\right) = -2 + \frac{\pi^2}{6} + \frac{6}{\pi^2} = -2 + \sum_n n^{-2} + \sum_n \mu(n) n^{-2}$

Edit: Also, I think it's considered uncouth 'round these parts to double post. If you want to add something and your post is the last one in the thread, you can just edit it.
Jerry Bona wrote:The Axiom of Choice is obviously true; the Well Ordering Principle is obviously false; and who can tell about Zorn's Lemma?

Orkimond
Posts: 53
Joined: Tue Oct 14, 2008 4:04 am UTC

Re: Reciprocal Prime Sums

Hmmmmm, thanks, I need to think bout that then, cheers boy