Reciprocal Prime Sums

[Orkimond]

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

[jaap]

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]

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

[jaap]

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]

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

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

Next time, do it yourself. LaTeX isn't that hard.

[Orkimond]

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

\sum_{p} \frac1p diverges asymptotically to \ln{\ln{x}} + O(1)
\sum_{n\geq 1} \frac1n diverges asymptotically to \ln{x} + O(1)

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

[Orkimond]

antonfire, you are a beautiful man

[jaap]

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]

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]

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 \prod_p \left(1-\frac1p\right), and that product goes to 0.

[Orkimond]

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]

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]

Like in the other sum, the n'th partial sum of that thing is one minus the n'th partial product of \prod_{p} \left(1-p^{-2}\right). However, we know that \prod_{p} \frac{1}{1-p^{-2}} = \sum_{n\geq 1} n^{-2} = \zeta(2) = \frac{\pi^2}{6}. So, that sum converges to 1-\frac{6}{\pi^2}.

[Orkimond]

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]

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 \omega(c) 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]

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, \sum_{ab=n} \mu(b) = 0, we have

\phi(n) = \begin{cases}
0 & n = 1 \\
1+\mu(n) & n \neq 1
\end{cases}

That is, \phi(n) 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.

[Orkimond]

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