Reciprocal prime number infinite series

By Charlie Bupp (P4556) on Sunday, June 3, 2001 - 01:16 am :

I understand that the reciprocal prime number series, that is:

(1/2) + (1/3) + (1/5) + (1/7) + (1/11) + (1/13) ... (1/p)

diverges as the terms approach to infinity, as calculated by Euler.
Can anyone help me out as to why this is? I would REALLY appreciate some help telling why this is a diverging series, as I am giving a speech on the interesting properties of primes. THANKS!

Charlie

By Michael Doré (Michael) on Sunday, June 3, 2001 - 12:40 pm :

OK. Let's assume that the series doesn't diverge to infinity, but is bounded above by L.

To start, we need the result that for |x| < 1:

-ln(1 - x) = x + x² /2 + x³ /3 + ...

If 0 < x < = 1/2 we know that:

x² /2 + x³ /3 + ... < (x² + x³ + ...)/2 = x² /2 x[1 + x + x² + ...] < = x² < x

So -ln(1 - x) < 2x for 0 < x < = 1/2.

Substitute in x = 1/2, 1/3, ...:

-ln(1 - 1/2) < 2/2
-ln(1 - 1/3) < 2/3
-ln(1 - 1/5) < 2/5
...
-ln(1 - 1/p) < 2/p

Add all these together:

-ln((1 - 1/2)(1 - 1/3)...(1 - 1/p)) < 2(1/2 + 1/3 + ... + 1/p) < 2L

Therefore:

(1 - 1/2)(1 - 1/3)...(1 - 1/p) > e^-2L

for all p. But e^-2L > 0 so assuming your series converges, we have deduced that:

(1 - 1/2)(1 - 1/3)(1 - 1/5)...(1 - 1/p)

is bounded below by a positive number. Writing K = e^2L we have:

(1 - 1/2)^-1 (1 - 1/3)^-1 ...(1 - 1/p)^-1 < K

for all prime p.

So:

(1 + 1/2 + 1/2² + ...)(1 + 1/3 + 1/3² + ...)...(1 + 1/p + 1/p² + ...) < K

for all prime p.

Can you see why this can't possibly be true? If you remember unique prime factorisation you should be able to derive a contradiction from this, which would then prove that our original assumption that the series was bounded above is false. As the series is positive it would then follow that it diverges to infinity as required. Write back if you have trouble finishing off the proof.

By Charlie Bupp (P4556) on Monday, June 4, 2001 - 03:14 am :

Sorry, but I'm still a little lost. I understand what you're saying but I'm unsure as to where you are going with it, and how you get the last step:

(1 + 1/2 + 1/22 + ...)(1 + 1/3 + 1/32 + ...)...(1 + 1/p + 1/p² + ...) < K

Thank you very much for your help!

Charlie

By Olof Sisask (P3033) on Monday, June 4, 2001 - 12:21 pm :

The last step comes from 1 + 1/x + 1/x² + 1/x³ ... = 1/(1-x), i.e. the formula for the sum to infinity of a geometric progression - are you familiar with this? Otherwise I'd be happy to explain it.

Regards,
Olof.

By Charlie Bupp (P4556) on Tuesday, June 5, 2001 - 04:31 pm :

Okay, I get the last step, thanks for the explanation. What I don't understand is how one proves that this resulting product does not converge. Thanks again for your help everyone!

Charlie

By Kerwin Hui (Kwkh2) on Tuesday, June 5, 2001 - 05:38 pm :

First, note that for any natural

n

, we have a unique prime factorisation of

n

, say

$n = p_{a_{1}}^{b_{1}} \dots p_{a_{s}}^{b_{s}}$

Now, we have supposed that the infinite product

$\underset{p prime}{Π} (1 - p^{- 1})^{- 1}$

is bounded above by $K$ . We have also found that this expression is equivalent to

$\underset{p prime}{Π} (1 + p^{- 1} + p^{- 2} + \dots)$

is bounded above by $K$ .

Now expand out the bracket, we see that we get to the infinite sum

$\sum_{r = 1}^{\infty} r^{- 1}$

This is the harmonic series, which is divergent. (Do you know how to prove this?), contradiction.

Kerwin

By Charlie Bupp (P4572) on Tuesday, June 5, 2001 - 07:22 pm :

One last question (I promise!!) I am not really familiar with sigma notation, and don't know exactly what you mean when you say "expand out the bracket." I know how to prove that the harmonic series is divergent, so once again all I need is a little clearing up about the middle step. Thanks again everyone for all your time!

Charlie

By Kerwin Hui (Kwkh2) on Tuesday, June 5, 2001 - 08:42 pm :

OK, let

p_{1}

p_{2}

, ... be all primes. Then we have the product

$(1 + p_{1}^{- 1} + p_{1}^{- 2} + \dots) (1 + p_{2}^{- 1} + p_{2}^{- 2} + \dots) \dots = \sum p_{1}^{a_{1}} p_{2}^{a_{2}} \dots$

Where the sum is taken over all (infinite) sequences of non-negative integers $(a_{1}, a_{2}, \dots)$ . We recognise this as the harmonic series and so it must diverge.

Kerwin

By Charlie Bupp (P4556) on Wednesday, June 6, 2001 - 12:48 am :

Thanks again Kerwin, but I don't understand how you can convert the product into a sigma notation. If you could break the last step down for me, I would appreciate it greatly. Finally, when you say "harmonic series" I assume you are referring to the series:

1/2 + 1/3 + 1/4.... etc

Am I right?

Charlie

By Kerwin Hui (Kwkh2) on Wednesday, June 6, 2001 - 01:21 am :

In my last message, the last sum should be

$\sum p_{1}^{- a_{1}} p_{2}^{- a_{2}} \dots$

To see this, consider, for the time being, the product

$(1 + p_{1}^{- 1} + p_{1}^{- 2} + \dots) (1 + p_{2}^{- 1} + p_{2}^{- 2} + \dots)$

Since the series in both brackets are absolutely convergent, we can multiply out the bracket and get

$(1 + p_{1}^{- 1} + p_{1}^{- 2} + \dots) + p_{2}^{- 1} (1 + p_{1}^{- 1} + p_{1}^{- 2} + \dots) + p_{2}^{- 2} (1 + p_{1}^{- 1} + p_{1}^{- 2} + \dots) + \dots$

which is the sum $\sum p_{1}^{- a_{1}} p_{2}^{- a_{2}}$ where the sum is taken over all ordered pairs of non-negative integers $(a_{1}, a_{2})$ .

Now it is an easy step to verify that, for any positive integer $N$ , the product

$(1 + p_{1}^{- 1} + p_{1}^{- 2} + \dots) (1 + p_{2}^{- 1} + p_{2}^{- 2} + \dots) \dots (1 + p_{N}^{- 1} + p_{N}^{- 2} + \dots)$

is the sum

$\sum p_{1}^{- a_{1}} p_{2}^{- a_{2}} \dots p_{N}^{- a_{N}}$

where the sum is taken over all strings of $N$ non-negative integers $(a_{1}, a_{2}, \dots, a_{N})$ .

Now let $N$ tend to infinity (as we assume the product is bounded above). This gives the required sum, which is $1 + 1 / 2 + 1 / 3 + 1 / 4 + \dots$

Kerwin

By Charlie Bupp (P4556) on Wednesday, June 6, 2001 - 02:45 am :

Kerwin:

You are a GOD! Thank you ever so much!

Charlie

Whilst the editor has every respect for Kerwin's mathematical abilities, she begs to disagree on the above statement!