Prove that this double infinite product diverges?

Question

Prove that this double infinite product diverges

Π Π (1 + (1/Prime(n))^m)

where Prime(n) is the nth prime number, and where n, m = 1 to ∞

Answer 1

There are probably a number of ways to do this, but here is what I thought of:

First, we need a lemma:

Lemma 1: If a_i > 0, then (1 + a_1)(1 + a_2)...(1 + a_n) > 1 + a_1 + a_2 + ... + a_n
Proof: Multiply out the product on the LHS, and you get the RHS + some terms (which are > 0)

With this in mind, we can compare the double product above with the following:

∏ ∑ (1/Prime(n))^m where the ∏ is over n (and goes from 1 to infinity) and the sum is over m (and goes from 0 to infinity).

Because of the lemma, if this diverges, then the original double product diverges (this is a comparison test).

But now think about what this new product of sums is. If you multiply it out, you will get (the reciprocal of) every combination of every power of of every prime. That is, you'll get the sum of the reciprocals of all integers (make sure you see this is true).

So the new product of sums actually equals:

∑ (1/n)

where n goes from 1 to infinity, which we know diverges (and so, by comparison, the original product does, too).

Answer 2

This product expands as

Π Π (1 + (1/Prime(n))^m) = Sum 1 / A[m2,m3,m5,m7,..]

where vector M = |mp> consists of integer powers
greater or equal to zero, and
A(M) = 2^m2 * 3^m3 * 5^m5 7^m7...* p^mp *...

Since A(M) is one-to-one corespondence between
vectors M and positive integers (Euclid, see the link below)

Π Π (1 + (1/Prime(n))^m) = Sum[n=1...inf] 1/n

Answer 3

It suffices to show that the inner product diverges when m=1. So the question is, show that the product of (1+1/p_n) diverges, as n ranges over natural numbers. It is an exercise that this product diverges if and only if the infinite sum of 1/p_n diverges, as n ranges over natural numbers. This, however, is a classical fact that dates back to Euler.

Claim: Let {a_n} be a sequence of positive real numbers. prod(1+a_n,n,1,infty) converges if and only if sum(a_n,n,1,infty) converges.

(Here, the notation is evident. For example, prod(1+a_n,n,1,infty) means the product of the terms 1+a_n as the variable n ranges from 1 to infinity.)

Proof: If prod(1+a_n,n,1,infty) converges, then simply taking the partial products and expanding them shows that sum(a_n,n,1,m) <= prod(1+a_n,n,1,m) <= prod(1+a_n,n,1,infty). Since the a_n's are positive, the sum sum(a_n,n,1,infty) converges if and only if the sequence of partial sums is bounded, and we just showed that these partial sums are in fact bounded by the number prod(1+a_n,n,1,infty).

If sum(a_n,n,1,infty) converges, then recall the simple inequality exp(x) >= 1+x for x>=0. Also note that the sequence of partial products of prod(1+a_n,n,1,infty) is an increasing sequence (again, the a_n's are positive), so to show that it converges, it suffices to show that the sequence of partial products is bounded. Taking the exponential of the convergent sum and using the above inequality shows that
exp(sum(a_n,n,1,infty)) >= exp(sum(a_n,n,1,m)) >= prod(1+a_n,n,1,m)
and this holds for every m. Hence, the partial products are bounded and we are done.

Claim: sum(1/p_n,n,1,infty) diverges, where p_n is the n-th prime number.

Proof: (1-1/(p_n))^(-1) is the sum of a geometric series. Namely,
(1-1/(p_n))^(-1)=sum(1/(p_n)^s,s,0,infty). Now, consider the partial product
prod((1-1/(p_n))^(-1),n,1,m).
By unique factorization of integers into products of primes (which are necessarily smaller or equal to the number itself), and expanding out this partial product via the geometric series above, we see that
prod((1-1/(p_n))^(-1),n,1,m) >= sum(1/n,n,1,m).
This sum is a partial sum for the harmonic series, which we know diverges as m goes to infinity. Hence, the product also diverges as m goes to infinity. Finally, recall the elementary inequality
x > (1/2) log(1/(1-x)) for 2>x>0. Applying this inequality to the partial products above shows that
(1/2) log( prod((1-1/(p_n))^(-1),n,1,m) ) =
(1/2) sum(log((1-1/(p_n))^(-1)), n,1,m) <
(1/2) sum(1/p_n,n,1,m).
But as we saw above, (1/2) log( prod((1-1/(p_n))^(-1),n,1,m) ) goes to infinity as m goes to infinity. Hence, sum(1/p_n,n,1,m) goes to infinity as m goes to infinity, as needed.

It is fun to notice that the rest of your double product actually converges. You can prove that using the facts that I mention here.