Surprising Connections to the Zeta Function

In yesterday’s entry we saw that the infinite sum of reciprocals of the unique digit sequences associated with the general polynomial equation (x – 1)ⁿ = 0 is related solely to the value of n and given by the simple expression (n – 1)/(n – 2).

So once again to illustrate the unique digit sequence associated with (x – 1)⁵ = 0 is

1, 5, 15, 35, 70, 126, 210, … and the corresponding infinite sum of reciprocals i.e.

1 + 1/5 + 1/35 + 1/70 + 1/126 + 1/210 + … = (5 – 1)/(5 – 2) = 4/3.

Now there is a surprisingly important significance to these results which may not be immediately apparent.

In discussions on the Riemann Hypothesis, I have repeatedly drawn attention to the importance of the Zeta 2 (which fully complements the well-known Zeta 1, i.e. Riemann) zeta function.

I was at pains in those discussions how each individual term both in the sum over (all) natural numbers and product over primes infinite expressions can in turn be expressed by a Zeta 2 infinite series.

So for example in the Zeta 1 (Riemann) function where s = 2, the sum over natural numbers is given as

1 + 1/2² + 1/3² + 1/4² + … = 1 + 1/4 + 1/9 + 1/16 and the product over primes,

= 4/3 * 9/8 * 25/24 * 49/48 * … = π²/6.

However each of these individual terms, both for the sum over natural numbers and product over primes expressions respectively, can be expressed by an infinite Zeta 2 function.

Therefore the Zeta 1 (Riemann) function can equally be expressed as 1) the overall sum of an infinite sequence of Zeta 2 functions and 2) the overall product of an infinite sequence of Zeta 2 functions.

So in general terms, the (infinite) Zeta 2 function is given as

ζ₂(s) = 1 + s + s² + s³ + …   = 1/(1 – s)

So, to illustrate the 1^st term i.e. 1, in the sum over natural numbers can be given as,

ζ₂(s) – 1, where s = 1/2 (i.e. 1/(1² + 1).

Then the next term i.e. 1/4 can be given as ζ₂(s) – 1, where s = 1/5 (i.e. 1/(2² + 1).

In this manner, the denominator can always be expressed as the (square of a prime + 1).

In similar fashion, all the remaining terms can be expressed in the form of ζ₂(s) – 1, with s given an appropriate value, based on the square of consecutive primes.

Then the 1^st term, i.e. 4/3, in the product over natural numbers expression can be given as

ζ₂(s), where s = 1/4 (i.e. 1/2²).

Then the 2^nd term, i.e. 9/8 can be given as ζ₂(s), where s = 1/9 (i.e. 1/3²).

And again in like manner, all the remaining terms can be expressed in the form of ζ₂(s), with s given an appropriate value (as the reciprocal of successive squares of a prime).

However just as the Zeta 1 (Riemann) function can be given two expressions as a sum over natural numbers and product over primes respectively, likewise the Zeta 2 can be given two similar complementary expressions.

So far however, we have only considered the product version of this (both as a product over primes and product over natural numbers respectively).

Therefore when again to illustrate, we take for the Zeta 1 (Riemann) function the product over primes expression (for s = 2), i.e. 4/3 * 9/8 * 25/24 * 49/48 * …, each of these individual terms can equally be expressed as an infinite sum of product over primes terms, entailing the Zeta 2 function.

Thus 4/3 = 1 + 1/2² + 1/2⁴ + 1/2⁸  + …

i.e. 1 + 1/(2 * 2) + 1/(2 * 2 * 2 * 2) + 1/(2 * 2 * 2 * 2 * 2 * 2 * 2 * 2) + …

So for the 1st term here (4/3), the prime number 2 is involved; then for the next term  (9/8) the prime number 3 is involved in the denominator and so on indefinitely over all the primes in this manner.

Then again with respect to the Zeta 1 (Riemann) function, we take the sum over natural numbers expression (for s = 2) i.e. 1 + 1/4 + 1/9 + 1/16 + …, each of these individual terms can equally be expressed as an infinite sum of products over a natural number entailing the Zeta 1 function.

Thus the 1^st term, i.e. 1 =  {1 + 1/2 + 1/2² + 1/2³ + …} – 1

= {1 + 1/(1² + 1)  + 1/(1² + 1)² + 1/(1² + 1)³ + …} – 1.

So the 1^st natural number 1 is used here in the denominator.

The 2^nd term i.e. 1/4 = {1 + 1/5 + 1/5² + 1/5³ + …} – 1

= {1 + 1/(2² + 1) + 1/(2² + 1)² + 1/(2² + 1)³ + …} – 1.

So the 2^nd natural number 2 is used here in the denominator.

The 3^rd term i.e. 1/9 = {1 + 1/10 + 1/10² + 1/10³ + …} – 1

= {1 + 1/(3² + 1) + 1/(3² + 1)² + 1/(3² + 1)³ + …} – 1.

So the 3^rd natural number 3 is used here in the denominator. And we can continue on indefinitely in this manner with all the natural numbers appearing in the denominator terms.

However in both cases here, we are obtaining the infinite sum of product terms, entailing with respect to the denominator all the primes and natural numbers respectively.