^{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 + 1/2

^{2}+ 1/3^{2}+ 1/4^{2 }+ … = 1 + 1/4 + 1/9 + 1/16 and the product over primes,
= 4/3 * 9/8 * 25/24 * 49/48 * … = π

^{2}/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

ζ

_{2}(s) = 1 + s + s^{2}+ s^{3}+ … = 1/(1 – s)
So, to illustrate the 1

^{st}term i.e. 1, in the sum over natural numbers can be given as,
ζ

_{2}(s) – 1, where s = 1/2 (i.e. 1/(1^{2 }+ 1).
Then the next term i.e. 1/4 can be given as ζ

_{2}(s) – 1, where s = 1/5 (i.e. 1/(2^{2 }+ 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 ζ

_{2}(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
ζ

_{2}(s), where s = 1/4 (i.e. 1/2^{2}).
Then the 2

^{nd}term, i.e. 9/8 can be given as ζ_{2}(s), where s = 1/9 (i.e. 1/3^{2}).
And again in like manner, all the remaining terms can be
expressed in the form of ζ

_{2}(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 + 1/2

^{2 }+ 1/2^{4 }+ 1/2^{8 }+ …
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 + 1/2^{2 }+ 1/2^{3 }+ …} – 1
= {1 + 1/(1

^{2 }+ 1) + 1/(1^{2 }+ 1)^{2}+ 1/(1^{2 }+ 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 + 1/5^{2 }+ 1/5^{3 }+ …} – 1
= {1 + 1/(2

^{2 }+ 1) + 1/(2^{2 }+ 1)^{2}+ 1/(2^{2 }+ 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 + 1/10^{2 }+ 1/10^{3 }+ …} – 1
= {1 + 1/(3

^{2 }+ 1) + 1/(3^{2 }+ 1)^{2}+ 1/(3^{2 }+ 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.

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