Now
empirical information had long suggested to me, that where s is odd that the sum
= .25 and where s is even, the sum = .75.

For example
where s = 3, 5, 7 and 9, ζ

_{1}(s) – 1 = .20205…, .03692…, .00834… and .002008… respectively.
So the sum
of these values = .2493…, which is already very close to .25. And as ∑{ζ

_{1}(s) – 1} where s ≥ 2 = 1, this would imply that the corresponding sum for even values of s = .75.
However, we now have a ready means using the approach of
yesterday to conveniently prove this result.

So once again we have,

_{1}(2) – 1 = 1/2

^{2 }+ 1/3

^{2 }+ 1/4

^{2 }+ … = .64493…

ζ

_{1}(3) – 1 = 1/2

^{3 }+ 1/3

^{3 }+ 1/4

^{3 }+ … = .20203…

ζ

_{1}(4) – 1 = 1/2

^{4 }+ 1/3

^{4 }+ 1/4

^{4 }+ … = .08232…

ζ

_{1}(5) – 1 = 1/2

^{5 }+ 1/3

^{5 }+ 1/4

^{5 }+ … = .03692…

… … …

… … …

Now if we
concentrate on vertical columns associated with odd numbered values of s (>
2), again we can see that can be defined in Zeta 2 terms in the form of simple
geometric series.

So the 1st
series = 1/2

^{3 }+ 1/2^{5 }+ 1/2^{7 }+ … = (1/8)/1 – 1/4) = 1/8 * 4/3 = 1/6
The 2

^{nd}series = 1/3^{3 }+ 1/3^{5 }+ 1/3^{7 }+ … = (1/27)(1– 1/9) = 1/27 * 9/8 = 1/24
The 3

^{rd}series = 1/4^{3 }+ 1/4^{5 }+ 1/4^{7 }+ … = (1/64)(1– 1/16) = 1/64 * 16/15 = 1/60
The 4

^{th}series = 1/5^{3 }+ 1/5^{5 }+ 1/5^{7 }+ … = (1/125)(1– 1/25) = 1/125 * 25/24 = 1/120
So the sum of all the vertical columns = corresponding sum
of all horizontal rows

= 1/6 + 1/24 + 1/60 + 1/120 + …

= 1/6(1 + 1/4 + 1/10 + 1/20 + …)

And the numbers inside the brackets correspond to the
reciprocals of the tetrahedral numbers i.e. the unique number sequence
associated with (x – 1)

^{4}= 0
Now 1 + 1/4 + 1/10 + 1/20 + … = 3/2

Therefore 1/6(1 + 1/4 + 1/10 + 1/20 + …) = 1/6 * 3/2 = 1/4.

So ∑{ζ

_{1}(s) – 1} where s is odd (>1) = 1/4 (.25).
Thus ∑{ζ

_{1}(s) – 1}where s is even (> 0) = 3/4 (.75).
There are many fascinating number patterns in evidence.

The series used to prove that ∑{ζ

_{1}(s) – 1} = 0 for all s (≥ 1) is
1/2 + 1/6 + 1/12 + 1/20 + 1/30 + …

= 1/2(1 + 1/3 + 1/6 + 1/10 + 1/15 + … )

And the 1/2 (outside the brackets) corresponds with the 1

^{st}term of the series (inside the brackets).
So the numbers inside the brackets correspond to the
reciprocals of unique number sequence associated with (x – 1)

^{n }= 0 (where s = 3),
Therefore = 1/2 * 2/1 = 1 (i.e. 1/1

^{2}).
Then the series used to prove that ∑{ζ

_{1}(s) – 1} where s is odd (>1) = 1/4, where s is odd (>1) is,
1/6(1 + 1/4 + 1/10 + 1/20 + 1/35 +…).

So the number outside the brackets corresponds to the 2

^{nd}term in the original series i.e.
1/2 + 1/6 + 1/12 + 1/20 + 1/30 + …, while the numbers inside,
correspond to the reciprocals of the unique digit sequence associated with (x –
1)

^{n }= 0 (where s = 4)
= 1/6 * 3/2 = 1/4 (i.e. 1/2

^{2}).
If we continue on this manner the next number to be placed
outside the bracket = 1/12 (the 3

^{rd}term of the original series) and the numbers inside the bracket will then represent the reciprocals of the unique number sequence associated with (x – 1)^{n }= 0 (where s = 5) = 1/12 (1 + 1/5 + 1/15 + 1/35 + 1/70 + …) = 1/12 * 4/3 = 1/9 (i.e. 1/3^{2}).
Using just one more example to illustrate the next number to
be placed outside the brackets = 1/20 (the 4

^{th}term of the original series) and the numbers inside corresponds to the reciprocals of the unique number sequence associated with (x – 1)^{n }= 0 (where s = 6)
= 1/20(1 + 1/6 + 1/21 + 1/56 + 1/126 + …) = 1/20 * 5/4 =
1/16 (i.e. 1/4

^{2}).
Note that there is a unique feature to the nature of these
products.

Take the last one for example where we have 1/20 * 5 * 1/4.
Now if we replace the last fraction (1/4) by its reciprocal we obtain 1/20 * 5
* 4 = 1. So instead of dividing 5 by 4 (the sum of the series inside the
brackets) we multiply 5 * 4 the resulting product = 1

And this will always be the case! So in the next example, 1/30 * 6/5 = 1/25 (i.e. 1/5

^{2}). and 1/30 * 6 * 5 = 1!
There is also another interesting feature. For example in
the first case,

1/6 * 3/2 = 1/4 (i.e. 1/2

^{2}). However if alternatively, we divide 1/6 by 3/2,
(1/6)/(3/2) = 1/6 * 2/3 = 1/9 (i.e. 1/3

^{2}) . And again this feature is universal, where in the former case 1/n^{2 }(n = 1, 2, 3,...) results and in the latter case 1/(1 + n)^{2}.
Thus as we have seen, 1/20 * 5/4 = 1/16 (i.e. 1/4

^{2}). And, 1/20 * 4/5 = 1/25 (i.e. 1/5^{2}).
By progressing in the manner above we generate the series,

1/1

^{2}+ 1/2^{2 }+ 1/3^{2 }+ 1/4^{2 }+ … i.e. ζ_{1}(2) i.e. the Riemann Zeta function for s = 2.
So we started by demonstrating the close links as between
the Zeta 1 (Riemann) as terms in horizontal rows and the Zeta 2 function as the
corresponding terms in the vertical columns.

And now we have been able to demonstrate very close links as
between the alternative Zeta 2 function - as the sums of reciprocals
corresponding to the unique number sequences associated with the general
polynomial expression (x – 1)

^{n }= 0 - and the Zeta 1 (Riemann) function.
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