^{n}= 0 (where n is an integer > 1) - what I refer to more simply as the alternative Zeta 2 function - can be shown to have fascinating connections with the corresponding Zeta 1 (Riemann) function.

In
particular, I showed how to generate the respective terms of the Zeta 1
function i.e. ζ

_{1}(s), for s = 2 from the alternative Zeta 2 function.
However, it is time now to attempt to generalise this
procedure for to all values of ζ

_{1}(s) where s is an integer ≥ 1.
Indeed the procedure appears trivial where s = 1.

So ζ

_{1}(1) = 1 + 1/2 + 1/3 + 1/4 + ….
And this equates directly with the sum of reciprocals of the
unique digit sequence for (x –
1)

^{2}= 1 + 1/2 + 1/3 + 1/4 + ….
Thus for simplicity we will refer to this latter expression
of the same harmonic series as Alt ζ

_{2}(2) i.e. where n in the polynomial expression (x – 1)^{n}, from which the reciprocals of the unique number sequence involved is derived = 2.
However there is an alternative way of relating the two zeta
expressions, which will be very useful for deriving the terms of ζ

_{1}(s), where s > 1.
ζ

_{1}(1), as the sum of the terms of the harmonic series, diverges to infinity.
However we can express each term of this expression as half the
sum of all Alt ζ

_{2}(s), where s > 3.
So Alt ζ

_{2}(3) = 2/1; Alt ζ_{2}(4) = 3/2; Alt ζ_{2}(5) = 4/3; ; Alt ζ_{2}(6) = 5/4 and so on.
Therefore ζ

_{1}(1) = 1/2{∑Alt ζ_{2}(s)} for s ≥ 3.
So Alt ζ

_{2}(2) = 1/2{∑Alt ζ_{2}(s)} for s ≥ 3.
Then as we saw yesterday to obtain the appropriate
“conversion” for the terms of ζ

_{1}(2),
we then use the respective terms of Alt ζ

_{2}(3) multiplied by 1/2. Now 1/2 in this context can be more precisely expressed as 1/2!
So 1/2!{Alt ζ

_{2}(3)} = 1/2(1 + 1/3 + 1/6 + 1/10 + 1/15 + …)
= 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + …

We then multiply each successive term by the corresponding
sum of infinite series values for Alt ζ

_{2}(3), Alt ζ_{2}(4), Alt ζ_{2}(5), Alt ζ_{2}(6), …
So 1/2{Alt ζ

_{2}(3)} = 1/2 * 2/1 = 1 i.e. (1/1^{2})
1/6{Alt ζ

_{2}(4)} = 1/6 * 3/2 = 1/4 i.e. (1/2^{2})
1/12{Alt ζ

_{2}(5)} = 1/12 * 4/3 = 1/9 i.e. (1/3^{2})
1/20{Alt ζ

_{2}(6} = 1/20 * 5/4 = 1/16 i.e. (1/4^{2})
… … ….

And we can continue on in this general manner to generate
the further terms in the infinite series for ζ

_{1}(2).
Then to generate the respective terms of ζ

_{1}(3), we move on to consideration of the respective terms of Alt ζ_{2}(4) i.e. 1 + 1/4 + 1/10 + 1/20 + 1/35 + …
However in this case each individual term is now multiplied
by 1/3! = 1/6.

So we now have 1/6 + 1/24 + 1/60 + 1/120 + 1/210 + …

Each of these individual terms is now successively
multiplied by Alt ζ

_{2}(4), Alt ζ_{2}(5),
Alt ζ

_{2}(6), Alt ζ_{2}(7) … (Note that Alt ζ_{2}(3) is now omitted).
Now the value of Alt ζ

_{2}(4)= 3/2. So we multiply the 1st term by this value; however we also multiply by (3 – 2)/2 = 1/2.
So in general terms we multiply each successive term by,

(n – 1)/(n – 2)} * (n
– 3)/(n – 2)

Therefore using the 1st term, 1/6 * 3/2 * 1/2 = 1/8 (i.e. 1/2

^{3}).
Then with the 2

^{nd}term we have 1/24 * 4/3 * 2/3 = 1/27 (i.e. 1/3^{3}).
With the 3

^{rd}term, we have 1/60 * 5/4 * 3/4 = 1/64 (i.e. 1/4^{3}).
Then with the 4

^{th}, we have 1/120 * 6/5 * 4/5 = 1/125 (i.e. 1/5^{3}).
… …

Though the derivation might appear arbitrary, once the
correct procedure is used, it works universally to generate all further terms.

Also this procedure preserves unique hidden number patterns.

For instance the 1

^{st}case 1/6 * 3/2 * 1/2 = 1/2^{3}.
Now if we replace 3/2 with 3 * 2 we obtain 1/6 * (3 * 2) *
1/2 = 1/2.

Alternatively if we replace 1/2 by 1 * 2, we get 1/6 * 3/2 *
2 = 1/2

And this pattern will universally hold.

For example in the last case we have 1/120 * 6/5 * 4/5
= 1/5

^{3}.
If we replace 6/5 by 6 * 5 we get 1/120 * (6 * 5) * 4/5 =
1/5

Then if we replace 4/5 by 4 * 5 we get 1/120 * 6/5 * 4 * 5 =
1/5.

So in this way we generate all the terms of the original
harmonic series (except 1).

Admittedly it does get more cumbersome attempting to
generate terms for ζ

_{1}(s), when s > 2.
However a structured pattern for doing this exists in every
case.

For example to create the individual terms of ζ

_{1}(4) - less the 1^{st}two terms - we move to consideration of the corresponding terms of Alt ζ_{2}(5) i.e.
1 + 1/5 + 1/15 + 1/35 + 1/70 + …

We now multiply each of these terms by 1/4! = 1/24 to obtain

1/24 + 1/120 + 1/360 + 1/840
+ 1/1680 + …

Alt ζ

_{2}(5) = 4/3.
We now use a more complex conversion factor based on this
value which in general terms is given as,

(n – 1)/(n – 2) * (n
– 3)/(n – 2) * (n – 4)/(n – 2)

So for Alt ζ

_{2}(5), this gives 4/3 * 2/3 * 1/3
So we using the 1st term, 1/24 we obtain 1/24 * 4/3 * 2/3 *
1/3 = 1/81 (i.e. 1/3

^{4})
Once again there are lovely hidden number patterns preserved
in this formulation.

So replacing 4/3 by 4 * 3 we get 1/24 * (4 * 3) * 2/3 * 1/3
= 1/9 (i.e. 1/3

^{2}).
Then replacing 2/3 by 2 * 3 we get 1/24 * 4/3 * (2 * 3) *
1/3 = 1/9 (i.e. 1/3

^{2}).
Finally replacing 1/3 by 1 * 3 we get 1/24 * 4/3 * 2/3 * (1
* 3) = 1/9 (i.e. 1/3

^{2}).
And a similar pattern presents itself with all subsequent
“conversions” so that we have the emergence of ζ

_{1}(2), less the 1^{st}two terms.
Just to illustrate further, using the 2

^{nd}term 1/120,
we obtain 1/120 * 5/4 * 3/4 * 2/4 = 1/256 (i.e. 1/4

^{4}).
Using the 3

^{rd}term 1/360, we obtain,
1/360 * 6/5 * 4/5 * 3/5 = 1/625 (i.e. 1/5

^{4}).
Then using the 4

^{th}term 1/840, we obtain
1/840 * 7/6 * 5/6 * 4/6 = 1/1296 (1/6

^{4}).
And again we can continue in this manner to generate the
further terms for

ζ

_{1}(4).
I will just finish by setting up the “conversion” to
generate the computable term

i.e. the 4

^{th}term of ζ_{1}(5).
Here we use the next series for Alt ζ

_{2}(6) i.e.
1 + 1/6 + 1/21 + 1/56 + 1/126 + …

Each of these individual terms is now multiplied by 1/5! =
1/120 to obtain

1/120 + 1/720 + 1/2520 + 1/6720 + 1/15120 + …

The general conversion factor now contains 4 product terms

i.e. (n – 1)/(n – 2) * (n – 3)/(n – 2) * (n – 4)/(n – 2) *
(n – 5)/(n – 2)

So when n = 6, this conversion factor = 5/4 * 3/4 * 2/4 *
1/4.

Therefore using the 1

^{st}term, we get,
1/120 * 5/4 * 3/4 * 2/4 * 1/4 = 1/1024 = (i.e. 1/4

^{5}).
Again we can illustrate even more impressively this time,
the number magic preserved within this formulation.

So replacing 5/4 with 5 * 4, we obtain,

1/120 * (5 * 4) * 3/4 * 2/4 * 1 /4 = 1/64 (i.e. 1/4

^{3}).
Then replacing 3/4 with 3 * 4 we obtain

1/120 * 5/4 * (3 * 4) * 2/4 * 1/4 = 1/64 (i.e. 1/4

^{3}).
Replacing 2/4 with 2 * 4 we obtain

1/120 * 5/4 * 3/4 * (2 * 4) * 1/4 = 1/64 (i.e. 1/4

^{3}).
Finally replacing 1/4 with 1 * 4 we obtain,

1/120 * 5/4 * 3/4 * 2/4 * (1 * 4) = 1/64 (i.e. 1/4

^{3}).
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