ζ(4) = 1/1

^{4 }+ 1/2^{4 }+ 1/3^{4 }+ 1/4^{4 }+ … = π^{4}/90
and when s
= 2,

ζ(2) = 1/1

^{2 }+ 1/2^{2 }+ 1/3^{2 }+ 1/4^{2 }+ … = π^{2}/6, so that
(1/1

^{2 }+ 1/2^{2 }+ 1/3^{2 }+ 1/4^{2 }+ …)^{2 }= π^{4}/36
Therefore
(1/1

^{2 }+ 1/2^{2 }+ 1/3^{2 }+ 1/4^{2 }+ …)^{2}/(1/1^{4 }+ 1/2^{4 }+ 1/3^{4 }+ 1/4^{4 }+ …)^{ }= 90/36 = 5/2
However we
have already seen that this can be written as a Dirichlet L-series i.e.

1/1

^{2}+ 2/2^{2}+ 2/3^{2}+ 2/4^{2}+ 2/5^{2}+ 4/6^{2}+ 2/7^{2}+ 2/8^{2}+ 2/9^{2}+ 4/10^{2}+ … = 5/2
So ζ(2)

^{2}/ζ(4) = 1/1^{2}+ 2/2^{2}+ 2/3^{2}+ 2/4^{2}+ 2/5^{2}+ 4/6^{2}+ 2/7^{2}+ 2/8^{2}+ 2/9^{2}+ 4/10^{2}+ …
Thus we have expressed the square of one Dirichlet series
(the Riemann zeta function where s = 2) divided by another Dirichlet series
(the Riemann zeta function where s = 4) by yet another Dirichlet series (where
s = 2).

And this relationship can equally be expressed as a product
over primes expression.

So ζ(2)

^{2}/ζ(4) = (4/3 * 9/8 * 25/24 * 49/48)^{2}/(16/15 * 81/80 * 625/624 * 2301/2300)
= 5/3 * 10/8 * 26/24 * 50/48 * …

However in
the terms that I employ, this collective relationship (for both expressions) relates
solely to the L1 function.

So the
question now arises as to how each individual term can be expressed as a
corresponding (infinite) L2 function.

Thus with
respect to the geometric series type expression, if we take the first term,
i.e. 5/3, we can express this as the product of two terms i.e. 5/4 * 4/3.

And then
each of these terms in turn can be expressed in the standard manner (related to
the first prime 2).

So 5/4 = 1
+ 1/(2

^{2 }+ 1)^{1}+ 1/(2^{2 }+ 1)^{2}+ 1/(2^{2 }+ 1)^{3}+ …
And 4/3 = 1
+ (1/2

^{2})^{1}+ 1/(1/2^{2})^{2}+ 1/(1/2^{2})^{3}+ …
So just as
we have expressed the L1 function as the quotient of two other functions, now
in complementary fashion, we have expressed the L2 as the product of two
functions.

We could
also express 5/3 directly as an L2 (geometric series) function in the following
way,

5/3 = 1 +
{2/(2

^{2 }+ 1)}^{1}+ {2/(2^{2 }+ 1)}^{2}+ {2/(2^{2 }+ 1)}^{3}+ …
So again
just as we can express the quotient of the two L1 functions in a direct manner
equally we can express the product of the two L2 functions equally in a direct
L2 manner.

Finally
this equally applies to the Alt L2 function.

So once
again 5/3 = 5/4 * 4/3

And each of
these can be expressed as an Alt 2 function.

So 5/4 = 1 + 1/6 + 1/21 + 1/ 56 + …

And 4/3 = 1
+ 1/5 + 1/15 + 1/35 + …

So 5/3
thereby represents the product of both these Alt L2 functions.

However
again - as with the geometric series form - a direct Alt L2 expression can be
found.

Here let
5/3 = (n – 1)/(n – 2)

So n = 3.5
which acts as the denominator of the 2

^{nd}term.
As the 1

^{st}term is always 1 then this implies that that the 2^{nd}term is 1/3.5 = 2/7.
Then to get
the 3rd term multiply 7/2 by (3.5 + 1)/2 = 7/2 * 9/4 and get the reciprocal
i.e. 8/63. Then for the next term multiply 63/8 by (3.5 + 2)/3 and get
reciprocal = 16/231, and continue on in this fashion in each case adding an
additional 1 to both numerator and denominator of the multiplying number.

So the
first 4 terms of the series are

1 + 2/7 +
8/63 + 16/231 + … = 5/3.

And then with respect to the sum over the integers expression each individual term can again be expressed both in an L2 (geometric series) and Alt L2 fashion by a similar approach (where 1 is subtracted in each case).

So for example the 7th term, i.e. 2/7

^{2 }= 2/49.
Thus we obtain the L2 and Alt L2 functions for 51/49 (and then finally adjust by subtracting 1 in each case).

The L2 function is {1 + [2/(7

^{2 }+ 2)]^{1 }+ [2/(7^{2 }+ 2)]^{2 }+ [2/(7^{2 }+ 2)]^{3 }+ ...} – 1
The Alt L2 function is {1 + 2/53 + 8/2915 +16/55385 + ...} – 1.

^{ }^{ }^{}
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