In fact this is equally true with respect to each term of the Zeta 1 sum over natural numbers expression.

In general terms,

ζ(s

_{1}) = 1/1

^{s1}+ 1/2

^{s1}+ 1/3

^{s1}

^{ }+ 1/4

^{s1 }+ …

For example when s

_{1}= 2,

ζ(2) = 1/1

^{2}+ 1/2^{2}+ 1/3^{2 }+ 1/4^{2 }+ … = 1 + 1/4 + 1/9 + 1/16 + …
Now each individual term can be expressed in a Zeta 2 manner.

In general terms,

ζ(s

Thus ζ(s

To convert each term in the Zeta 1 expression to the corresponding Zeta 2 format,

let s

So, as in our example, when n = 2 and s

s

Then in this case, ζ(s

For the next term in our example, s

So s

Therefore we can universally express all the individual variable terms (i.e. other than 1) in the Zeta 1 (Riemann) function, for all positive integers of s, through corresponding Zeta 2 expressions, (i.e. strictly ζ(s

In principle, it is possible to extend these notions for negative integers of s in the Zeta 1 (Riemann) function.

So when again for example, s

ζ(– 2) = 1/1

Then, when s

s

So, – ζ(s

Thus we have now replicated the corresponding term in the Zeta 1 (Riemann) function.

We can also express the product over primes expression for each individual term in the Zeta 1 (Riemann) function through a corresponding Zeta 2 expression.

So the general formula for the product over primes expression of the Zeta 1 function is given as

So with s

1/(1 – 1/2

= (– 1/3) * (– 1/8) * (– 1/25) * ....

So with p = 2 and s1 = – 2, the 1st term in the product over primes expression is given as – 1/3.

Then to express each term as a Zeta 2 infinite series, we let s

Thus again with p = 2 and s1 = – 2,

ζ(s

= 1/(1 – 4) = – 1/3

_{2}) = 1 + s_{2}^{1}+ s_{2}^{2 }+ s_{2}^{3}+ ...Thus ζ(s

_{2}) – 1 = s_{2}^{1}+ s_{2}^{2 }+ s_{2}^{3}+ ...To convert each term in the Zeta 1 expression to the corresponding Zeta 2 format,

let s

_{2}= 1/(1 + n^{s1}), where n = 1, 2, 3, ...So, as in our example, when n = 2 and s

_{1}= 2, ,s

_{2}= 1/(1 + 2^{2}), = 1/(1 + 4) = 1/5Then in this case, ζ(s

_{2}) – 1 = 1/5 + 1/5^{2 }+ 1/5^{3 }+... = 1/4.For the next term in our example, s

_{1}= 2, and n = 3.So s

_{2}= 1/(1 + 3^{2}) = 1/10, with ζ(s_{2}) – 1 = 1/10 + 1/10^{2 }+ 1/10^{3 }+... = 1/9.Therefore we can universally express all the individual variable terms (i.e. other than 1) in the Zeta 1 (Riemann) function, for all positive integers of s, through corresponding Zeta 2 expressions, (i.e. strictly ζ(s

_{2}) – 1, expressions).In principle, it is possible to extend these notions for negative integers of s in the Zeta 1 (Riemann) function.

So when again for example, s

_{1}= – 2,ζ(– 2) = 1/1

^{–}^{2}+ 1/2^{–}^{2}+ 1/3^{–2 }+ 1/4^{–}^{2}^{ }+ … = 1 + 4 + 9 + 16 + …^{}^{}Then, when s

_{2}= – 1/(1 + n^{s1}) with n = 2 and s_{1}= 2,s

_{2}= 1/(1 + 2^{–}^{2}) = – 1/(1 + 1/4) = 5/4So, – ζ(s

_{2}) = – {5/4 + (5/4)^{2 }– (5/4)^{3}+ ...} = – (1 – 5/4) = 1/4 = 4, where ζ(s2) for negative values = 1 – {ζ(s_{2}) – 1}.Thus we have now replicated the corresponding term in the Zeta 1 (Riemann) function.

We can also express the product over primes expression for each individual term in the Zeta 1 (Riemann) function through a corresponding Zeta 2 expression.

So the general formula for the product over primes expression of the Zeta 1 function is given as

∏(1/(1 – 1/p

^{s1})So with s

_{1}= – 2, we obtain,1/(1 – 1/2

^{–}^{2}) * 1/(1 – 1/3^{–}^{2}) * 1/(1 – 1/5^{–}^{2}) * ...= (– 1/3) * (– 1/8) * (– 1/25) * ....

So with p = 2 and s1 = – 2, the 1st term in the product over primes expression is given as – 1/3.

Then to express each term as a Zeta 2 infinite series, we let s

_{2 }= 1/p^{s1}.Thus again with p = 2 and s1 = – 2,

ζ(s

_{2}) = 1 + 4 + 4^{2 }+ 4^{3}+ ...= 1/(1 – 4) = – 1/3

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