Yesterday I illustrated just one important example of the intimate relationship exists as between the Zeta 1 and Zeta 2 Functions.

Once again - though this web-page does not ideally lend itself to sophisticated mathematical notation - I denote the Zeta 1 as ζ1(s) and the Zeta 2 as ζ2(s)respectively.

We saw then that that ζ1(1)i.e. the harmonic series = 1 + 1/2 + 1/3 + 1/4 + .... can be expressed in terms of a sum of Zeta 2 series where the value of s (for the Zeta 2) ranges over the reciprocals of all the natural numbers (except 1).

So the complementarity here relates to the fact that the value of s in the Zeta 1 (= 1) is related to all other natural numbers (except 1) with respect to the Zeta 2.

Thus ζ1(1) = ∑{ζ2(s)- 1} where s represents the reciprocal of 2, 3, 4, ....

Now a fascinating reverse form of complementarity also exists as between the Zeta 2 and the Zeta 1 Functions.

When s = 1

ζ2(1) = 1 + 1 + 1 + 1 +.....

Therefore for a finite range of values (i.e. for s = 1, up to n)

ζ2(1) = n

Now with respect to the corresponding Zeta 1 Function

ζ1(2) = (π^2)/6 = 1.6449340668...

ζ1(3) = 1.2020569032...

ζ1(4) = (π^4)/90 = 1.0823232337...

ζ1(5) = 1.0369277551...

ζ1(6) = (π^6)/945 = 1.0173430619...

ζ1(7) = 1.0083492774...

ζ1(8) = (π^8)/9450 = 1.0040773561...

ζ1(9) = 1.0020083928...

ζ1(10)= (π^10)/93555 = 1.0009945751...

and so on

Now if we subtract 1 from each value of the Zeta 1 and sum up the remaining values the total sum (for all natural number values of s > 1) = 1.

Indeed, when we sum up the first 10 values (with again 1 subtracted from each value) the total already converges very closely on 1 i.e. .9990146221...

Therefore when we obtain the total for all these Zeta 1 Functions over the finite range (this time from s = 2 up to n)

the answer converges on (n - 1) + 1 = n

So just as we have shown that,

ζ1(1) = ∑{ζ2(s)- 1} where s represents the reciprocal of 2, 3, 4, .... ,

Now in like reverse manner where the value of s with respect to Zeta 2 can range from 2 to n (with no finite upper limit on n)

ζ2(1) ~ ∑{ζ1(s)

Now in the former expression we subtracted 1 from the sum of each of the Zeta 2 values.

We do not do the same in the case of each Zeta 1 value as there is a lagged nature between both Functions differing by 1.

For example in the case where s = 0 with respect the the Zeta 1,

ζ1(0) = 1 + 1 + 1 + 1 +...

However this corresponds exactly with the Zeta 2 where now s = 1!

i.e. ζ2(1) = 1 + 1 + 1 + 1 +...

In fact other fascinating aspects with respect to Zeta 1 values can be briefly illustrated here!

Again when we subtract 1 from the sum of the Zeta 1 Function (ranging over the natural numbers from 2 upwards),

the sum of even numbered values converges to .75!

Again using the values listed above for the first 5 even values (up to 10), the relevant sum is

.6449340668 + ..08232332337 +.01734306198 + .00407735619 + .0009945751 = .74967229717

So the answer here has already converged very closely on .75!

This implies that the sum of the odd numbered values (excluding s = 1) converges on .25!

It equally implies that the alternating series (where each even is balanced by its succeeding odd term)

i.e. ζ1(2) - ζ1(3) + ζ1(4) - ζ1(5) +..... = .5

It is well known that with respect to the Zeta 1, for the sum of the harmonic series,

ζ1(1)~ log n + γ (where γ is the Euler-Mascheroni Constant = .5772156649..)

Fascinatingly γ is directly connected with the Zeta 1 Function (for all natural number values starting with 2) in the following manner

i.e. γ = ζ1(2)/2 - ζ1(3)/3 + ζ1(4)/4 - ζ1(5)/5 + ....

Therefore

ζ1(1)~ log n + ζ1(2)/2 - ζ1(3)/3 + ζ1(4)/4 - ζ1(5)/5 + ....

So,

log n ~ ζ1(1) - ζ1(2)/2 + ζ1(3)/3 - ζ1(4)/4 + ζ1(5)/5 - .... where the Zeta Functions are summed over a finite range of terms.

Alternatively,

γ ~ ζ1(1) – {ζ1(2)/2 + ζ(13)/3 + ζ1(4)/4 + ζ1(5)/5 +…..} again when summed to a finite n with approximation improving as n increases.

Thus,

log n ~ ζ1(2)/2 + ζ1(3)/3 + ζ1(4)/4 + ζ1(5)/5 +….. when summed to a finite n with approximation improving as n improves.

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