Some Number Magic

It struck me when I had finished the last entry that a ready means thereby existed for solving another interesting feature regarding the sum of the Riemann Zeta functions (less 1) for both odd and even integer values of s (> 1).

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, ζ₁(s) – 1 = .20205…, .03692…, .00834… and .002008… respectively.

So the sum of these values = .2493…, which is already very close to .25. And as ∑{ζ₁(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,

ζ₁(2) – 1       = 1/2² + 1/3² + 1/4² +  …       =  .64493…
ζ₁(3) – 1       = 1/2³ + 1/3³ + 1/4³ +  …       =  .20203…
ζ₁(4) – 1       = 1/2⁴ + 1/3⁴ + 1/4⁴  + …        =  .08232…
ζ₁(5) – 1       = 1/2⁵ + 1/3⁵ + 1/4⁵  + …        =  .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 = 1/2³ + 1/2⁵ + 1/2⁷ + …  = (1/8)/1 – 1/4) = 1/8 * 4/3            = 1/6

The 2^nd series = 1/3³ + 1/3⁵ + 1/3⁷ + …  =  (1/27)(1– 1/9)  = 1/27 * 9/8           = 1/24

The 3^rd series  = 1/4³ + 1/4⁵ + 1/4⁷ + …  = (1/64)(1– 1/16)  = 1/64 * 16/15     = 1/60

The 4^th series = 1/5³ + 1/5⁵ + 1/5⁷ + …  =  (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 ∑{ζ₁(s) – 1} where s is odd (>1) = 1/4 (.25).

Thus ∑{ζ₁(s) – 1}where s is even (> 0) = 3/4 (.75).

There are many fascinating number patterns in evidence.

The series used to prove that ∑{ζ₁(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. 1/1²).

Then the series used to prove that ∑{ζ₁(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. 1/2²).

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. 1/3²).

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. 1/4²).

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. 1/5²). 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. 1/2²). However if alternatively, we divide 1/6 by 3/2,

(1/6)/(3/2) = 1/6 * 2/3 = 1/9 (i.e. 1/3²) . And again this feature is universal, where in the former case 1/n² (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. 1/4²). And, 1/20 * 4/5 = 1/25 (i.e. 1/5²). 

By progressing in the manner above we generate the series,

1/1² + 1/2² + 1/3² + 1/4² + … i.e. ζ₁(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.