## Thursday, September 28, 2017

### 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 2nd series = 1/33 + 1/35 + 1/37 + …  =  (1/27)(1– 1/9)  = 1/27 * 9/8           = 1/24
The 3rd series  = 1/43 + 1/45 + 1/47 + …  = (1/64)(1– 1/16)  = 1/64 * 16/15     = 1/60
The 4th 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 1st 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 2nd 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 3rd 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 4th 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.