## Monday, September 18, 2017

### Ratios and Sums of Reciprocals

In previous entries, I have shown the first 9 terms (as an illustration) in the respective unique number sequences for the 9 equations from (x – 1)1 = 0 to (x – 1)9 = 0 .
 1 1 1 1 1 1 1 1 1 1 2 3 4 5 6 7 8 9 1 3 6 10 15 21 28 36 45 1 4 10 20 35 56 84 120 165 1 5 15 35 70 126 210 330 495 1 6 21 56 126 252 462 792 1287 1 7 28 84 210 462 924 1716 3003 1 8 36 120 330 792 1716 3432 6435 1 9 45 165 495 1287 3003 6435 12870

Now we can always derive the unique number sequences for (x – 1)n + 1 from the corresponding sequences for (x –1)n through application of the fact that the kth term in the former = the sum of the first k terms in the latter sequence.

Therefore as we can see in the unique number sequence for (x – 1)3 i.e. 1, 3, 6, 10, …, the first term 1, represents the sum of the first single term in the corresponding sequence for sequence for (x – 1)2 i.e. 1, 2, 3, 4, …; the second term, 3 then represents the sum of the first 2 terms in the corresponding sequence i.e. 1 + 2; the 3rd term, 6 represents the sum of the first 3 terms in the previous sequence i.e. 1 + 2 + 3; the fourth term, 10 then represents the sum of the first 4 terms in the previous sequence i.e. 1 + 2 + 3 + 4 and so on indefinitely.

However the limitation of this procedure is that we must already know the unique number sequence corresponding to (x – 1)n to be able to calculate the corresponding sequence for (x – 1)n + 1.

However there is a simple way to calculate independently the unique number sequence corresponding to (x – 1) n for any given n.

The bais for this calculation is that in general terms the ratio of the (k + 1)th to the kth term
i.e. (k + 1)th/ kth = (k + n – 1)/k

So for example in the sequences above, when n = 4, the unique digit sequence for (x – 1)4
is given by 1, 4, 10, 20, 35, 56, 84, 120, 165,…

So if for example we take k = 6 then the  ratio of the 7th to the 6th term i.e. 7th/6th = (6 + 4 – 1)/6 = 9/6 (i.e. 3/2).

And as we can see, this is indeed true for the 7th term = 84 and the 6th term = 56 and 84/56 = 3/2.

So aided with this simple general fact, regarding the ratio of successive terms, to illustrate, I will now calculate the unique number sequence corresponding to (x – 1)12.
Now the first term is - as always - 1.
Therefore the  2nd term  =1 * (1 + 12 – 1)/ 1 = 12.
The 3rd term then = 12 * (2 + 12 – 1)/2 = 12 * 13/2 = 78.
The 4th term = 78 * (3 + 12 – 1)/3 = 78 * 14/3 = 364.
The 5th term = 364 * (4 + 12 – 1)/4  = 364 * 15/4 = 1365.
The 6th term = 1365 * (5 + 12 – 1)/5 = 1365 * 16/5 = 4368.
The 7th term = 4368 * (6 + 12 – 1)/6 = 4368 * 17/6 = 12376.

So the unique digit sequence corresponding to (x – 1)12 is

1, 12, 78, 364, 1365, 4368, 12376, …

Now we already know that the 12 roots of this equation are 1.

However if we attempt to approximate these roots through the ratio of (k + 1)th/kth terms, we must include a great number of terms so as to get a valid approximation.

So ultimately (k + 1)th/kth  term  ~ 1 (when k is sufficiently large).

I then looked at the sum of reciprocals for these unique number sequences to find that an interesting general pattern was at work.

Clearly from a conventional perspective, the sums of reciprocals of the numbers associated with the first two sequences for (x – 1)1 and (x – 1)2 respectively, diverge.

However the sum of reciprocals of the sequence, corresponding to (x – 1)3
i.e. 1 + 1/3 + 1/6 + 1/10 + 1/15 + …  converges to 2.

In fact a general result can be given for all such convergent sequences with respect to the sums of reciprocals of all the unique number sequences associated with (x – 1)3 where n ≥ 3.

This result in fact depends solely on the value of n (as the dimensional power or index) and is given simply as (n – 1)/(n – 2).

So for example, the sum of reciprocals of the next number sequence, corresponding to
(x – 1)4, i.e.

1 + 1/4 + 1/10 + 1/20 + 1/35 + …, converges to (4 – 1)/(4 – 2) = 3/2.

And the sum of reciprocals corresponding to the number sequence uniquely associated with
(x – 1)12, that we earlier calculated i.e.

1 + 1/12 + 1/78 + 1/364 + 1/1365 + 1/4368 + 1/12376 + … thereby converges to (12 – 1)/(12 – 2)  = 11/10 = 1.1.

In fact the sum of the first 7 terms above = 1.09994…, which is already very close to the postulated answer.