Formal-Numeric Series

```It's interesting to consider formal associations of different
numerical series with each other according to their partial sums
rather than according to their terms.  Specifying the partial sums
implies an ordering that removes infinitely many degrees of freedom
from the evaluation of a series.  The importance of this for
conditionally convergent numerical series (such as 1 - 1/2 + 1/3 -
1/4 + ...) is obvious, but this condition is also significant for
associations between formal-numeric divergent series.  As an example,
consider the natural series

1 + 2 + 3 + 4 + ...

which is obviously divergent.  If we disregard the implied partial
sums we can re-partition this "quantity" in infinitely many ways,
such as

1  +  2  +  3  +   4   +  5  +  6  +  7  +   8 +  ....

= (1) + (2) + (3) + (1+3) + (5) + (6) + (7) + (3+5) + ...

= (1) + (2) + (3+1) + (3+5) + (6+7+3) + (5+..) + ...

=  1  +  2   +  4   +   8   +   16  +  ...

On this basis it's clear that the natural series is "equivalent" to
any other divergent series, such as the geometric series of powers
of 2, so this kind of "formal equivalence" is not particularly
interesting.  However, we can impose a bit more structure on our
definition of equivalence of formal-numeric series if we require
equality of partial sums.  In the example above, the only occurrences
of equal partial sums for the natural series 1+2+3+... and the
geometric series 1+2+4+... are

1 = 1
1+2 = 1+2
1+2+3+4+5 = 1+2+4+8
1 + 2 + 3 + ... + 89 + 90  =  1 + 2 + 4 + 8 +... + 1024 + 2048

Thus, only a finite number of partial sums of these two formal-
numeric series are equal.  But what about other bases?  In general,
since the nth partial sum of the natural sequence is n(n+1)/2, and
the mth partial sum of the geometric sequence 1 + b + b^2 + ...  is
(b^m - 1)/(b-1), we have a case of equal partial sums for every pair
of integers n,m such that  n(n+1)/2 = (b^m - 1)/(b-1).  For any
given b, solutions of this Diophantine equation are usually quite
rare.  A few examples in small integers (b,n,m) are (2,90,12),
(23,159,4), and (26,37,3).

An exception is b=9, for which there are infinitely many
solutions. This occurs because we have n(n+1)/2 = (9^m - 1)/8,
which can be written  4n^2 + 4n + 1 = 9^m, and since both sides
are squares we have  2n+1 = 3^m , and so n = (3^m - 1)/2, which
gives an integer solution for every integer m.  Thus, EVERY
partial sum of 1 + 9 + 81 + 729 +...  equals a partial sum
of the natural sequence.  [Is there a name for this kind of
relationship between formal-numeric series, i.e., if every
partial sum of S1 is a partial sum of S2)?]

In general, the geometric series in the base b will have all of
its partial sums equal to triangular numbers (i.e., partial sums
of the natural series) if and only if b and (b-1)/2 are both
squares.  Thus we need integers n,m such that

n^2  =  2m^2 + 1

which is simply a Pell equation, with the infinite sequence of
solutions
n      m
---    ---
1      0
3      2
17     12
99     70
etc.

Therefore, the geometric series has only triangular partial sums
if and only if the base b has one of the values 9, 289, 9801, and
so on.
```