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.

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