## Signs and Sums in a Sequence

In a sequence of numbers, the sum of any successive three terms is positive. Is it possible for the total of all the terms to be negative? [Shapovalov, p. 15].

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Solution

### References

1. A. Shapovalov, The Principle of Narrow Passages (in Russian), ISBN 5-94057-234-0, 2006, pp 24. In a sequence of numbers, the sum of any successive three terms is positive. Is it possible for the total of all the terms to be negative?

### Solution

The answer (yes or no) depends on the length (say N) of the sequence. If N is divisible by 3, then the answer is an unqualified No. Indeed, in this case, the total consists of the sum of the terms taken by three.

Let's consider the case where N = 1 (mod 3). We may split the sequence into successive triples in a variety of ways each, time leaving a single term which of necessity must be negative. This makes the terms in positions 1, 4, 7, ..., N negative. The rest may be positive. Assume all the negative terms equal -x, x > 0, and all positive terms equal y, y > 0. We have two inequalities to satisfy:

-x + 2y > 0,
-Kx + (N - K)y < 0,

Where K = [N/3] + 1, the brackets being a symbol for the floor function. Note that K > N/3, so that N - K <2N/3 and, hence, 1 < (N - K)/K < 2. So, if we can find x and y that satisfy

 (1) 2y > x and x < [(N - K)/K]y

the problem will be solved. But there's a plenty of pairs (x, y) that satisfy (1), e.g., (4, 3) works for N = 4 and (9, 5) for N ≤ 25.

If N = 2 (mod 3) the consideration does not flow that smoothly. However, solutions do appear to exist. E.g., for N = 8, we have the sequence -8, 4, 5, -2, -2, 5, 4, -8; for N = 11, the sequence -10, 4, 7, -5, -1, 7, -1, -5, 7, 4, -10.

The problem is actually simpler than it sounds. W. McWorter observed that the sequence -1, -1, 2.1, -1 and the sequence -1, -1, 2.1, .1, -1 say that, when N is not divisible by 3, the answer is yes. We can even extend these sequences by appending K copies of the triple -1, 2+e , -1, where e is a positive number such that Ke < .8, providing sequences of any length N > 3 not divisible by 3 such that the answer is still yes.

Nathan Bowler has also remarked that if n is not 0 mod 3 we may easily find a sequence for which all sums of 3 successive terms are positive, but the sum of all the terms is negative.

First we take an obvious simplifying condition: The sequence must be periodic of period 3. Say the first 3 terms are a, b and c. We must have a + b + c > 0, so take a + b + c = 1.

Suppose now that n = 3k + 1. Then we also require a + k < 0. So take a = -k - 1. Now we just need b + c = k + 2. So let, for example, b = k, c = 2.

On the other hand, if n = 3k + 2, we need a + b + k < 0. So take a = -k, b = -1, c = k + 2 (for example). The method obviously generalises for similar problems, such as when the sum of every successive set of 4 terms must be positive.

### Remark

In a circular arrangement of numbers, the situation is entirely different. If we add the sums of all the successive triples, the result will be three times the total, so that the latter is necessarily positive if all such sums are. 