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Author: Isaac Zylstra Calvin College Ratio Series Ratio series are series for which the ration an+1 / an is bounded, either above or below by a constant have interesting properties. 1 an+1 If ≤ < 1 for every n, then the set of 2 an selective sums is [0,s]. an+1 1 If 0 < ≤ , then the set of selective sums is an 2 totally disconnected. Sets of Selective Sums of Infinite Series What is a selective sum? ∞ π=1 ππ be a series. A number x is a selective sum of the series = ∞ π=1 ππππ, where {cn} is a sequence consisting only of 0's Let if x and 1's. That is, a selective sum is a sum of the terms of a subsequence of the sequence {an}. The set of selective sums of an infinite series is the set of all possible selective sums for the series ∞ π=1 ππ. Representations Different sets of selective sums have varying numbers of representations for each of its selective sums. If an > Rn for all n, then every selective sum has a unique representation. If an < Rn for infinitely n, and there are infinitely many n such that the selective sum x π π is in both [ π=1 ππππ , π=1 ππππ + Rn+1] and [ ππ=1 ππππ + an + 1, ππ=1 ππππ + Rn], then there are infinitely many representations of x as a selective sum. Construction of the set One way of constructing the set is to do a Cantor-like construction. This construction is called "Cantor-like" because the construction is similar to the way the Cantor Set is constructed. ∞ π=1 ππ Alternating Series as a convergent series of strictly positive terms and strictly decreasing terms. π+1 Then ∞ (−1) ππ is our alternating series. π=1 ∞ ∞ Suppose that π=1 π2π−1 = s1 and π=1 π2π = s 2. an+1 1 If ≥ for all n, then the set of selective of an √2 the alternating series is [-s2, s1] an+1 1 If < , then S is equal to the difference of an √2 two totally disconnected sets. Let ∞ π=1 ππ be a convergent series with sum s, and with terms that are strictly positive and strictly decreasing. Let Rn = ∞ π=π+1 ππ . Begin with the closed interval [0,s], and let F1 = [0, R1] ∪ [s-R1, s] = [0, R1 ∪ [a1, s]. Then let Fn+1 be the union of all intervals [x, x+Rn] and [y-Rn, y], where [x, y] is an interval in Fn. Then ∞ ∞ πΉ is equal to the set of selective sums of the series π=1 π π=1 ππ. Five Rules This construction leads to a set of rules which are the basis for our results. These rules are: 1. If an ≤ Rn for all n, then the set of selective sums is [0, ∞ π=1 ππ] 2. If an < Rn for infinitely many n, but an > Rn for finitely many n, then the set of selective sums is a union of intervals on [0, ∞ π=1 ππ]. 3. If an > Rn for infinitely many n, and an < Rn for infinitely many n, it is possible to get a union of totally disconnected sets, but beyond that, we still don't know (more on this below). 4. If an > Rn for infinitely many n, but an < Rn for finitely many n, then the set of selective sums is an union of totally disconnected sets on [0, ∞ π=1 ππ]. 5. If an > Rn for all n, then the set of selective sums is totally disconnected on [0, ∞ π=1 ππ]. Open Questions We were unable to solve all of the problems we can across. 3 problems in particular seemed interesting: 1. Rule 3, as mentioned earlier: Can an interval could be contained with the set of selective sums of a series with an > Rn for infinitely many n, and an < Rn for infinitely many n? 2. Uncountably infinite representations: Is it possible to construct a series whose set of selective sums contains a selective sum with an uncountable infinite number of representations? 3. 2-dimensional selective sums: What are the properties of selective sums of series of the ∞ form π=1 πππ₯ + πππ¦?