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I'm wondering about a simple question that has multiple possible variants depending on a few parameters. The prototypical one would be:

Does there exist an infinite series such that any two subseries are linearly independent over $\mathbb{Q}$?

Assume that the two subseries (sums of subsets of terms from the original series) in question are to be summed in order of increasing index - this allows one to put in place a restriction on absolute or conditional convergence. One can add a combination of restrictions from {finite, cofinite, infinite} to each of the two subseries, e.g. "any finite subseries $\circ$ and cofinite subseries $\bullet$." Lastly, one can replace linear independence with algebraic independence - much tougher, I imagine - or replace indepenence with de$\text{}$pendence.

Are there any general results on these sorts of questions?

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An infinite series of what? Rational numbers, or reals? Reals is easy -- just take a transcendence basis for $\mathbb R/\mathbb Q$ and select $\aleph_0$ random elements from it to make your series. –  Henning Makholm Sep 30 '11 at 2:28
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@Henning: I was thinking reals. I see how choosing any countable number of elements from a transcendence basis (such that their sum converges) would make any two finite or two cofinite subsums linearly independent, but I don't see how that holds between generally infinite subseries. Intuitively, since a sum of rationals takes elements from the equivalence class represented by $0\in\mathbb{R}/\mathbb{Q}$, but as an infinite series can evaluate to any kind of real number, I figure that an infinite sum's transcendence properties aren't determined by those of its terms - is my impression wrong? –  anon Sep 30 '11 at 2:38
    
Hmm, I hadn't grasped that you were talking about independence as sums rather than as mere sequences. Sorry. –  Henning Makholm Sep 30 '11 at 2:46
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It seems worth pointing out the example François Dorais gives of the series $\sum\dfrac{1}{n!}$ with a linearly independent family of subseries indexed by $\mathbb R$. While not all subseries are considered there, the conclusion of the whole set being linearly independent is much stronger. –  Jonas Meyer Jul 26 '13 at 8:07
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A partial answer to this question is that the answer to the boxed question is "yes", assuming that the subseries are taken to be distinct and nonempty. I will construct a sequence $r_1$, $r_2$, $\dots$ of positive real numbers such that $\sum_j r_j$ is convergent, and, for any choice of $k$, subsets $S_1$, $S_2$, $\dots$, $S_k$ of $\{1,2,3,\ldots\}$, and $\alpha_1$, $\dots$, $\alpha_k\in\Bbb Q$, $$ \sum_{1\le i\le k} \alpha_i \sum_{j\in S_i} r_j=0 \ \ \ \Rightarrow\ \ \ \sum_{1\le i\le k} \alpha_i 1_{S_i}=0, \ \ \ \ \ \ \ (*) $$ where $1_T$ is the indicator function of $T$. This means that there are no $\Bbb Q$-linear relationships between the subsums of the $r_j$s apart from those forced to exist by relationships such as $$ (r_1+r_2)+(\sum_{j\ge 3} r_j)=\sum_j r_j, $$ $$ (\sum_{j{\rm\ odd}} r_j)+(\sum_{j{\rm\ even}} r_j)=\sum_j r_j, $$ and so on.

The sequence is defined similarly to the example mentioned by Jonas Meyer above. Let $T_1$, $T_2$, $\dots$ be any partition of $\{1,2,3,\dots\}$ into countably many subsets, each countably infinite. Then, set $$ r_j:=\sum_{\ell\in T_j} \frac{1}{\ell!}, \qquad j=1, 2, 3, \dots. $$ To prove $(*)$, assume that the left-hand side of $(*)$ holds. We can clear denominators from the $\alpha_i$s to find that, for some $a_1$, $\dots$, $a_k\in\Bbb Z$, $$ \sum_{1\le i\le k} a_i \sum_{j\in S_i} r_j=0, $$ which can be rewritten as $$ \sum_\ell \frac{C_\ell}{\ell!}=0, \ \ \ \ {\rm where}\ \ \ \ C_\ell:=\sum_{1\le i\le k} a_i 1_{S_i}(\phi(\ell)), \ \ \ \ \ \ \ (**) $$ and $\phi$ is the function taking any $\ell\in\{1,2,3,\dots\}$ to the unique $j$ satisfying $\ell\in T_j$. Now, $|C_\ell|$ is bounded above by $R:=\sum_{1\le i\le k} |a_i|$. Assume that $s\ge R$; then multiplying $(**)$ by $s!$ gives $$ \frac{C_{s+1}}{s+1} +\frac{C_{s+2}}{(s+1)(s+2)} +\frac{C_{s+3}}{(s+1)(s+2)(s+3)} + \cdots\in\Bbb Z, $$ but the magnitude of the difference of the left-hand side of this expression from $C_{s+1}/(s+1)$ is less than $R/(s+1)^2<1/(s+1)$. It follows that $C_{s+1}/(s+1)$ must be an integer, so $C_{s+1}=0$ whenever $s\ge R$. Since each $T_j$ is infinite, each $T_j$ must contain some member exceeding $R$, so $\phi$ must have full range when restricted to $\{R+1,R+2,R+3,\dots\}$. It follows that $\sum_{1\le i\le k} a_i 1_{S_i}=0$, proving $(*)$.

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Perfect. ${}{}$ –  anon Jan 31 at 4:13
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