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Is there any sequence $\{a_n:n\in \mathbb N\}$ with each non-zero terms s.t. $\displaystyle \sum_{n=1}^{\infty}a_n=0$?

I'm finding an infinite linearly dependent set which has no finite linearly dependet subset. If I could find the above then I could say so.

Edit: sorry I wrote first 'strictly positive' terms instead of 'non-zero'.

Mini_me
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  • In your current form, yes - but the easiest examples have finite sums which also add to zero. Consider any convergent series $\sum_n b_n$ - e.g. $b_n=\frac1{n^2}$ - and then let $a_{2n} = b_n, a_{2n+1} = -b_n$. – Steven Stadnicki Sep 18 '17 at 05:17
  • There is a positive answer to your 'root' question - is there an infinite set summing to zero with no linearly dependent finite subset - but the best examples I know are conditionally convergent, so there are very thorny questions of what 'infinite linear dependence' means. There are certainly absolutely convergent examples, but I imagine those are more complicated still. – Steven Stadnicki Sep 18 '17 at 05:23
  • @StevenStadnicki Here 'infinite linear dependent' set means a linearly dependent subset which has infinite elements of some vector space X – Mini_me Sep 18 '17 at 05:29

3 Answers3

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No.If there were one as you said then $0 < a_1 \le \sum a_n = 0$, contradiction.

DeepSea
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No, there is not such sequence. If there were a sequence of nonnegative terms $a_n$ with one strictly positive term $a_{n_0}$ then we would have $$ \sum_{n=1}^\infty a_n\geq a_{n_0}>0. $$

[For the updated question] Yes: If $\{a_n:n\geq1\}$ is any sequence of non-zero terms such that $S:=\sum_{n=1}^\infty a_n\neq0$ then the sequence $\{b_n:n\geq1\}$ defined by $b_1=S,\ b_n=a_{n-1}$ is a sequence of non-zero terms and its sum is $0$.

Jonatan B. Bastos
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Let convergence be understood in the $\|\cdot\|_\infty$ norm.

Consider the familiy of sequences $$\mathcal{F} = \left\{-\left(\underbrace{0, 0, \ldots, 0}_{n-1}, \frac{1}{2^n}, 0, 0, \ldots\right) : n \in \mathbb{N}\right\} \cup \left\{\left(1, \frac{1}{2}, \frac{1}{4}, \ldots, \frac{1}{2^n}, \frac{1}{2^{n+1}}, \ldots\right)\right\} \subseteq c_0$$

Notice that every finite subset of $\mathcal{F}$ is linearly independent, since the last sequence has infinite support, and the others have finite disjoint supports.

Also, $$\sum_{x \in \mathcal{F}}x = \left(1, \frac{1}{2}, \frac{1}{4}, \ldots, \frac{1}{2^n}, \frac{1}{2^{n+1}}, \ldots\right) - \sum_{n=1}^\infty\left(\underbrace{0, 0, \ldots, 0}_{n-1}, \frac{1}{2^n}, 0, 0, \ldots\right) = 0$$

since:

$$\left\|\left(1, \frac{1}{2}, \frac{1}{4}, \ldots, \frac{1}{2^n}, \frac{1}{2^{n+1}}, \ldots\right) - \sum_{n=1}^N\left(\underbrace{0, 0, \ldots, 0}_{n-1}, \frac{1}{2^n}, 0, 0, \ldots\right)\right\|_\infty$$

$$=\left\|\left(1, \frac{1}{2}, \frac{1}{4}, \ldots, \frac{1}{2^n}, \frac{1}{2^{n+1}}, \ldots\right) - \left(1, \frac{1}{2}, \frac{1}{4}, \ldots, \frac{1}{2^{N}}, 0, 0, \ldots\right)\right\|_\infty$$

$$=\left\|\left(0, 0, \ldots, 0, \frac{1}{2^{N+1}}, \frac{1}{2^{N+2}}, \ldots\right)\right\|_\infty = \frac{1}{2^{N+1}}\xrightarrow{N\to\infty} 0$$

This property is sometimes called $\omega$-independence:

$$\sum_{n=1}^\infty\alpha_n a_n = 0 \implies \alpha_n = 0, \,\forall n\in\mathbb{N}$$

Thus, $\mathcal{F}$ is (finitely) linearly independent, but it is not $\omega$-independent.

mechanodroid
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  • An even better example might be the set ${\bar{a}n} = {\langle a{n,i}\rangle}$ with $a_{n,n}=2^{-n}$ and $a_{n,n+1}=-2^{-(n+1)}$ (along with a $a_{-1} = \langle -1,0,0,\ldots\rangle$); each member is finitely supported, no finite linear combination is zero, but the infinite sum converges to zero. – Steven Stadnicki Sep 18 '17 at 16:04