Consider a (Hausdorff and complete) locally convex topological vector space $V$ and a countable subset $(v\_k)\_{k=1}^\infty \subset V$ of non-zero vectors.



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> $(\*)$ Under what conditions on this subset are we guaranteed the existence of a sequence of positive real numbers $(\alpha\_k)\_{k=1}^\infty$, such that the series $\sum\_{k=1}^\infty \alpha\_k v\_k$ converges (that is, the sequence of its partial sums) converges to a vector $v \in V$?
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The positivity of the coefficients, $\alpha\_k > 0$ for each $k$, is the crucial aspect of the question.


On the one hand, if $V$ is a Banach space, the answer is always Yes. We can always choose $\alpha\_k = c\_k / \|v\_k\|$, where $(c\_k)\_{k=1}^\infty$ is any summable sequence of positive numbers.


On the other hand, consider the space $c\_{00}$ of finite sequences of real numbers, turned into a complete locally convex space in the usual way. The the subset $(v\_k)\_{k=1}^\infty$, where the $v\_k$ is the sequence that is zero everywhere except the $k$-th place, is an example for which such a sequence of $(\alpha\_k)\_{k=1}^\infty$ does not exist. This is clear because here a necessary and sufficient condition for the convergence of $\sum\_{k=1}^\infty \alpha\_k v\_k$ is that only finitely many $\alpha\_k$ are non-zero.


So maybe I could ask the question in a slightly different way. Is there a well-studied (joint?) property of the space $V$ and the subset $(v\_k)\_{k=1}^\infty$ that is sufficient (hopefully also necessary) for the answer to $(\*)$ to be Yes? I was hoping that the answer would be related to one of the standard theorems of functional analysis, but I haven't recognized the right one yet.



Existence of a countable linear combination with positive coefficients
As a partial answer, note the following necessary (and often sufficient) criteria.


Let $E$ be a (real or complex) topological vector space. Let us say that a sequence $\{x\_n\}\_{n=1}^\infty$ in $E$ *can be made summable/bounded/to converge to zero* if there is a sequence $\{\alpha\_n\}\_{n=1}^\infty$ of strictly positive real scalars such that $\{\alpha\_n x\_n\}\_{n=1}^\infty$ is summable/bounded/converges to zero.



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> **Proposition.** *Consider the following statements:*
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> *$\text{(a)}$. $\{x\_n\}\_{n=1}^\infty$ can be made summable;*
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> *$\text{(b)}$. $\{x\_n\}\_{n=1}^\infty$ can be made to converge to zero;*
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> *$\text{(c)}$. $\{x\_n\}\_{n=1}^\infty$ can be made bounded.*
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> *In every topological vector space $E$ one has*
> $$\text{(a)} \Longrightarrow \text{(b)} \Longleftrightarrow \text{(c)}.$$
> *If $E$ is locally convex and sequentially complete, then*
> $$\text{(a)} \Longleftrightarrow \text{(b)} \Longleftrightarrow \text{(c)}. $$
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> *Proof.* $\text{(a)} \Longrightarrow \text{(b)}$. Suppose that $\sum\_{k=1}^\infty \alpha\_k x\_k$ converges to $y \in E$. Let $s\_n := \sum\_{k=1}^n \alpha\_k x\_k$ denote the $n$-th partial sum. Then $\lim\_{n\to\infty} s\_n = \lim\_{n\to\infty} s\_{n+1} = y$, so by continuity of addition
>  $$ \lim\_{n\to\infty} \alpha\_{n+1} x\_{n+1} = \lim\_{n\to\infty} s\_{n+1} - s\_n = y - y = 0. $$
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> $\text{(b)} \Longrightarrow \text{(c)}$. [Every convergent sequence is bounded](https://math.stackexchange.com/q/1471005/).
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> $\text{(c)} \Longrightarrow \text{(b)}$. If $\{\alpha\_n x\_n\}\_{n=1}^\infty$ is bounded, then $\lim\_{n\to\infty} \alpha\_n \beta\_n x\_n = 0$ for every sequence $\{\beta\_n\}\_{n=1}^\infty$ of scalars converging to zero (see e.g. Rudin, Theorem 1.30).
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> Assume now that $E$ is locally convex and sequentially complete. $\text{(b)} \Longrightarrow \text{(a)}$. Suppose that $\lim\_{n\to\infty} \alpha\_n x\_n = 0$. Define $s\_n := \sum\_{k=1}^n \frac{\alpha\_k}{2^k} x\_k$. Use local convexity to prove that $\{s\_n\}\_{n=1}^\infty$ is Cauchy, and use sequential completeness to conclude that the series $\sum\_{k=1}^\infty \frac{\alpha\_k}{2^k} x\_k$ converges. $\Box$
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Some remarks:


1. Each of the properties $\text{(a)}$, $\text{(b)}$ and $\text{(c)}$ carries over to all weaker (i.e. coarser) topologies, even those with different dual spaces. Interestingly, in locally convex spaces, property $\text{(c)}$ only depends on the dual pair.
2. If the topology of $E$ is given by a metric $d$, then every sequence $\{x\_n\}\_{n=1}^\infty$ satisfies $\text{(b)}$. To prove this, note that for every $z \in E$ we have $\lim\_{k\to\infty} \frac{1}{k}z = 0$ (by continuity of scalar multiplication). Equivalently: for every $\varepsilon > 0$ there exists some $K\_\varepsilon \in \mathbb{N}\_1$ so that $d(0,\frac{1}{k} z) < \varepsilon$ for all $k \geq K\_\varepsilon$. Using this, we choose a sequence $\{\alpha\_n\}\_{n=1}^\infty$ of strictly positive real scalars so that $d(0,\alpha\_n x\_n) < \frac{1}{n}$ for all $n\in\mathbb{N}\_1$. Then $\lim\_{n\to\infty} \alpha\_n x\_n = 0$.
3. The answer of Robert Israel follows from the previous remark and the Proposition.
4. There are also non-metrizable tvs where every sequence satisfies $\text{(b)}$ (take a metrizable space and pass to a weaker topology). Likewise, there exist non-Fréchet tvs where every sequence satisfies $\text{(a)}$. Interesting [follow-up question](https://mathoverflow.net/q/355466/120251): if every sequence satisfies $\text{(b)}$ (resp. $\text{(a)}$), then is there necessarily a finer topology which is metrizable (resp. Fréchet)?


