I know there is a definition of "sumable family" in topological vector spaces as follows: the family of vectors $(v_\alpha)_{\alpha \in \Gamma}$ possesses for sum the vector $v$ if for every neighbourhood $\mathrm{N}$ of zero there exists a finite subset $\mathrm{F} \subset \Gamma$ such that $v - \sum\limits_{\alpha \in \mathrm{K}} v_\alpha \in \mathrm{N},$ regardless what finite set $\mathrm{K} \subset \Gamma$ was as long as $\mathrm{F}$ belonged (as a subset) to $\mathrm{K}.$ When dealing with normed spaces, this condition is weaker than absolute convergence for the case $\Gamma = \Bbb N.$ Is there a condition of "equal strenght" or stronger than absolute convergence?
Summable vs absolutely summable
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sequences-and-series
normed-spaces
divergent-series
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0The condition of equal strength to absolute convergence is the condition of absolute convergence. Stronger? Sure: all but finitely many terms are zero. – 2017-01-27
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0But every tautology is of equal strength of the definition it describes. I want something meaningful that translates to the original case when appropriate. For instance, the problem with the previous definition is that what is not in $\mathrm{F}$ is uncontrolled (in the case of a series is that there is no control on the tail). So, my guess goes as what to impose in the tail of the series (however, there is no tail in the case of families). If you need further clarification, I'm happy to do so. – 2017-01-27