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  "body": "\n## Definition\n\nLet $X$ be a [[normed vector space|00A0RPAQ8A]] over $\\R$ or $\\C$.\nLet $I$ be a set, and let $(x_i)_{i \\in I}$ be a family of elements of $X$.\nFor each [[finite subset|0BBTH3YSZA]] $J \\subseteq I$ put $s_J = \\sum_{i \\in J} x_i$.\nThe family is summable with sum $s \\in X$ if the following holds.\nFor every $\\varepsilon \u003e 0$ there is a finite set $J_0 \\subseteq I$ with $\\norm{s_J - s} \u003c \\varepsilon$ for every finite $J$ with $J_0 \\subseteq J \\subseteq I$.\nThen one writes $s = \\sum_{i \\in I} x_i$.\n\n## Remarks\n\nEquivalently, the [[net|4DVJQJV0DG]] $(s_J)$, indexed by the finite subsets of $I$ ordered by inclusion, converges to $s$.\nThe sum is unique when it exists.\nThe same definition works in any Hausdorff abelian topological group, with neighbourhoods of $s$ in place of balls; see [[summation in topological groups|TXS2H001R4]].\nIf all $x_i$ are nonnegative reals, the family is summable if and only if the finite sums $s_J$ are bounded, and then $s = \\sup_J s_J$.\nExample: the family $(2^{-m} 3^{-n})_{(m, n) \\in \\N \\times \\N}$ is summable with sum $3$.\nIts finite sums are bounded by $\\sum_{m \\le M} 2^{-m} \\sum_{n \\le M} 3^{-n} \\le 2 \\cdot \\tfrac32$, and the squares $\\set{0, \\ldots, M}^2$ give sums tending to $3$.\nNon-example: the family $((-1)^n / (n+1))_{n \\in \\N}$ is not summable, although the series $\\sum (-1)^n/(n+1)$ converges.\nAny finite set $J_0$ can be enlarged by many even indices $n$, which adds more than $1$ to $s_J$, because $\\sum 1/(2k+1)$ diverges.\nIn a Banach space, the family is summable if and only if for every $\\varepsilon \u003e 0$ there is a finite $J_0$ with $\\norm{s_K} \u003c \\varepsilon$ for every finite $K$ disjoint from $J_0$.\nA summable family in a normed space has at most countably many nonzero terms.\n\n## Explanations\n\n### L1 human\n\nThe sum of a family is the limit of its finite partial sums, where the finite batches of terms grow in every direction at once.\nThis removes any dependence on the order of the terms.\n",
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