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  "body": "\n## Statement\n\nFor each natural number $n$, let $(a_k(n))_{k \\ge 0}$ be a sequence of complex numbers.\nAssume that:\n\n1. for each $k$, the limit $a_k = \\lim_{n \\to \\infty} a_k(n)$ exists;\n2. there are nonnegative real numbers $M_k$ with $\\abs{a_k(n)} \\le M_k$ for all $k$ and $n$, and $\\sum_k M_k \u003c \\infty$.\n\nThen the series $\\sum_k a_k(n)$ and $\\sum_k a_k$ [[converge absolutely|0W3K3QPEWT]], and\n$$\\lim_{n \\to \\infty} \\sum_{k=0}^{\\infty} a_k(n) = \\sum_{k=0}^{\\infty} a_k.$$\n\n## Remarks\n\nGiven $\\varepsilon \u003e 0$, choose $K$ with $\\sum_{k \u003e K} M_k \u003c \\varepsilon$.\nThe tails beyond $K$ of both series are then smaller than $\\varepsilon$, since $\\abs{a_k} \\le M_k$ too.\nThe finitely many terms with $k \\le K$ converge, so the difference of the two sums is less than $3\\varepsilon$ for large $n$.\nExample: for real $x$ put $a_k(n) = \\binom{n}{k} x^k / n^k$ for $k \\le n$ and $a_k(n) = 0$ for $k \u003e n$.\nThen $\\sum_k a_k(n) = (1 + x/n)^n$, $a_k(n) \\to x^k/k!$, and $\\abs{a_k(n)} \\le \\abs{x}^k / k!$.\nSo $(1 + x/n)^n \\to \\sum_k x^k / k! = e^x$.\nWithout the bound of clause 2 the conclusion fails: $a_k(n) = 1$ for $k = n$ and $0$ otherwise gives sums equal to $1$ but limits $a_k = 0$.\n\n## Explanations\n\n### L1 human\n\nTannery's theorem allows you to take a limit inside an infinite sum.\nIt needs one fixed summable bound for all the terms, uniformly in $n$.\nThen no mass can escape to infinity, so the limit of the sums is the sum of the limits.\n\n### L2 human\n\nTannery's theorem is the [[dominated convergence theorem|BYNV84PF86]] for the counting measure on $\\N$, but it predates measure theory and has a short direct proof.\nA typical use is to pass from finite products or binomial expansions to their limits, as in the proof that $(1 + x/n)^n \\to e^x$.\nCompare the [[Weierstrass M-test|0NX5NWAFC9]], which uses the same kind of bound to get uniform convergence of a series of functions.\n",
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