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  "body": "\n## Statement\n\nLet $\\sum_{n \\ge 0} a_n$ be a series of complex numbers.\nAssume that it is [[Cesàro summable|N8ZGG7EW6J]] to $s$.\nAssume also that $a_n = O(1/n)$ in [[big O notation|0RYTGBJAGN]], that is, there is a constant $K$ with $n \\abs{a_n} \\le K$ for all $n \\ge 1$.\nThen $\\sum_{n \\ge 0} a_n$ [[converges|0CVVKY7GY2]] and its sum is $s$.\n\n## Remarks\n\nLet $s_n$ be the partial sums and $\\sigma_n$ their Cesàro means.\nFor $m \u003e n$ one has the identity\n$$s_n = \\frac{(m+1)\\sigma_m - (n+1)\\sigma_n}{m - n} - \\frac{1}{m - n} \\sum_{k=n+1}^{m} (s_k - s_n).$$\nThe first term tends to $s$ when $m \\approx (1 + \\varepsilon) n$ and $n \\to \\infty$.\nThe second term is at most $\\max_{n \u003c k \\le m} \\abs{s_k - s_n} \\le K \\frac{m - n}{n} \\approx K\\varepsilon$, so $\\limsup \\abs{s_n - s} \\le K\\varepsilon$ for every $\\varepsilon \u003e 0$.\n\nThe Tauberian condition cannot be dropped.\nThe series $1 - 1 + 1 - \\cdots$ is Cesàro summable to $\\frac12$ and diverges; here $n \\abs{a_n} = n$ is unbounded.\nFor real $a_n$, the one-sided condition $n a_n \\ge -K$ is already enough.\n\n## Explanations\n\n### L1 human\n\nAveraging the partial sums can give a value to a series that diverges.\nHardy's theorem says this cannot happen when the terms shrink at least like $1/n$.\nThen a Cesàro sum is an honest sum.\n\n### L2 human\n\nThis is one of the first Tauberian theorems in which the condition $n a_n \\to 0$ of [[Tauber's theorem|8SK3GQ4S0M]] is relaxed to $n a_n = O(1)$.\nThe analogous result for Abel summation is [[Littlewood's Tauberian theorem|Z35TD4ZZRD]], which is harder, because Abel summation is stronger than Cesàro summation.\nHardy's theorem is used to show that a Fourier series of a function of bounded variation converges wherever its Cesàro means do, since its coefficients are $O(1/n)$.\n",
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