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  "body": "\n## Statement\n\nLet $x = (x_n)_{n \\ge 1}$ be a [[bounded sequence|0A8JNBGT2Z]] of real numbers and let $L$ be a real number.\nThen $x$ is [[almost convergent|8HTHB1QPZY]] to $L$ if and only if\n$$\\lim_{p \\to \\infty} \\frac{1}{p} \\sum_{k = n}^{n + p - 1} x_k = L \\quad \\text{uniformly in } n \\ge 1.$$\n\n## Remarks\n\nProof idea for \"if\": let $\\varphi$ be a [[Banach limit|SYF2ND6J97]] and $\\varepsilon \u003e 0$.\nBy shift invariance, $\\varphi(x)$ equals $\\varphi$ of the sequence of block averages $y_n = \\frac{1}{p} \\sum_{k=n}^{n+p-1} x_k$.\nFor $p$ large, $\\abs{y_n - L} \\le \\varepsilon$ for all $n$, and positivity of $\\varphi$ then gives $\\abs{\\varphi(x) - L} \\le \\varepsilon$.\nFor \"only if\", one shows with the [[Hahn–Banach theorem|QB5AE3VYZZ]] that Banach limits take every value between the lower and upper limits of the uniform block averages.\n\nExample: for $x = (1, 0, 1, 0, \\dots)$, any block of $p$ consecutive terms has sum $\\floor{p/2}$ or $\\ceil{p/2}$, so the averages tend to $1/2$ uniformly in $n$.\n\nThe uniformity in $n$ is necessary.\nLet $x_n = 1$ when $4^k \\le n \u003c 4^k + k$ for some integer $k \\ge 1$, and $x_n = 0$ otherwise.\nAmong $x_1, \\dots, x_N$ there are at most about $(\\log_4 N)^2$ ones, so the averages starting at $n = 1$ tend to $0$.\nBut the block of length $k$ starting at $4^k$ has average $1$, so the convergence is not uniform in $n$ and $x$ is not almost convergent.\n\n## Explanations\n\n### L1 human\n\nA bounded sequence is almost convergent to $L$ exactly when the average of any long run of consecutive terms is close to $L$, no matter where the run starts.\nThis turns a definition about abstract functionals into a concrete test.\n\n### L2 human\n\nThe definition of almost convergence quantifies over all Banach limits, which exist only by the Hahn–Banach theorem and cannot be written down.\nLorentz's theorem replaces this by a computable condition on block averages.\nIt also shows that almost convergence is a summability method in its own right.\nFor bounded sequences it lies between ordinary convergence and Cesàro summability, because the case $n = 1$ of the condition is Cesàro convergence.\n",
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