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  "body": "\n## Example\n\nFor $x \u003e 0$ let\n$$F(x) = \\int_0^\\infty \\frac{e^{-t}}{1 + x t}\\,dt.$$\nThen, as $x \\to 0^+$,\n$$F(x) \\sim \\sum_{n \\ge 0} (-1)^n n!\\, x^n = 1 - x + 2x^2 - 6x^3 + \\cdots,$$\nand this series is a [[divergent series|0JWMJSTBSG]] for every $x \\ne 0$.\n\nCheck of the expansion: for $N \\ge 0$ and $t \\ge 0$,\n$$\\frac{1}{1 + x t} = \\sum_{n=0}^{N} (-x t)^n + \\frac{(-x t)^{N+1}}{1 + x t}.$$\nIntegrating against $e^{-t}$ and using $\\int_0^\\infty t^n e^{-t}\\,dt = n!$ gives\n$$F(x) - \\sum_{n=0}^{N} (-1)^n n!\\, x^n = R_N(x), \\qquad \\abs{R_N(x)} \\le x^{N+1} \\int_0^\\infty t^{N+1} e^{-t}\\,dt = (N + 1)!\\, x^{N+1}.$$\nSo the error is $O(x^{N+1})$ for each $N$, which is the condition of a [[Poincaré asymptotic series|2HVC5G0H5K]] at $0$.\n\nCheck of divergence: for $x \\ne 0$ the ratio of consecutive terms is $(n + 1) \\abs{x} \\to \\infty$, so the terms do not tend to $0$.\n\n## Remarks\n\nFor a fixed small $x$ the terms $n!\\, x^n$ first decrease and then grow, and the smallest one is near $n \\approx 1/x$.\nStopping there, which is [[optimal truncation|1YR53GZEPT]], gives an error of order $\\sqrt{2\\pi/x}\\, e^{-1/x}$ by Stirling's formula, which is extremely small for small $x$.\nThe function $F$ is the Borel sum of the series, and $F(x) = \\frac1x e^{1/x} E_1(1/x)$, where $E_1$ is the [[exponential integral|EV6ZDQ0NDY]].\n\n## Explanations\n\n### L1 human\n\nThe series $1 - x + 2x^2 - 6x^3 + 24x^4 - \\cdots$ has coefficients that grow like factorials, so it diverges for every nonzero $x$.\nYet for $x = 0.1$ its first ten or so terms give $F(0.1)$ to about three decimal places.\nEach partial sum is a good approximation; only the infinite sum is meaningless.\nThis is typical of the series that come out of asymptotic analysis, and it is why one studies asymptotic series and not only convergent ones.\n",
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