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  "body": "\n## Statement\n\nThe number\n$$\\zeta(3) = \\sum_{n=1}^{\\infty} \\frac{1}{n^3}$$\nis [[irrational|0DZXAXEJQ9]].\nHere $\\zeta$ is the [[Riemann zeta function|1XWBY4N0VT]].\n\n## Remarks\n\nThe proof constructs two sequences of rational numbers $a_n$ and $b_n$, where the $b_n$ and the $2 d_n^3 a_n$ are integers, with $d_n$ the [[least common multiple|007NPQX34K]] of $1, 2, \\dots, n$.\nThe differences $b_n \\zeta(3) - a_n$ are nonzero and tend to $0$ so fast that $2 d_n^3 (b_n \\zeta(3) - a_n)$ still tends to $0$.\nIf $\\zeta(3) = p/q$, then $2 q d_n^3 (b_n \\zeta(3) - a_n)$ would be a nonzero integer tending to $0$, which is impossible.\nA useful series in the proof is\n$$\\zeta(3) = \\frac{5}{2} \\sum_{n=1}^{\\infty} \\frac{(-1)^{n-1}}{n^3 \\binom{2n}{n}}.$$\nFor even arguments much more is known: $\\zeta(2k)$ is a rational multiple of $\\pi^{2k}$, so it is transcendental, as in the [[Basel problem|0DMY75D6FV]].\nIt is not known whether $\\zeta(5)$ is irrational, and it is not known whether $\\zeta(3)$ is transcendental.\n\n## Explanations\n\n### L1 human\n\nThe sum $1 + 1/8 + 1/27 + 1/64 + \\cdots$ of the reciprocals of the cubes is about $1.202$.\nApéry's theorem says that this number is not a fraction.\nUnlike the sum of the reciprocals of the squares, which equals $\\pi^2/6$, no closed form for it is known.\n\n### L2 human\n\nApéry's sequences satisfy a three-term recurrence with polynomial coefficients, and their growth controls the speed of approximation.\nLater work gave other proofs, for example with Beukers' integrals over the unit cube.\nRivoal proved that infinitely many of the values $\\zeta(2k+1)$ are irrational, and Zudilin proved that at least one of $\\zeta(5), \\zeta(7), \\zeta(9), \\zeta(11)$ is irrational.\n",
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