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  "body": "\n## Example\n\nLet $\\tau$ be a point of the [[upper half-plane|X6XGGH5FPW]] and let $k \\ge 3$ be an integer.\nThe [[Eisenstein series|EYREVY31BN]] $G_k(\\tau) = \\sum_{(m,n) \\ne (0,0)} (m\\tau + n)^{-k}$ is a [[double series|PAXCH4FFSV]] indexed by $\\Z^2 \\setminus \\set{(0,0)}$.\nIt converges [[absolutely|0W3K3QPEWT]].\nCheck: since $1$ and $\\tau$ are linearly independent over $\\R$, there is $c \u003e 0$ with $\\abs{x\\tau + y} \\ge c \\max(\\abs{x}, \\abs{y})$ for all real $x, y$.\nFor each integer $r \\ge 1$ there are exactly $8r$ pairs $(m, n)$ with $\\max(\\abs{m}, \\abs{n}) = r$.\nSo $\\sum \\abs{m\\tau + n}^{-k} \\le \\sum_{r \\ge 1} 8r \\, (cr)^{-k} = 8 c^{-k} \\sum_{r \\ge 1} r^{1-k}$, which is finite because $k - 1 \\ge 2$.\nBy absolute convergence, the sum does not depend on the order of the terms.\nSumming by rows, by columns or over growing squares gives the same value.\nFor odd $k$ the terms for $(m,n)$ and $(-m,-n)$ cancel and $G_k = 0$, so the interesting cases are the even $k \\ge 4$.\n\nFor $k = 2$ the same count gives the divergent bound $\\sum 8r(cr)^{-2}$, and indeed the double series is not absolutely convergent.\nSumming first over $n$ and then over $m$ still gives a value, but this function is not a modular form.\nThe order of summation matters here.\n\n## Explanations\n\n### L1 human\n\nThe Eisenstein series adds up one term for each point of a two-dimensional grid, so it is naturally a double series.\nCounting the grid points on squares of growing size shows that the terms decay fast enough when $k \\ge 3$.\nThen the order in which the terms are added does not matter, and that is what makes the function well behaved.\n",
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