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  "body": "\n## Statement\n\nLet $(a_n)_{n \\ge 1}$ be a [[decreasing|0KN22AAW5R]] real sequence with $a_n \\ge 0$ for all $n$.\nLet $u_0 \u003c u_1 \u003c u_2 \u003c \\cdots$ be positive integers, and put $\\Delta u_k = u_{k+1} - u_k$.\nAssume that there is a constant $M$ with $\\Delta u_{k} \\le M \\, \\Delta u_{k-1}$ for all $k \\ge 1$.\nThen $\\sum_{n=1}^{\\infty} a_n$ [[converges|0CVVKY7GY2]] if and only if $\\sum_{k=0}^{\\infty} \\Delta u_k \\, a_{u_k}$ converges.\n\n## Remarks\n\nThe proof cuts the series into blocks $u_k \\le n \u003c u_{k+1}$ of $\\Delta u_k$ terms each.\nSince $(a_n)$ decreases, the block sum lies between $\\Delta u_k \\, a_{u_{k+1}}$ and $\\Delta u_k \\, a_{u_k}$.\nThe upper bound shows that convergence of the condensed series gives convergence of $\\sum a_n$.\nThe lower bound and $\\Delta u_{k+1} \\le M \\Delta u_k$ show that $\\Delta u_{k+1} a_{u_{k+1}}$ is at most $M$ times the block sum, which gives the converse.\nWith $u_k = 2^k$ this is the [[Cauchy condensation test|0B96KMDDD1]].\nWith $u_k = k^2$ for $k \\ge 1$ the condensed series is $\\sum (2k+1) a_{k^2}$, and the condition holds since $(2k+3)/(2k+1) \\le 3$.\nFor $a_n = 1/n^p$ with $p \u003e 0$ this gives $\\sum (2k+1)/k^{2p}$, which converges exactly when $2p - 1 \u003e 1$, that is $p \u003e 1$.\n\n## Explanations\n\n### L1 human\n\nFor a decreasing series, you may keep only a sparse set of terms, provided you weight each kept term by the length of the gap after it.\nThe gaps may grow, but not too fast from one gap to the next.\nThe sparse weighted series and the original series then converge or diverge together.\n\n### L2 human\n\nThe Cauchy condensation test uses the gaps $2^k$.\nSchlömilch's version lets you choose gaps adapted to the series, such as the squares, which helps when $a_{k^2}$ has a simpler form than $a_{2^k}$.\nThe hypothesis that consecutive gaps have bounded ratio is what makes the lower bound work.\nThe proof is the same comparison of block sums as in the Cauchy case.\n",
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