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  "body": "\n## Statement\n\nLet $(a_n)_{n \\ge 1}$ be a decreasing real sequence with $a_n \\ge 0$ for all $n$.\nThen $\\sum_{n=1}^{\\infty} a_n$ converges if and only if the condensed series $\\sum_{k=0}^{\\infty} 2^k a_{2^k}$ converges.\n\n## Remarks\n\nThe proof groups the terms from $2^k$ to $2^{k+1} - 1$. Since the sequence decreases, $\\sum_{n=1}^{\\infty} a_n \\le \\sum_{k=0}^{\\infty} 2^k a_{2^k} \\le 2 \\sum_{n=1}^{\\infty} a_n$.\nWith $a_n = 1/n^p$ it gives the [[p-series test|0ZNSSK8ZK1]].\nWith $a_n = 1/(n \\ln n)$ for $n \\ge 2$, the condensed series is a multiple of $\\sum 1/k$, so $\\sum 1/(n \\ln n)$ diverges.\n\n## Explanations\n\n### L1 human\n\nFor a decreasing sequence, the terms between $2^k$ and $2^{k+1}$ are all about the same size as $a_{2^k}$.\nSo only one term out of each such block matters, weighted by the size $2^k$ of the block.\n",
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