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  "body": "\n## Statement\n\nLet $(a_n)_{n \\ge 1}$ be a sequence of positive real numbers such that $\\sum a_n$ [[diverges|0NBRP6093V]].\nLet $s_n = a_1 + \\cdots + a_n$ be the [[partial sums|0NB643K60Q]], and let $\\alpha$ be a real number.\nThen $\\sum_{n=1}^{\\infty} a_n / s_n^{\\alpha}$ [[converges|0CVVKY7GY2]] if $\\alpha \u003e 1$, and diverges if $\\alpha \\le 1$.\n\n## Remarks\n\nFor $\\alpha = 1$ and $m \u003e n$, the terms with index from $n+1$ to $m$ satisfy $\\sum_{k=n+1}^{m} a_k / s_k \\ge (s_m - s_n)/s_m$.\nSince $s_m \\to \\infty$, this tends to $1$ as $m \\to \\infty$, so the [[Cauchy criterion|06PVXMY0TH]] fails.\nFor $\\alpha \u003c 1$, one has $s_n \\ge 1$ for large $n$, so $a_n / s_n^{\\alpha} \\ge a_n / s_n$.\nFor $\\alpha \u003e 1$, one has $a_n / s_n^{\\alpha} \\le \\int_{s_{n-1}}^{s_n} t^{-\\alpha} \\, dt$ for $n \\ge 2$, and these integrals add up to at most $\\int_{s_1}^{\\infty} t^{-\\alpha} \\, dt \u003c \\infty$.\nExample: with $a_n = 1$, one has $s_n = n$, and the theorem says that $\\sum 1/n^{\\alpha}$ converges exactly when $\\alpha \u003e 1$.\n\n## Explanations\n\n### L1 human\n\nStart from any divergent series of positive terms.\nDividing each term by the partial sum so far gives a new series that still diverges, but more slowly.\nDividing by a slightly higher power of the partial sum makes it converge.\n\n### L2 human\n\nThe theorem shows that there is no slowest divergent series: from any divergent series of positive terms, $\\sum a_n / s_n$ diverges, and its terms are much smaller than $a_n$.\nIt also gives a supply of convergent series near the boundary.\nSo no single comparison series can decide all series of positive terms, which is one reason why there is a whole scale of [[convergence tests|FJKV01J9F1]].\n",
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