= Power-law comparison from a refined ratio bound
{title2=$a_{n+1}/a_n\le1-c/n\Longrightarrow a_n=O(n^{-c})$}
Suppose $a_n>0$ and, for all sufficiently large $n$, $a_{n+1}/a_n\le1-c/n$ with fixed $c>0$. Choose an initial index $N>c$. Since $\log(1-x)\le-x$ for $0<x<1$, summing logarithmic ratios gives $\log(a_n/a_N)\le-c\sum_{k=N}^{n-1}1/k\le-c\log(n/N)$. Thus $a_n\le a_NN^c n^{-c}$. For $c>1$, the <comparison test for series> with a <P-series> proves convergence. This supplies a useful refinement when a limiting ratio of one makes the ordinary <ratio test> inconclusive.
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