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Nov 14, 2013 at 23:38 comment added Kevin P. Costello If you want something non-asymptotic that's easily certifiable, you can expand $k^{15}f(k+\frac{1}{k}-\frac{1}{k^2}-\frac{3}{k^3}+\frac{4}{k^4}+\frac{7}{k^5})$ in powers of $(k-5)$. Every coefficient is negative, so $f$ must be less than $0$ there.
Nov 14, 2013 at 2:55 history answered Jacques Carette CC BY-SA 3.0