Home Competitions Putnam 1949 B2
Answer either (i) or (ii):
(i) Prove that
∑ n = 2 ∞ cos ( log log n ) log n \sum_{n=2}^{\infty} \frac{\cos (\log \log n)}{\log n} n = 2 ∑ ∞ log n cos ( log log n ) p > 0 , a > 0 p>0, a>0 p > 0 , a > 0 a c − b 2 > 0 a c-b^{2}>0 a c − b 2 > 0 ∫ − ∞ ∞ ∫ − ∞ ∞ d x d y ( p + a x 2 + 2 b x y + c y 2 ) 2 = π p − 1 ( a c − b 2 ) − 12 \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{d x d y}{\left(p+a x^{2}+2 b x y+c y^{2}\right)^{2}}=\pi p^{-1}\left(a c-b^{2}\right)^{-12} ∫ − ∞ ∞ ∫ − ∞ ∞ ( p + a x 2 + 2 b x y + c y 2 ) 2 d x d y = π p − 1 ( a c − b 2 ) − 12
Dataset text PutnamGAP . Quoted with source.
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