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Today's calculation of Integral 155

Source: Kyoto University entrance exam/science 2000

September 27, 2006
calculusintegrationtrigonometrylimitcalculus computations

Problem Statement

The sequence {cn}\{c_{n}\} is determined by the following equation. cn=(n+1)01xncosπx dx (n=1, 2, ).c_{n}=(n+1)\int_{0}^{1}x^{n}\cos \pi x\ dx\ (n=1,\ 2,\ \cdots). Let λ\lambda be the limit value limncn.\lim_{n\to\infty}c_{n}. Find limncn+1λcnλ.\lim_{n\to\infty}\frac{c_{n+1}-\lambda}{c_{n}-\lambda}.