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Today's calculation of Integral 183

Source: Tokyo University entrance exam/Science, Problem 2, 2007

February 25, 2007
calculusintegrationlimittrigonometrycalculus computations

Problem Statement

Let n≥2n\geq 2 be integer. On a plane there are n+2n+2 points O, P0, P1, ⋯PnO,\ P_{0},\ P_{1},\ \cdots P_{n} which satisfy the following conditions as follows. [1] ∠Pk−1OPk=πn (1≤k≤n), ∠OPk−1Pk=∠OP0P1 (2≤k≤n).\angle{P_{k-1}OP_{k}}=\frac{\pi}{n}\ (1\leq k\leq n),\ \angle{OP_{k-1}P_{k}}=\angle{OP_{0}P_{1}}\ (2\leq k\leq n). [2] OP0‾=1, OP1‾=1+1n.\overline{OP_{0}}=1,\ \overline{OP_{1}}=1+\frac{1}{n}. Find lim⁡n→∞∑k=1nPk−1Pk‾.\lim_{n\to\infty}\sum_{k=1}^{n}\overline{P_{k-1}P_{k}}.
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