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847

Part of 2012 Today's Calculation Of Integral

Problems(1)

Today's calculation of Integral 847

Source: 2012 Tohoku University entrance exam/Science, 2nd exam

9/12/2012
Consider a right-angled triangle with AB=1, AC=3, ∠BAC=π2.AB=1,\ AC=\sqrt{3},\ \angle{BAC}=\frac{\pi}{2}. Let P1, P2, ⋯⋯ , Pn−1 (n≥2)P_1,\ P_2,\ \cdots\cdots,\ P_{n-1}\ (n\geq 2) be the points which are closest from AA, in this order and obtained by dividing nn equally parts of the line segment ABAB. Denote by A=P0, B=PnA=P_0,\ B=P_n, answer the questions as below.
(1) Find the inradius of △PkCPk+1 (0≤k≤n−1)\triangle{P_kCP_{k+1}}\ (0\leq k\leq n-1).
(2) Denote by SnS_n the total sum of the area of the incircle for △PkCPk+1 (0≤k≤n−1)\triangle{P_kCP_{k+1}}\ (0\leq k\leq n-1).
Let In=1n∑k=0n−113+(kn)2I_n=\frac{1}{n}\sum_{k=0}^{n-1} \frac{1}{3+\left(\frac{k}{n}\right)^2}, show that nSn≤3π4InnS_n\leq \frac {3\pi}4I_n, then find the limit lim⁡n→∞In\lim_{n\to\infty} I_n.
(3) Find the limit lim⁡n→∞nSn\lim_{n\to\infty} nS_n.
calculusintegrationgeometryinradiuslimittrigonometrycalculus computations