HMMT 二月 2003 · CALC 赛 · 第 6 题
HMMT February 2003 — CALC Round — Problem 6
题目详情
- For n an integer, evaluate ( ) 1 1 1 √ √ √ lim + + · · · + . 2 2 2 2 n →∞ 2 2 n − 0 n − 1 n − ( n − 1)
解析
- For n an integer, evaluate ( ) 1 1 1 √ √ √ lim + + · · · + . 2 2 2 2 n →∞ 2 2 n − 0 n − 1 n − ( n − 1) Solution: π/ 2 1 1 1 √ √ Note that = · , so that the sum we wish to evaluate is just a Riemann 2 2 i n 2 n − i 1 − ( ) n sum. Then, ( ) ∫ n − 1 1 ∑ 1 1 1 π 1 − 1 √ √ lim = dx = [sin x ] = . 0 2 n →∞ i 2 n 2 0 1 − x 1 − ( ) i =0 n