HMMT 二月 2003 · CALC 赛 · 第 6 题
HMMT February 2003 — CALC Round — Problem 6
题目详情
英文原题
- 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
英文解析
- 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
2 nin − i
1 − ( )
sum. Then,n
( )
∫
n − 1
∑1
1 1 1 π − 11
√ √
lim = dx = [sin x ] = .
n →∞ i 20
n 22
1 − x 0
1 − ( )
i =0 n