Question

请问这个极限怎么求n求极限lim(n->∞）∑sin(k/n^2),其中n^2代表n的平方k=1

Answer 1

u(n) = ∑sin(k/n^2) = sin(1/n^2) +sin(2/n^2) + .+ sin(n/n^2)

sin(1/n^2) + sin(n/n^2) = 2 sin [(n+1)/(2n^2)] cos [(n-1)/(2n^2)] 和差化积

sin(2/n^2) + sin[(n-1)/n^2] = 2 sin [(n+1)/(2n^2)] cos [(n-3)/(2n^2)]

……

当n为偶数时,

u(n) = 2 sin [(n+1)/(2n^2)] { cos [(n-1)/(2n^2)] + cos [(n-3)/(2n^2)] + …… + cos [1/(2n^2)] }

=> n * sin [(n+1)/(2n^2)] cos [(n-1)/(2n^2)] ≤ u(n)≤ n * sin [(n+1)/(2n^2)]

当n->∞时,n * sin [(n+1)/(2n^2)] cos [(n-1)/(2n^2)] -> 1/2

n * sin [(n+1)/(2n^2)] -> 1/2

=> 当n->∞时,u(n) -> 1/2

当n为奇数时,类似可求.

Lim ∑sin(k/n^2) = 1/2