lim cos(π/n)*lim∑i/(n^2 + i)
=1*lim∑i/(n^2 +i)
=lim∑i/(n^2 + i)
因为：
i/(n^2+n) ≤ i/(n^2 + i) ≤ i/(n^2 +1) 注：1≤ i ≤ n
所以,
∑i/(n^2 +n) ≤∑i/(n^2 + i) ≤∑i/(n^2 +1)
(1+2+┈+n)/(n^2 +n) ≤ ∑i/(n^2+i) ≤ (1+2 +┈ + n)/(n^2 + 1)
n(n+1)/[2*n(n+1)] ≤ ∑i/(n^2+i) ≤ n(n+1)/[2*(n^2 +1)]
1/2≤ ∑i/(n^2+i) ≤ (1 + 1/n)/[2*(1+1/n^2)] 注：分子、分母同除以 n^2
则：
1/2 ≤ lim∑i/(n^2+i) ≤ lim (1+0)/[2*(1+0)] = 1/2
所以,lim∑i/(n^2+i) = 1/2
因此,这道题的极限也等于 1/2.
