n趋近于无穷大,求limn(1/(n2+1)+1/(n2+2)+...+1/(n2+n)=

Question

n趋近于无穷大,求limn(1/(n2+1)+1/(n2+2)+...+1/(n2+n)=

Answer 1

n^(1/n) = e^ln(n^(1/n))=e^((1/n)ln n)=e^((ln n)/n)
当n趋近于无穷大时,(ln n)/n是∞/∞型,可以用洛必达法则：
lim(ln n)/n = lim (ln n)'/(n)' =lim (1/n)/1 =lim(1/n)
当n->∞时,1/n->0. 所以 limn^(1/n) = lim[e^((ln n)/n)] = e^0 =1