lim[√(1+2+...n)-√(1+2+...+n-1)]
n→∞
=lim{√[(n+1)n/2]-√[(n-1+1)(n-1)/2]}(根号下等差求和)
n→∞
=(1/√2)lim{√[(n+1)n]-√[n(n-1)]}
n→∞
=(1/√2)lim(√n)[√(n+1)-√(n-1)]
n→∞
=(1/√2)lim(√n)[√(n+1)-√(n-1)][√(n+1)+√(n-1)]/[√(n+1)+√(n-1)](分子有理化)
n→∞
=(1/√2)lim(√n)×2/[√(n+1)+√(n-1)]
n→∞
=(√2)lim(√n)/[√(n+1)+√(n-1)]
n→∞
=(√2)lim1/[√(1+1/n)+√(1-1/n)]
n→∞
=(√2)×1/[√(1+0)+√(1-0)]
=(√2)/2