n2n1n1的精選

1.用比值審斂法或根值審斂法判別（n/(2n+1))^(n+1)的斂散性a(n+1)=[n/(2n+1)]^(n+1)。a(n+1)=[1/(2+1/n)]^(n+1)→1/2^(n+1)→0。數列{a(n)}收斂。【1】比值判定。a(n+1)/a(n)=[n/(2n+1)]^(n+1)/[(n-1)/(2n-1)]^n=[n/(2...