That's right. a[n] = (n？ + 1)^( 1/3)/(n？ +2)，lim { n→∞a[n]/( 1/n^(4/3))= lim { n→∞}( 1+ 1/n？ )^( 1/3)/( 1+2/n？ ) = 1.

That is, a[n] and 1/n (4/3) are equivalent infinitesimals, and the convergence of ∑a[n] can be known from the convergence of ∑ 1/n (4/3) (p- series).

B. It is easy to see that the general term of the series is greater than 1, so it cannot converge to 0, so the series diverges.

C. for n > 2, the absolute value of the general term of the series n/(ln (1+n)) 8 > n/ln (1+n) >; 1.

The general term cannot converge to 0, so the series diverges.

D this series is equal to the series ∑ (- 1) n/n and ∑ 1/n? The sum of.

The former is a staggered series, and the absolute value of the general term 1/n monotonically decreases to zero, and its convergence is known by Leibniz discriminant method.

The latter is that the series of P > P- 1 also converges (integral discrimination method).

The sum of two convergent series still converges.