Question

6. Investigate the behavior (convergence or divergence) of Σ a" if (b) as =

Answer 1

a_n = (squareroot n + 1 - squareroot n) (squareroot n + 1 + squareroot n)/squareroot n + 1 + squareroot n = n + 1 - n/squareroot n + 1 + squareroot n = 1/squareroot n + 1 + squareroot n a_n = 1/squareroot n + 1 + squareroot n (a_n + 1/a_n) = 1/squareroot n + 2 + squareroot n + 1/1/squareroot n + 1 + squareroot n = squareroot n + 1 + squareroot n/squareroot n + 2 + squareroot n + 1 = squareroot 1 + 1/n + squareroot 2/squareroot 2 + 2/n + squareroot 1 + 1/n lim_n rightarrow infinity (a_n + 1/a_n) = lim_n rightarrow infinity squareroot 1 + 1/n + squareroot 1/squareroot 1 + 2/n + squareroot 1 + 1/n = 2/2 = 1 sigma^infinity_n = 2 a_n We cannot say this is convergent or divergent a_n = 1/squareroot n 1/squareroot 1 + 1/n + 1 Let the auxiliary series be sigma V_n