Question

(1 point) We will determine whether the series n3 + 2n an - is convergent or divergent using the Limit Comparison Test (note that the Comparison Test is difficult to apply in this case). The given series has positive terms, which is a requirement for applying the Limit Comparison Test. First we must find an appropriate series bn for comparison (this series must also have positive terms). The most reasonable choice is ba - (choose something of the form 1/mp for some number p, so that Σ bn is a p-series). Evaluate the limit below - as long as this limit is some finite value c>0, then either both series or both series diverge. lim From what we know about p-series we conclude that the series Σ bn is "divergent") Finally, by the Limit Comparison Test we conclude that the series 2 an is "divergent") an and bn converge 1-00 (enter "convergent" or (enter "convergent" or

Answer 1

Sigma a_n - sigma^infinity_n = 1^5 squareroot n^3 + 2 n/squareroot 4 n^3 + 2 n choose b_n = 1/n^-3/5 + 3/2 = 1/n^9/10 sigma b_n = sigma^1/n^9/10 is divergent since sigma 1/n^8 = sigma 1 = n^9/10 Hr P = 9/1 < 1 by p-series test and lim_n rightarrow infinity a_n/b_n = lim_n rightarrow infinity (^5 squareroot n^3 + 2 n/squareroot 4 n^3 + 2 n)/(1/n^9/10) = lim_n rightarrow infinity (n^3/5 squareroot 1 + 2/n^2/n^3/2 squareroot 4 + 2/n^2 middot n^9/infinity) lim_n rightarrow infinity (n^-9/10 middot n^9/10 squareroot 1 + 2/n^2/squareroot 4 + 2/n^2) = 1/squareroot 4 = 1/2 (notequalto 0 and finite) lim_n rightarrow infinity a_n/b_n = 1/2 By comparison test, sigma a_n and sigma b_n be have a like is sigma a_n is divergent.