Test for convergence the series 3.6.9 ... 3n (i) E x" (x > 0), 7.10.13 ... (3n + 4) I1 = | 2.4.6 ... (2n +2) x-! 3.5.7 ... (2n + 3) (ii) E (x > 0). n = 1

(4) n=2 3n n3x5m (4) Determine if the series converges, enter CON when converging, and DIV when diverging. - J ............. +

Test the series for convergence or divergence. E (-1)n (3n – 5) 2n + 5 n = 1 Evaluate the following limit. (If the quantity diverges, enter DIVERGES.) (3n - 5) lim (-1)" 2n + 5 n- 00 Since lim (-1)" (3n - 5) --Select--- ---Select-- n- 00 2n + 5 -Select-- Need Help? exists and equals 0 exists and does not equal 0 Watch It does not exist

Please solve and show work. Thank you.

(-1)"n(z-4)" 3. Find the radius of convergence and interval of convergence for the power series n-1 2n

C and D

3 2 (a) Use the Comparison Test to show that the series n=1 4n3n²+3 converges. (You may use the fact n°-1 1 that the series -1 converges if k > 1 and diverges if 0 < k < 1.) n=1 3n (b) Use the Ratio Test to decide whether the series -1 4√n converges or diverges. converges. (-1)" (c) Use the Alternating Series Test to show that the series n=1 2n-1