Question

1. Answer the following questions. Justify your answers. (a) (3 marks) For what values of x does the series 1 + 22x 2 + 24x 4 + 26x 6 · · · + 22nx 2n + · · · converge? (b) (9 marks) Is the following series absolutely convergent, conditionally convergent or divergent? i. X∞ n=1 2 √ n − 2 √ n + 1 ii. X∞ n=1 arctan(n) n2 + 1 iii. X∞ k=1 (−1)k k

1. Answer the following questions. Justify your answers, (a) (3 marks) For what values of r does the series 1 +222 +2° +226. +22n 22n +... converge? (b) (9 marks) Is the following series absolutely convergent, conditionally convergent or divergent? 2 2 Vn+1 arctan(n) ii. n2+1 (-1) k i. na n=1 111 k=1

Answer 1

8 W an am W1 nar Ratio test: if, lins ay si then Ean converges mtoo amt am myoo amt! <1, then an diverges n gan xan فه Jan 2-2 more an amt! (1H) 2(6+1) २२३ : lim am the converges, when we are all MD amith € 34 9 2 2+1) S+ Go n+ su Sutl) nal lim (2-0) no since, series has a sum 2 and all teams are positive therefore series is absolutely convergenth. b ann lim ✓ tant x x +1 dac bao let u tank du - 1 da hol tant b lim u du b- 1/4 tant b lim boo N4

& tanta lim ((tant b3 - † (2) (9'- ( (: lim dan b-tan D-T T(4-5 12( 34) - Se byo since, series has a sum and all terms are positive, therefore series is absolutely convergent. pl Leibnitz test: Ž (H)an is an alternating teries, then this series is converges if, sans is decresing sequence lim anco ha here, k is an alternating series, so let's apply, teibnitz test, * = is clearly decreting sey (since k is increasing limo ntah NA L29. & tuk converges. Hence, by leibuita test, but a sok J = 2 is diverges (from ptest} p=151) hence, series tijk is conditionally convergent ni 1 p test: converges. of pr1, otherwise diverges (psi) ha np