Question

Determine whether the given series are absolutely convergent, conditionally convergent or divergent. (same answers can be used multiple times)

Determine whether the given series are absolutely convergent, conditionally convergent or divergent. (-1)"(2n +3n2) 2n2-n is n=1 M8 M8 M8 (-1)"(n +2) 2n2-1 is absolutely convergent. divergent conditionally convergent. n=1 (-1)" (n+2) 2n2-1 is n = 1

Answer 1

(-1)(2n +312) 2n²_n nel an= (2n+3 n3) 2narni anti (-1)+1 ( 2 6n +1) + 3 (n+1)2] 2 (n+1) ² (n+1) (-199+' [2(0+1) +3 (n+1)?) 2 (n+1)2 - (n+1) hi hui anti an nto Ella (2n +312) 2n²_n 2n²_n homi (-1)"+" [2(n+1) + 3 (n+1)%] 2 (n+1)2-(n+1) n00 Eur (2n+3 m2) X Luni n> (0"(+)'[(n+1){2+3(n+1)] (n+1) [2(n+1)-1] m (271-1) (-1(n (2+3n)) Luni X n(2n-1) (1)"(n) (39+2) (1)".(-) (n+1) ( 3n+s) (n+1( 2n+1) -1) (30+5)(2n-1) (2n+1) (3n+2) huni n>

hini (30+5)(1-21) (217+1 )(3n+2) о - и (uz-1)(stus) zu hui (2n+1)(3n+2) 72 3nts n -) (120) u hem no (20+) ( 3n+2 u 3+5lm - Luni ( ?-?n) (2+ 'n) (3+4) no himit as n oo then henri (3+oxs) (1-2x0) Þ0 (2+1x0)(3+2x0) (3) (1) (2) (3) mlo NI . The giren series is absolutely convergent. gro

(3) Z (-1)"(n +2) 2n²1 n=1 On = (-1)" (+2) 2n2-) L = lim anti an lim (-1)(n+1+2) 2(n+1)-1 (-1)" (n+2) 2n2-1 n+1 = lim = (-1)(n+1+2) 2(n+1)-1 x 2n?) n-> (-1)" (n+2) 11 lim (-1)(-1)'(n+3) * 2(n+1)-1 29?-1 (-1)"(n+ 2) lim -(n+3) (2n-1) n->co [2 (n+1)2-1]. (n+2) (n+3)(2021) lim 53 (Dividing NY & D by no) n100 2(n+1)-1)(n+2)

- (n+3) (2n2-1) - lim n2 no 2(n+1) 21 = lim И со 2 n+1 1 + 2 h2 -) (42) -(+ ) (2-6) n) lim - (+2)(2-6) no ( 2 (1+4)*72) (+) - (1+0) (2-0) (As no (2(1+0)-0(1+0) 1 +)2 2(1). = 1 3 - H=1 L1 idest is in conclusive. lie. The exists both convergent and divergent senes that satisfy this case. 1 - 1 Thus the giren Senes converges but not absolutely kut lim anti lim noo an . The giren series is conditionaly convergent