Question

2. Determine the convergence and divergence of the following series. (n + 3)! 3!n! 31

Answer 1

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Ź (m+3)! E am = 31 m) 30 an= (m+3)! 31m 30 Now anti= (n+1+3)! 3! (MU! 3+) (replacem byn+i) - Anti 9 (m+4)! 31 (M+)3.3 Ratio test Consides series { an. Detine L=lim ant) nos an Then @ it LCI series convergent.

Now anti and an Х it Lol, then series is divergent. ③ il L=1, test is inconclusive. (m+4)! 31 (m+1)! 3" 3 (using 6 (m+3)) 31 M!3n = (m+4) ) 31m! 3M 3! (n+1)/3? 3 (m+3).) (m+4). (m+3)) Х m! (m+1).m.3 (m+3)). m!= ndn-1)! (m+4) (m+) 3 ant) (as mi (n+4) 3 (n+1) an

anti Am (here (m+y) 3 (mt) may so for all nalo for all na >o for alln nu 1 (2+4) >o for all 3 (nt)) 1 (mt4) Imty) 3 (mt)) as 1x)= x if azo 1x1 = -x if xco So 3 inti) Now L=lim anti an 3 1 3 m) L= lim (m+4) 3 (mt) n-

lim (1+y) می (h n- A 30 (+1) - lim (1+) 3('+ (1+y) 3/1+ do) (1+0) 3(1to) if ktoj L - I 31+ Task As L= 1 / 3 = I () =) LCI 3. By ratio test, Series Ě (m+3)! =1 31230 is convergent