University:
Boston University
Course:
CAS MA 123 | Calculus I
Academic year:

2023
Views:

124

Pages:

2
Author:

Yung D

10.3 The Integral Test 58 Corollary : A series I an of nonnegative terms converges iff its partial sums (sn) are bounded from above. لله Exp ()^ = ½+()²+(½)+...+(2)+... L 1 M geomen's series with r=<1 note that since the serie convergs Luse onb Note that That is Sn 51 n = 1,2,3,... S2=+(½) 53= ++()+ Sn= 3 Exp The harmonic series is divergent. n= = 1++(+)+(++)+(1+½ 2 T 8 16 >= 8 > < The sequence of the partial Sums is not bounded above because we don't have Sn ≤ Sn+1. •Thus, the harmonic series diverges to oo. The process is very slow. That is after 178 million terms, its partial sum is 20. Tha The Integral Test" Consider the series San, where 00 an is a sequence of postive terms • an=f(n) is sit f is continuous, positive, decreasing on [x, 10) 00 Then the series an and the integral (f(x)dx both converges of k both diverges. Exp Does the following series converge / diverge? (anso as nsas) so it may converge f(x)= ½ is continae, positive, decreasing function on [1,00) له * by exp." 9 dy 1 = X lim dx == Dote mat 2-1 series & the converges by the 00 this is not sum integral test. Thus, 3 1 2 The p- series P n converges if p>1 diverges if p≤1 and d nal 00 ③ (anus) f(x) = ۱ x²+1 so it may converge is continour, positive, decreasing function on [1,00) " by exp 00 b X x²+1 lim tan'x 60 600 x²+1 =lim[tan'b-tani] = 2 1 Thus, the series 19 n²+1 converges by the integral test. n sin() diverges by the ith term lest lim an 100 lim sin th = lim sinx = 1 0 so it may converge 24-1 function on [1,00) Sfundx = lim de = lim & In/2x-1| = ±1m(26-1) = 00 lim 2.X-1 Thus, the series diverges by the Integral test. 0049 f(x) = (ano asn00) 2x-1 is continous, positive, decreasing