Lecture Note

10.3 The Integral Test 58 Corollary A series an of nonnegative terms converges iff n=1 its partial sums (sn) are bounded from above. Exp = 1/2 + (H) (1) + (1) geometric series with r=1

Exp Does the following series converge / diverge? 59 & \ E 1/2 (an as n os) So it may converge n=1 f(x)= X 2 is continous, positive, decreasing & & function on [1,00 b find x = S dx = lim " by exp.' X2 dx = = ED b & 2 2-1 note that I I X this is not sum 8 we don't know the Thus, the series E 1/2 converges by the integral test. sum. n=1 2 The p- series converges if p>1 and by n=1 diverges if PEI & 1 (an as n 00) so it may converge 3 L n - +1 \ n=1 f(x) = is continous, positive, decreasing X 2 +1 function on [1, 00) 8 D b fwidx dx S = = lim dx lim tan X 2 x +1 00 I b 8 I = lim fan b tan'i] = 1/1 2 - " 4 = lef b D converges the integral test. by Thus, the series 2 n2+1 n=1 th D E diverges by the n term test 4 n sin(h) sinth lim sinx = 1 #0 n=1 lim an lim = = n-so n n->00 o) so it may converge 00 \ 5 2 (an as n 2n- n=1 I f(x) = is continous, positive, decreasing 2x- I function on [1,00] & In(26-1) = as funds = lim dx lim 1/2 In/2x-11 lim = = is / 8 2X-1 b-300 Thus, the series diverges by the integral test.

The Integral Test

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