Question

(3) Check that the series converges with the Integral Test (if you haven't already in a previous problem), then approximate to an error less than 0.001 (a) Zaal nie (b) ke-k2 1 (c) En=1 Zn=1 e-n

Answer 1

solution: Griven, h2 nal E By integral so 7 dna (= 1 1 3 test, 11 C 0 + - ( converges) Hence, E 7 / 12 converges. n=1 b) E kiek? kal integ ral test, By let 2 Soxerado u= n2 du aenda -u du xda So 2 = 1 u du e when xa1 uzl x=x ; u=& - [edio 1 l - en te 3 [-ot - د (converges) qe e

Hence, the series I Ma I kienk? converges. k=1 (0) E n=1 I e-In Ja da let e By integral Test, Sir Ja -S, ekleden u= In du dx = 2 2 Jx g des Ju new limits: In de e-u du when x=1;u=1 x=0; uIN 13 لا et = 2 2 [-e-t te -2 [-o + e (1 2 6 C converges) Hence, the series E F e-Jъ converges. n=1