Please solve all parts *3. Problem 1 presented the two series (log n)- and (log n)-", ofwhich the first diverges while the second converges. The series1(log n)log n'which lies between these two, is analyzed in parts (a) and (b).(a) Show that e" /yu dy exists, by considering the series(e/n)".(b) Show that1(log n)logconverges, by using the integral test. Hint: Use an appropriatesubstitution and part (a).(c) Show that1(log n)log (log n)diverges, by using the integral test. Hint: Use the same substitutionas in part (b), and show directly that the resulting integral diverges.

Given : ∑n=2∞logn-k and ∑n=2∞logn-n two series where 1st diverges and 2nd converges. Integral test…

(a) Show that e /yu dy exists, by considering the series (e/n)". (b) Show that 1 (log n)los n converges, by using the integral test. Hint: Use an appropriate substitution and part (a). (c) Show that 1 4 (log n)ieg (log n) diverges, by using the integral test. Hint: Use the same substitution as in part (b), and show directly that the resulting integral diverges.

(a) Show that e /yu dy exists, by considering the series (e/n)". (b) Show that 1 (log n)los n converges, by using the integral test. Hint: Use an appropriate substitution and part (a). (c) Show that 1 4 (log n)ieg (log n) diverges, by using the integral test. Hint: Use the same substitution as in part (b), and show directly that the resulting integral diverges.

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Please solve all parts

*3. Problem 1 presented the two series (log n)- and (log n)-", of
which the first diverges while the second converges. The series
1
(log n)log n'
which lies between these two, is analyzed in parts (a) and (b).
(a) Show that e" /yu dy exists, by considering the series
(e/n)".
(b) Show that
1
(log n)log
converges, by using the integral test. Hint: Use an appropriate
substitution and part (a).
(c) Show that
1
(log n)log (log n)
diverges, by using the integral test. Hint: Use the same substitution
as in part (b), and show directly that the resulting integral diverges.

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