Show that #sum_(n=1)^oo 1/n# is divergent using the integral criterion ?

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Answer 1 · 2016-08-27T13:27:23

The series sum of #1/n# is divergent.

#1/n, n = 1,2,3, cdots# is a monotonic decrescent sequence and
#1/x# is a monotonic decrescent function for #x in [1,oo)#

Comparing

#sum_(k=1)^oo 1/k# and #int _1^oo dx/x# we know that

#int_1^oo (dx)/x le sum_(k=1)^oo 1/k#

#int _1^(alpha) dx/x = log_e(alpha)#

and

#lim_(alpha->oo)log_e(alpha) = oo#

so

#sum_(k=1)^oo 1/k = oo#