Use the integral method to determine if the series converge or diverge:?

Question

#sum_(n=1)^oo(1/(nlnn))#

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Answer 1 · 2018-06-18T22:30:03

The series:

#sum_(n=2)^oo 1/(nlnn)#

is divergent.

Consider the function;

#f(x) = 1/(xlnx)#

As:

(1) #f(x) >0 # for #x >1#

(2) #lim_(x->oo) f(x) = 0#

(3) #f'(x) = -(1+lnx)/(xlnx)^2 <0 # for #x>1#

so that #f(x)# is strictly decreasing in #(1,+oo)#.

(4) #f(n) = 1/(nlnn)# for #n>=2#

Then the convergence of the series:

#sum_(n=2)^oo 1/(nlnn)#

is equivalent to the convergence of the improper integral:

#int_2^oo dx/(xlnx)#

Now:

#int_2^oo dx/(xlnx) = lim_(t->oo) int_2^t dx/(xlnx)#

#int_2^oo dx/(xlnx) = lim_(t->oo) int_2^t (d(lnx))/lnx#

#int_2^oo dx/(xlnx) = lim_(t->oo) [ln(lnx)]_2^t #

#int_2^oo dx/(xlnx) = -ln(ln2)+ lim_(t->oo) (ln(lnt)) = +oo#

So the series is divergent.