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

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

1 Answer
Jun 18, 2018

The series:

sum_(n=2)^oo 1/(nlnn)n=21nlnn

is divergent.

Explanation:

Consider the function;

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

As:

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