How do you use the integral test to determine whether #int dx/(x+lnx)# converges or diverges from #[2,oo)#?

Question

1659 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-01T12:39:40

The integral:

#int_2^oo dx/(x+lnx)#

diverges.

As in the interval #x in [2,+oo)# the function:

#f(x) = 1/(x+lnx)#

is positive and decreasing, and:

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

#f(n) = 1/(n+lnn)#

based on the integral test the convergence of the improper integral:

#int_2^oo dx/(x+lnx)#

is equivalent to the convergence of the series:

#sum_(n=2)^oo 1/(n+lnn)#

Consider now the harmonic series:

#sum_(n=0)^oo 1/n = oo#

that we know to be divergent and apply the limit comparison test:

#lim_(n->oo) (1/(n+lnn))/(1/n) = lim_(n->oo) (1/(1+lnn/n)) = 1#

as the limit is finite the two series have the same character, then also the series

#sum_(n=2)^oo 1/(n+lnn)#

is divergent.