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

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Answer 1 · 2018-06-25T12:57:06

The integral

#int_2^oo dx/lnx#

is divergent.

Note that in the interval #x in[2,oo)# the function:

#f(x) = 1/lnx#

is:

1) Infinitesimal as #lim_(x->oo) f(x) = 0#

2) Positive as #f(x) >0 # for #x >1#

3) Decreasing. In fact #f'(x) = -1/(xln^2x) < 0#

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

So, based on the integral test, the convergence of the integral:

#int_2^oo dx/lnx#

is equivalent to the convergence of the series:

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

Now we can easily demonstrate that:

#lnn < n#

so that:

#1/ln n > 1/n#

and as we know that the harmonic series:

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

is divergent, we can conclude that:

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

is divergent by direct comparison, and hence also:

#int_2^oo dx/lnx#

is divergent.