How do you use the integral test to determine if #ln2/sqrt2+ln3/sqrt3+ln4/sqrt4+ln5/sqrt5+ln6/sqrt6+...# is convergent or divergent?

The integral test for convergence says that if we have a series:
#sum_(n=k)^oo f(n)#

Where #f# is positive, continuous and decreasing on the interval #[k,oo)#, the series converges if
#int_k^oo f(x)\ dx# converges.

We can write our sum like this:
#sum_(n=2)^oo ln(n)/sqrtn#

Unfortunately, our function isn't positive and decreasing on the interval #[2,oo)#, so we can't use the integral test just yet. We can however rewrite the sum like so:
#sum_(n=2)^7 ln(n)/sqrtn+sum_(n=8)^oo ln(n)/sqrtn#

The function is positive and decreasing on #[8,oo)#, which means we can use the integral test on the right sum.

This means that the sum diverges if the integral
#int_8^oo ln(x)/sqrtx\ dx# diverges.

To find the antiderivative, we begin by using integration by parts:
#int\ f(x)g'(x)\ dx=f(x)g(x)-int\ f'(x)g(x)\ dx#

Letting #f(x)=ln(x)# and #g'(x)=1/sqrtx#, we get:
#int\ ln(x)/sqrtx\ dx=2ln(x)sqrtx-int\ (2sqrtx)/x\ dx=#

#=2ln(x)sqrtx-2int\ x^(-1/2)\ dx=2ln(x)sqrtx-2*2sqrtx+C#

Now we can plug in the bounds of integration:
#int_8^ooln(x)/sqrtx\ dx=[2ln(x)sqrtx-4sqrtx]_8^oo=#

#=lim_(x->oo)(2ln(x)sqrtx-4sqrtx)-(2ln(8)sqrt8-4sqrt8)=oo#

Since the integral diverges, the sum also diverges.