Question

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

Answer 1

The sum diverges.

Explanation:

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.