Question

How do you use the integral test to determine if `Sigma lnn/n^3` from `[2,oo)` is convergent or divergent?

Answer 1

`sum_(n=2)^(+oo) lnn/n^3` is convergent

Explanation:

We can start noting that for `n=1`, `a_n = ln n/n^3=0`, so we can write the series as:

`sum_(n=1)^(+oo) lnn/n^3`

We now that this series is convergent iff the integral:

`int_1^(+oo) (lnx )/x^3dx`

is convergent as well.

We can calculate the integral by parts:

`int_1^(+oo) (lnx )/x^3dx = -1/2int_1^(+oo) lnx* d(1/x^(2)) = [-1/2 ln x/x^2]_1^(+oo) +1/2 int_1^(+oo)1/x*1/x^2dx = 0 + 1/2 int_1^(+oo)(dx)/x^3 = [-1/(4x^2)]_1^(+oo)=1/4`

and therefore the series is convergent.