Question

How do you test the series `(lnk)/(k^2)` for convergence?

Answer 1

The series:

`sum_(k=1)^oo lnk/k^2`

is convergent.

Explanation:

To test the convergence of the series:

`sum_(k=1)^oo lnk/k^2`

we can use the integral test, with test function:

`f(x) = lnx/x^2`

Verify that the hypotheses of the integral test theorem are satisfied:

(i) `f(x)` is positive in `[1,+oo)`

(ii) `(df)/dx = (1-2lnx)/x^3` is negative in `[1,+oo)`

so the function is monotone decreasing in the interval

(iii) `lim_(x->oo) f(x) = 0`

(iv) `f(k) = lnk/k^2`

So the convergence of the series is equivalent to the convergence of the integral:

`int_1^oo lnx/x^2dx`

We start solving the indefinite integral by parts:

`int lnx/x^2dx = int lnx d(-1/x) = -lnx/x +int dx/x^2=-lnx/x-1/x+C`

so:

`int_1^oo lnx/x^2dx = [-lnx/x-1/x]_1^oo`

`int_1^oo lnx/x^2dx = 1 -lim_(x->oo) lnx/x - lim_(x->oo) 1/x =1`

The integral converges, so the series is also proven to be convergent.