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

1 Answer
Feb 5, 2017

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.