How do you test the series Sigma 1/(ln(n!))Σ1ln(n!) from n is [2,oo)[2,∞) for convergence?
1 Answer
The series:
is divergent.
Explanation:
Given the series:
(1)
First we note that for
We can now analyse the convergence of the series:
(2)
which is easier to determine using the integral test.
We take as test function:
and as
and:
all the hypothes of the integral test are satisfied and the series (2) is convergent only if the integral:
is also convergent.
Now we have:
So the series (2) is divergent and then also the series (1) is divergent.