How do you test the series Sigma 1/(nlnn) from n is [2,oo) for convergence?
1 Answer
Mar 12, 2017
Explanation:
Note that:
d/(dx) ln x = 1/x
So:
d/(dx) ln(ln x) = 1/(ln x) * d/(dx) ln x
color(white)(d/(dx) ln(ln x)) = 1/(ln x) * 1/x
color(white)(d/(dx) ln(ln x)) = 1/(x ln x)
So:
int 1/(x ln x) dx = ln(ln x) + C
Then:
lim_(x->oo) ln(ln x) = oo
So by the integral test:
sum_(n=2)^oo 1/(n ln n)" " diverges.