How do you test the series #Sigma 1/(nlnn)# from n is #[2,oo)# for convergence?

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Answer 1 · 2017-03-12T12:47:35

#sum_(n=2)^oo 1/(n ln n)" "# diverges.

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.