Does $sum_(n=2)^oo1/(nln(n))$ converges ?

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Answer 1 · 2018-02-06T02:58:52

The series diverges.

Use the integral test. If we let $u =lnn$ , then $du = 1/ndn$ and $ndu = dn$

$I = int_2^oo 1/(nlnn) dn$

$I = int_2^oo 1/(n u) * ndu$

$I = int_2^oo 1/u du$

$I = [lnu]_2^oo$

$I = [ln(lnn)]_2^oo$

The limit $lim_(x->oo) ln(lnn)$ clearly equals $oo$ , therefore the series diverges.

Hopefully this helps!

Does sum_(n=2)^oo1/(nln(n))sum_(n=2)^oo1/(nln(n)) converges ?