Answer
Divergent
Work Step by Step
$\Sigma \frac{1}{nln(n)}$
$\lim\limits_{t \to \infty}\int_{1}^{t}\frac{1}{xln(x)}=\lim\limits_{t \to \infty}[ln(ln(x))]^{t}_{1}=ln(ln(\infty))-ln(ln(2)=\infty$
Divergent
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