Answer
Divergent
Work Step by Step
$\Sigma \frac{1}{nln(n)}$
$\int^{\infty}_{2} \frac{1}{xln(x)}dx=\lim\limits_{t \to \infty}\int^{\infty}_{2} \frac{1}{xln(x)}dx=\lim\limits_{t \to \infty}[ln(ln(x))]^{t}_{2}=\infty-ln(ln(2))=\infty$
Divergent
Calculus: Early Transcendentals 8th Edition
by Stewart, James
Published by
Cengage Learning
ISBN 10:
1285741552
ISBN 13:
978-1-28574-155-0
Chapter 11 - Section 11.3 - The Integral Test and Estimates of Sums - 11.3 Exercises - Page 726: 21
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