Answer
Converges for values $p>1$, diverges $p\leq1$
Work Step by Step
Acts the same way as $\Sigma_{k=0}^\infty\frac{1}{k^p}$
Calculus: Early Transcendentals (2nd Edition)
by Briggs, Bill L.; Cochran, Lyle; Gillett, Bernard
Published by
Pearson
ISBN 10:
0321947347
ISBN 13:
978-0-32194-734-5
Chapter 8 - Sequences and Infinite Series - 8.4 The Divergence and Integral Tests - 8.4 Exercises - Page 638: 5
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