Answer
$|p| \gt 3$
Work Step by Step
$\Sigma^{\infty}_{n=1} (\frac{3}{p})^{n}$
By geometric series $|r| \lt 1$ for the sum to converge,
So $|\frac{3}{p}| \lt 1$
$|p| \gt 3$, This satisfies the geometric series for convergence
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 9 - Infinite Series - 9.3 Exercises - Page 610: 51
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