Answer
false
Work Step by Step
False; for example $\Sigma ^{\infty}_{k=0}2^{-k}$ converges to 2, but $\Sigma \frac{1}{2^{-k}}=\Sigma 2^k$ diverges.
Calculus, 10th Edition (Anton)
by Anton, Howard
Published by
Wiley
ISBN 10:
0-47064-772-8
ISBN 13:
978-0-47064-772-1
Chapter 9 - Infinite Series - 9.4 Convergence Tests - Exercises Set 9.4 - Page 630: 31
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