Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Section 11.2 - Series - 11.2 Exercises - Page 717: 77

Answer

The harmonic series diverges.

Work Step by Step

$e^{S_{n}} = exp (1 + \frac{1}{2} +\frac{1}{3} +\frac{1}{4} +...+ \frac{1}{n}) = e^{1} \times e^{\frac{1}{2}} \times e^{\frac{1}{3}} \times e^{\frac{1}{4}} ... e^{\frac{1}{n}}$ Since $e^{n} \gt 1+n$ $e^{1} \times e^{\frac{1}{2}} \times e^{\frac{1}{3}} \times e^{\frac{1}{4}} ... e^{\frac{1}{n}} \gt (1+1) \times (1+\frac{1}{2}) \times (1+\frac{1}{3}) \times (1+\frac{1}{4}) \times ... \times (1+\frac{1}{n})$ $e^{S_{n}} \gt (\frac{2}{1}) \times (\frac{3}{2}) \times (\frac{4}{3}) \times (\frac{5}{4}) \times ... \times (\frac{n+1}{n})$ $e^{S_{n}} \gt n+1$ $\ln e^{S_{n}} \gt \ln(n+1)$ $S_{n} \gt \ln (n+1)$ $\lim\limits_{n \to \infty}S_{n} \gt \lim\limits_{n \to \infty} \ln(n+1)$ Sum of harmonic series $\gt \infty$ Hence the harmonic series diverges.
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