University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.3 - The Integral Test - Exercises - Page 504: 5


Converges to $\frac{1}{2}e^{-2}$

Work Step by Step

Consider $f(x)=e^{-2x}$. This is positive, continuous and decreasing for $x \geq 1$ Now, take the integral test to find the convergence and divergence for the sequence. We have $\int_1^\infty e^{-2x} dx= \lim\limits_{a \to \infty} \int_1^a e^{-2x} dx$ or, $ \lim\limits_{a \to \infty} [\dfrac{-1}{2} e^{-2x}]_1^a=\lim\limits_{a \to \infty} [\dfrac{-1}{2} e^{-2a}+\frac{1}{2}e^{-2}]= \frac{1}{2}e^{-2}$ Hence, the sequence $\Sigma_{n=1}^\infty e^{-2n}$ converges to $\frac{1}{2}e^{-2}$
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