Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 9 - Infinite Series - Review Exercises - Page 677: 46

Answer

Converges

Work Step by Step

Here, we have the given series as $\dfrac{1}{ 5^{n}}$ Here, we can see that the function $F(n)=\dfrac{1}{ 5^{n}}$ shows a positive and continuous for $n \geq 1$. Next, we have $F(n)=\dfrac{1}{ 5^{n}} \implies F'(n)=\dfrac{-n}{5^{n+1}}$. We can see that $F'(n) \lt 0 $ for $n \gt 1$. This implies that the given function is decreasing. Thus, we need to apply the integral test. So, $\int_1^{\infty} \dfrac{dn}{5^n}=\lim\limits_{a \to \infty} \int_1^{a} \dfrac{dn}{5^n}$ or, $=\lim\limits_{a \to \infty} [-\dfrac{5^{-n}}{\ln 5}]_1^{\infty}$ or, $=\dfrac{1}{5 \ln (5)}$ Therefore, the series converges by the integral test.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.