Log In Start studying!

Select your language

Suggested languages for you:
Answers without the blur. Sign up and see all textbooks for free! Illustration


Essential Calculus: Early Transcendentals
Found in: Page 453
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

Answers without the blur.

Just sign up for free and you're in.


Short Answer

Determine whether the series is convergent or divergent:\(\sum\limits_{n = 0}^\infty {\frac{{1 + \sin n}}{{{{10}^n}}}} \).

The series \(\sum\limits_{n = 0}^\infty {\frac{{1 + \sin n}}{{{{10}^n}}}} \) is convergent.

See the step by step solution

Step by Step Solution

Comparing the series:

In this series, \(\sum\limits_{n = 0}^\infty {\frac{{1 + \sin n}}{{{{10}^n}}}} \)

Since, \(\sin n \le 1\)

So, \(\frac{{1 + \sin n}}{{{{10}^n}}} < \frac{2}{{{{10}^n}}}\)

Convergence of geometric series:

Now, the series \(\sum\limits_{n = 0}^\infty {\frac{2}{{{{10}^n}}}} \) is a geometric series with \(n = \frac{1}{{10}}\).

Since \(|n| < 1\)

So, \(\sum\limits_{n = 0}^\infty {\frac{2}{{{{10}^n}}}} \) series is convergent.

Comparison tests:

As \(\sum\limits_{n = 0}^\infty {\frac{2}{{{{10}^n}}}} \) series is convergent, therefore,

By comparison test, the series

\(\sum\limits_{n = 0}^\infty {\frac{{1 + \sin n}}{{{{10}^n}}}} \) is also convergent.

Hence, \(\sum\limits_{n = 0}^\infty {\frac{{1 + \sin n}}{{{{10}^n}}}} \)is also convergent.\(\)


Want to see more solutions like these?

Sign up for free to discover our expert answers
Get Started - It’s free

Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.

Sign up for free
94% of StudySmarter users get better grades.