5 sin n 11. Use the direct comparison test to show that converges. n! n-1 5 sin n 5. 5 sin n converges, so n-1 converges as well. n! n! n! * In n 12
5 sin n 11. Use the direct comparison test to show that converges. n! n-1 5 sin n 5. 5 sin n converges, so n-1 converges as well. n! n! n! * In n 12
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Could you show the missing pieces here? Why is 5/n! Greater then 5sin^2n/n!
![**11. Use the direct comparison test to show that the series ∑ (from n=1 to ∞) (5 sin² n / n!) converges.**
We observe that:
\[
\frac{5 \sin^2 n}{n!} \leq \frac{5}{n!}
\]
The series:
\[
\sum_{n=1}^{\infty} \frac{5}{n!}
\]
converges, so the series:
\[
\sum_{n=1}^{\infty} \frac{5 \sin^2 n}{n!}
\]
also converges as well.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F61b75115-d70f-4fe2-af93-2076876ad69a%2Faa9bede0-ca49-429a-a7c6-385e66a068cf%2Fddj9pe3_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**11. Use the direct comparison test to show that the series ∑ (from n=1 to ∞) (5 sin² n / n!) converges.**
We observe that:
\[
\frac{5 \sin^2 n}{n!} \leq \frac{5}{n!}
\]
The series:
\[
\sum_{n=1}^{\infty} \frac{5}{n!}
\]
converges, so the series:
\[
\sum_{n=1}^{\infty} \frac{5 \sin^2 n}{n!}
\]
also converges as well.
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