MAT136H1 Lecture 25: Series
40 views3 pages
![](https://s3.us-east-1.wasabisys.com/prealliance-avatars.oneclass.com/avatars/515914/small/RackMultipart20201118-71849-4b6192.png?1637711045)
![MAT136H1 Full Course Notes](https://new-docs-thumbs.oneclass.com/doc_thumbnails/list_view/2418954-class-notes-ca-utsg-mat-136h1-lecture21.jpg)
92
MAT136H1 Full Course Notes
Verified Note
92 documents
Document Summary
A n = a 1 + a 2 + a 3 + . 1 + 2 + 3 + 4 + 5 n=1 a i = a 1 + a 2 + a 3 + . + a n a n n=1 is convergent a n n=1 is divergent r. R = a r 2 n(1. S (1 a(1 r )2 ns = 1 r lim n n. S = lim a(1 r )2 n (1 r) n lim n n = therefore it diverges because summation goes to. S = a + a + a 2 + a 3 + . S = a + a + a 2 + . n. Ar n 1 n 1 + a n r (subtract these two) r n. S = a + a + a 2. S = 7 3 a = 7 = 3. S = lim n n n(n+1) = n.
Get access
Grade+
$40 USD/m
Billed monthly
Homework Help
Study Guides
Textbook Solutions
Class Notes
Textbook Notes
Booster Class
10 Verified Answers
Class+
$30 USD/m
Billed monthly
Homework Help
Study Guides
Textbook Solutions
Class Notes
Textbook Notes
Booster Class
7 Verified Answers