{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Day34-1432class - Math 1432 Dont forget online quizzes...

Info icon This preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon
Math 1432 Don’t forget online quizzes. THINK: Find the number to replace the question mark. 369542 is to 246359 as 172896 is to 268179 as 417638 is to ?
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Review: Geometric Series Test. The geometric series diverges if r 1 . If < r 1 , then the series converges to the sum = - 1 a S 1 r . The sum of the first n terms of a geometric series is ( 29 - = - n 1 a 1 r S 1 r . Ex. ( 29 = n n 1 3 1 2 .
Image of page 2
Basic Divergence Test If → ∞ n n a 0 lim , then the series = n n 1 a diverges. Ex. = - + 2 2 n 1 n 1 n n
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
p-Series Test: A series of the form = = + + + + + p p p p p n 1 1 1 1 1 1 n 1 2 3 n ... ... The p-series diverges if < 0 p 1. The p-series converges if p > 1. Ex. = 8 n 1 27 n = 2 n 1 3 1 n
Image of page 4
Integral Test : If f is positive , continuous , and decreasing for x 1 and ( 29 = n a n f , then = n n 1 a and ( 29 1 x dx f either both converge or both diverge. Ex. ( 29 = n 2 1 n n ln
Image of page 5

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Basic Comparison Test: If n a 0 and n b 0 and 1) If = n n 1 b converges and n n 0 a b , then = n n 1 a converges.
Image of page 6
Image of page 7
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern