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Math 2062

by: Austin Ledermeier

Math 2062 MATH1062

Austin Ledermeier
GPA 3.4
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About this Document

These notes contain sections 11.1-11.4 in calculus 2.
Calculus II
Dr. Jason Jasper
Class Notes
Calculus ll




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This 2 page Class Notes was uploaded by Austin Ledermeier on Sunday March 6, 2016. The Class Notes belongs to MATH1062 at University of Cincinnati taught by Dr. Jason Jasper in Spring 2016. Since its upload, it has received 31 views. For similar materials see Calculus II in Mathematics (M) at University of Cincinnati.

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Date Created: 03/06/16
11.1-11.4 11.1 Sequences Sequence- can be thought of as a list of numbers written in a definite order. a1, a2, a3,……. ax EX.) (n^2)from n=1 to n=infinity {1, 4, 9, 16…..} EX.)((-1)^n)/(n+3) { 1/5, -1/6, 1/7, -1/8} Notice that because of the negative 1 to the n the series alternates from positive to negative A sequence has a limit L. This shows what all the numbers are going to in the sequence. Lim as n goes to infinity of (an)=L EX.)lim as n goes to infinity of (1/n)=0 As n gets bigger the denominator is getting large and the denominator is staying the same so the number is getting close and close to 0 If the sequence has a limit then the sequence converges. Otherwise, the sequence diverges. Like the sequence n^2 as n gets large the numbers go to infinity making the sequence divergent. 11.2 Series Infinite series- is the sum of the terms of a sequence. Given a series the sum of (an) from n=1 to infinity… a1+ a2+ a3+….. let sn be the partial sum. If the sequence (sn) is convergent and the limit of sn exists as a real number then the series is convergent. EX.) the sum between n=1 and infinity of (n+1)/n^2… 2/1+3/4+4/9… So to see if this series is divergent or convergent we can take the limit as n goes to infinity of (n+1)/n^2 Because of hospitals rule saying we cant have infinity/infinity, we can take the derivative of the numerator and denominator and take the limit of that. This leaves us with the limit as n goes to infinity of 1/2n which is now seen as 0 so the series converges. 11.3 Integral test If an is positive, decreasing, and continuous on [1,infinity) for all n on the series an. Then an converges if and only if the integral from 1 to infinity of an converges to a number EX.) sum from n=1 to infinity of 1/squareroot(n)… Positive: yes because from 1 to infinity an will be positive Decreasing: Yes because when you find the derivative of an you get -1/2*squareroot(n^3). Continuous: Yes because the only problem would be when n is 0 or negative, which it isn’t on the interval So now the integral from 1 to infinity of an can be taken to see whether the series converges or diverges. So this comes out to be 2*squareroot(infinity)-2*squareroot(1) which is infinity. This is not a number so the integral diverges. So, the series 1/squareroot(n) Diverges Also, there is a shortcut for series that look like this previous one. These series are called p series. Where an= 1/x^p and when p is greater then 1 then the series converges and when its less then or equal to one then the series diverges 11.4 Comparison tests Direct comparison theorem Assume 0 <=an<=bn for all an 1.) if the sum from 1 to infinity of bn converges then the sum from 1 to infinity of an convergers 2.) if the sum from 1 to infinity of an diverges then the sum from 1 to infinity of bn diverges. Limit comparison theorem If you have two positive sequences an and bn (both going from 1 to infinity) and lim an/bn= c when c doesn’t equal infinity or 0 then either an and bn converge or diverge. EX.) the sum from 1 to infinity of 1/(2^(n)-n) applying the limit comparison theorem we see that 1/(2^n) kinda looks like the original. We will make the original an and the new one bn. So the we take the limit as n goes to infinity of bn/an. This turns out to be lim as n goes to infinity of (2^(n)-n)/2^n which then is the limit 1-(n/2^n). which is 1-0 So we know that bn converges so that mean an converges because of the limit comparison theorem that we just did. My study guides will have many more problems with solution.


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