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by: Zachary Hill

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# Calculus II Notes Week #12 MATH 1220

Marketplace > Tulane University > Mathematics (M) > MATH 1220 > Calculus II Notes Week 12
Zachary Hill
Tulane
GPA 3.88

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These notes cover the p-Test, the Leibniz Test, and use of the Ratio Test.
COURSE
Calculus II
PROF.
Benjamin Klaff
TYPE
Class Notes
PAGES
6
WORDS
CONCEPTS
Math, MATH 1220, Calculus, calculus II, Klaff
KARMA
25 ?

## Popular in Mathematics (M)

This 6 page Class Notes was uploaded by Zachary Hill on Sunday April 10, 2016. The Class Notes belongs to MATH 1220 at Tulane University taught by Benjamin Klaff in Spring 2016. Since its upload, it has received 12 views. For similar materials see Calculus II in Mathematics (M) at Tulane University.

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Date Created: 04/10/16
MATH 1220 Notes for Week #12  4 April 2016  ● Realize you can bound cos(nx)  where n is a positive integer above and below by [1,− 1]   ● Then this is bounded on [− R, R] when R = 1  cos(nx) ● Let fn(x) = n  on [− R, R], R > 0; can you bound  f (x) |nrom|above?   ● Let M  ne the upper bound; since cos(nx)  is bounded above by 1 ,  cos2nxshould be  n 1 bounded above by M = n n2   ∞ ● Does  ∑ 2 converge?  n=1n ● Took bad notes this day, but it was mostly just a setup for the other days’ notes; you  should get enough information from them      5 April 2016  The p­Test  ∞ ● Consider whether  ∑ n  converges  n=1 ∞ N ● First, look at ∫x2dx = lim ∫x2dx = lim[(− )N− (− )]1= lim(1 − ) =N1 − 0 = 1   1 N→∞ 1 N→∞ N→∞ N N ∞ ● Realize that  lim ∑ 12< lim ∫ 2dx < ∞  because  ∑ 12 is a right hand Riemann sum of the  N→∞ n=2n N→∞ 1x n=2n ∞ curve   2n [1, ∞); then  ∑ 12 converges  x n=2n ∞ 1 1 ∞ 1 ∞ 1 ● Since  ∑ n2= 1 + ∑ n2, then  ∑ n2  also converges  n=1 n=2 n=1 ∞ ● 1dx converges if p > 1  ∫xp 1 ● For p > 1 ∞ −p 1 −p+1 N 1 −p+1N 1 −p+1 −p+1 1 1 lim ∫ dx = lim [ −p+1 ∙ x ]1= −p+1 lim [x ]1= −p+1lim(N − 1 ) = −p+1(0 − 1) = p−1< ∞  N→∞ 1 N→∞ N→∞ N→∞ ∞ −1 ● If p = 1 lim x∫dx " = " ∞, meaning it does not converge; this also occurs if p < 1  N→∞ 1 ∞ 1 ● ∑  ns called the harmonic series  n=1 ● p­Test     p < 1  p = 1  p > 1  ∞ ∫ 1dx  Diverges.  Diverges.  Converges  xp 1 Evaluate.  Evaluate.  ∞ 1 Diverges by  Diverges by CT.  Converges by CT.  ∑ np  n=1 Comparison Theory  (CT).        6 April 2016  This is true, but a logical trap  ∞ n ● Use the ratio test to determine for which x,  ∑ x  converges.  n=0 ∞ |n+1| ● If n→∞ |an |= L < 1, then  ∑ a  nonverges  n=0 ∞ ● lim |n+| = lim |x| = |x| < 1, so when |x| < 1, then  ∑ x  converges  n→∞ |xn| n→∞ n=0 ● Logical trap because we only know to apply the ratio test because we assume that there  ∞ n will be some x so that  ∑ x  converges  n=0 ● We can also apply the root test  ∞ n ● If n→∞ √ =nL < 1, then  ∑ a  converges  n=0 ∞ ● lim |x| = |x| < 1 , so when |x| < 1, then  ∑ x  converges  n→∞ √ n=0 Ratio Test Proof for Geometric Series  N N n n ● Let − 1 < x < 1. S =N∑ x . xS = x N x .  n=0 n=0 0 1 2 N ● SN= (x + x + x + . . . + x )   ● xSN= (x)(x + x + x + . . . + x ) = (x + x + x + . . . + xN+1)  ●   S = (x + x + x + . . . + x )  N xS = (x + x + x + . . . + x N+1)  N     S − xS = 1 − x N+1   N N N+1 SN(1 − x) = 1 − x   1−xN+1 SN= 1−x   N+1 ● So  lim S =Nlim 1−x = ( 1 ) lim(1 − x N+1) = (1 )( lim 1 − lim xN+1 ) = (1 )(1 − 0) = 1  for  N→∞ N→∞ 1−x 1−x N→∞ 1−x N→∞ N→∞ 1−x 1−x any choice on − 1 < x < 1  N i 1 ● So  ∑ x = 1−x for − 1 < x < 1, and therefore converges  i=0       8 April 2016  Leibniz Test (Alternating series test)  ∞ ● Given  ∑ a , nf a  isndecreasing and  lim a = 0, thennthe alternating sum of a   n n=0 n→∞ converges.  Use the ratio test to investigate the convergence of these      Converges on (− R, R)   Investigate the convergence  (R = 0, R > 0, R " = " ∞)  when x = R and x =− R  ∞   ∑ n!x   Only converges when x = 0;  N/A since R = 0  n=0 therefore R = 0  ∞ n Converges when |x| < 1;  Diverges at x = 1  or x =− 1  ∑ nx   n=0 therefore R = 1  ∞ n ∑ n+1   Converges when |x| < 1;  Diverges at x = 1 ; converges  n=0 therefore R = 1  at x =− 1  ∞ ∑ x 2       n=0 (n+1)   ∞ ∑ n!x  n n=0 n+1 n+1 ● lim |n+1)!x |= lim |!(n+1)x |= lim |(n + 1)x| = |x| lim |n + 1| < 1 only if x = 0  n→∞ | n!xn | n→∞ | n!xn | n→∞ n→∞ ∞ n ∑ nx   n=0 (n+1)xn+1 (n+1)x |1+1| ● lim | nxn | = lim | n | = |x| lim |n1 |= |x| lim | 1n| = |x| < 1   n→∞ n→∞ n→∞ n→∞ | | ∞ ∞ ● ∑ n(1) = ∑ n  n=0 n=0 ∞ ● lim |+n |= |1| = 1 so  ∑ nx  does not converge at x = 1  n→∞ n=0 ∞ n ● ∑ n(− 1)   n=0 | n+1| ∞ ● lim |(n+1)(−1n | = lim |n+1)(−1| = lim |−n−1| = |− 1| = 1so  ∑ nx  does not converge at x =− 1  n→∞ | n(−1) | n→∞ | n | n→∞ | n | n=0 ∞ xn ∑ n+1   n=0 x+1 1 ● lim ||(nx1)| = lim |xn+1∙n+1 |= lim |(n+1)| = |x| lim |n+1 |= |x| lim |1+ n| = |x| < 1  n→∞ | n+1 | n→∞ |n+2 xn | n→∞ | n+2 | n→∞ |n+2| n→∞ |+ n| ∞ (1) ∞ 1 ● ∑ n+1 = ∑ n+1   n=0 n=0 1 ∞ |(n+1)+1 | 1 n+1 | |+1 | xn ●   n→∞m | n+1 | = n→∞ |n+2 ∙ 1 |= ln→∞ |n+2|= |1| = 1so  ∑ n+1  does not converge at x = 1  | | n=0 ∞ n ● ∑ (−1)   n+1 n=0 (−1)n (−1) (−1) ● Recognize   is decreasing and lim = lim n 1 = lim 0 = 0, so by the Leibniz test,  n+1 n→∞ n+1 n→∞ 1+n n→∞ 1+0 ∞ (−1) ∑ n+1  converges  n=0 ∞ ∑ xn 2  (n+1) n=0 ● Using the previous three as examples, try to solve this one

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