Absolute covergece Defiitio A series P a is called absolutely coverget if the series of absolute values P a is coverget. If the terms of the series a are positive, absolute covergece is the same as covergece. Example Are the followig series absolutely coverget? X ( ) + 3, X ( ). To check if the series P ( ) + 3 check if the series of absolute values P P 3 ( ) + 3 Sice P is absolutely coverget, we eed to is coverget. is a p-series with p = 3 >, it coverges ad therefore is absolutely coverget. To check if the series P ( ) check if the series of absolute values P Sice P P ( ) 3 is absolutely coverget, we eed to is coverget. is a p-series with p =, it diverges ad therefore is ot absolutely coverget. Aette Pilkigto Lecture 28 :Absolute Covergece, Ratio ad root test
Coditioal covergece Defiitio A series P a is called coditioally coverget if the series is coverget but ot absolutely coverget. Which of the series i the above example is coditioally coverget? Sice the series P ( ) + is absolutely coverget, it is ot 3 coditioally coverget. Sice the series P ( ) last day to show this), but the series of absolute values P coverget, the series P is coverget (used the alteratig series test is ot ( ) is coditioally coverget. Aette Pilkigto Lecture 28 :Absolute Covergece, Ratio ad root test
Absolute cov. implies cov. Theorem If a series is absolutely coverget, the it is coverget, that is if P a is coverget, the P a is coverget. (A proof is give i your otes) Example Are the followig series coverget (test for absolute covergece) X ( ) + 3, X si() 4. Sice P ( ) + is absolutely coverget, we ca coclude that this 3 series is coverget. To check if the series P si() is absolutely coverget, we cosider the 4 series of absolute values P. si() 4 Sice 0 si(), we have 0 si(). 4 4 Therefore the series P coverges by compariso with the covergig p-series P Therefore the series P coverget. si() 4. 4 si() 4 is coverget sice it is absolutely Aette Pilkigto Lecture 28 :Absolute Covergece, Ratio ad root test
The Ratio Test This test is useful for determiig absolute covergece. Let P a be a series (the terms may be positive or egative). a Let L = lim + a. If L <, the the series P a coverges absolutely (ad hece is coverget). If L > or, the the series P a is diverget. If L =, the the Ratio test is icoclusive ad we caot determie if the series coverges or diverges usig this test. This test is especially useful where factorials ad powers of a costat appear i terms of a series. (Note that whe the ratio test is icoclusive for a alteratig series, the alteratig series test may work. ) Example Test the followig series for covergece X ( ) 2! lim a + a 2 = lim + (+)! 2 = lim = 0 <. + 2! Therefore, the series coverges. Aette Pilkigto Lecture 28 :Absolute Covergece, Ratio ad root test
Example 2 Ratio Test Let P Let L = lim a + a a be a series (the terms may be positive or egative).. If L <, the the series P a coverges absolutely. If L > or, the the series P a is diverget. If L =, the the Ratio test is icoclusive. Example 2 Test the followig series for covergece lim a + a X ( ) 5 (+) 5 = lim + = + lim 5 5 lim ( + /) = 5 <. Therefore, the series coverges. 5 = Aette Pilkigto Lecture 28 :Absolute Covergece, Ratio ad root test
Example 3 Example 3 a lim + a lim Test the followig series for covergece P! (+) = lim + (+)! +! = lim + (+)(+) = lim x. = limx + x lim x + x x = limx e x l(+/x) = e limx x l(+/x). lim x x l( + /x) = lim x l(+/x) /x = (L Hop) lim x lim x (+/x) =.! = (+)! /x 2 (+/x) /x 2 = x a Therefore lim + a = limx + x = e = e > ad the series P diverges.! Aette Pilkigto Lecture 28 :Absolute Covergece, Ratio ad root test
Example 4 Example 4 Test the followig series for covergece P ( ) 2 We kow already that this series coverges absolutely ad therefore it coverges. (we could also use the alteratig series test to deduce this). Lets see what happes whe we apply the ratio test here. a lim + (+) a = lim 2 2 = lim 2 + = 2 lim +/ =. Therefore the ratio test is icoclusive here. Aette Pilkigto Lecture 28 :Absolute Covergece, Ratio ad root test
The Root Test Root Test Let P a be a series (the terms may be positive or egative). If lim p a = L <, the the series P a coverges absolutely (ad hece is coverget). If lim p a = L > or lim p a =, the the series P a is diverget. If lim p a =, the the Root test is icoclusive ad we caot determie if the series coverges or diverges usig this test. Example 5 Test the followig series for covergece P ( ) 2 + lim p r 2 a = lim + = 2 lim = + lim 2 = +/ 2 > Therefore by the th root test, the series P ( ) 2 + diverges. Aette Pilkigto Lecture 28 :Absolute Covergece, Ratio ad root test
Example 6 Root Test For P a. L = p lim a. If L <, the the series P a coverges absolutely. If L > or, the the series P a is diverget. If L =, the the Root test is icoclusive. Example 6 Test the followig series for covergece P lim p r a = lim /2 < 2+ Therefore by the th root test, the series P 2+ = lim 2+ = lim 2+/ = 2+ coverges. Aette Pilkigto Lecture 28 :Absolute Covergece, Ratio ad root test
Example 7 Root Test For P a. L = p lim a. If L <, the the series P a coverges absolutely. If L > or, the the series P a is diverget. If L =, the the Root test is icoclusive. Example 7 Test the followig series for covergece P lim p r l a = lim = l lim (L Hop) lim x /x = 0 < Therefore by the th root test, the series P l. = lim x l x x = l coverges. Aette Pilkigto Lecture 28 :Absolute Covergece, Ratio ad root test
Rearragig sums If we rearrage the terms i a fiite sum, the sum remais the same. This is ot always the case for ifiite sums (ifiite series). It ca be show that: If a series P a is a absolutely coverget series with P a = s, the ay rearragemet of P a is coverget with sum s. It a series P a is a coditioally coverget series, the for ay real umber r, there is a rearragemet of P a which has sum r. Example The series P ( ) is absolutely coverget with 2 P ( ) = 2 ad hece ay rearragemet of the terms has sum 2. 2 3 3 Aette Pilkigto Lecture 28 :Absolute Covergece, Ratio ad root test
Rearragig sums It a series P a is a coditioally coverget series, the for ay real umber r, there is a rearragemet of P a which has sum r. P Example Alteratig Harmoic series coverget, it ca be show that its sum is l 2, ( ) is coditioally 2 + 3 4 + 5 6 + 7 8 + 9 + ( ) + = l 2. Now we rearrage the terms takig the positive terms i blocks of oe followed by egative terms i blocks of 2 2 4 + 3 6 8 + 5 0 2 + 7 = 2 4 + 3 6 8 + 5 0 2 + 7 = 4 2 2 + 3 4 + 5 6 + 7 8 + 9 + ( ) +... ) = l 2. 2 Obviously, we could cotiue i this way to get the series to sum to ay umber of the form (l 2)/2. Aette Pilkigto Lecture 28 :Absolute Covergece, Ratio ad root test