ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS. Absolute Covergece Give ay series a we ca cosider a related series a a + a 2 +... + a +... whose terms are the absolute values of the terms of the origial series Defiitio.. A series a is called absolutely coverget if the series of absolute values a is coverget. Notice that if a is a series with positive terms the a a ad absolute covergece is the same as covergece. ( ) 4 Example.2. The series 2 + 4 4 4 +... is absolutely coverget 4 sice ( ) 4 4 is a coverget p-series with p 4 >. Example.. We kow that the alteratig harmoic series is coverget. However, it is ot absolutely coverget sice the correspodig series of absolute values is the harmoic series which is diverget as this is a p-series with p : ( ) ( ) From this example, we should defie a ew defiitio of covergece Defiitio.4. A series a is called coditioally coverget if it is coverget but ot absolutely coverget. The alteratig harmoic series is coditioally coverget, ad so there are series that are coditioally coverget but ot absolutely coverget. Absolute covergece is a desirable property as the ext theorem will show: Theorem.5. If a series a is absolutely coverget, the it is coverget. Example.6. Determie if the series cos 2 cos 2 + cos 2 2 2 + cos 2 is coverget or diverget.
2 ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS Proof. This series has positive ad egative terms but it is t alteratig as we kow it, i.e., i the maer ( ). We ca apply the Compariso Test to the series of absolute values though cos As cos for all, the 2 cos 2. cos 2 2 This series is a coverget p-series, with p 2 > ad so the cos / 2 is coverget by the Compariso Test, ad the origial series is absolutely coverget, implyig covergece by the above theorem. Example.7. Show the series 2 ( ) ++ Proof. The series of absolute values are compariso with the Harmoic series, lim 2 ++ 2 ++ lim + + > 0, is coverget or ot. which diverges by the limit the series diverges by the Limit Compariso Test sice the harmoic series diverges. The origial series does ot coverge absolutely, but it is coditioally coverget. Writig f(x) x2 x +x+ the f (x) x(x x 2) (x + x + ) 2 this is egative for large x, ad so b 2 ++ is evetually decreasig, by the Alteratig Series Test the origial series is coverget. 2. The Ratio Test As a helpful tool for determiig absolute covergece, we have the test: Theorem 2.. The Ratio Test a i) If lim + a L < the the series a is absolutely coverget, ad hece coverget. a ii) If lim + a L > or a + lim a the the series a is diverget. a iii) lim + a the Ratio Test is icoclusive, meaig we caot say aythig about the covergece or divergece of a. Remark 2.2. To uderstad the third part of the Ratio Test, cosider the coverget series, 2 a + a (+) 2 2 ( + ) 2 + 2, as 2
ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS whereas for the diverget series we have: a + a (+) ( + ) + 2, as Ad so if lim a + /a the series a might coverge or it might diverge. I this case, some other test must be employed to prove covergece. Example 2.. Test the series ( ) for absolute covergece. Proof. Usig the ratio test with a ( ) : a + ( ) + (+) a + ( + ) ( ) + + ( + ) <. By the Ratio Test, the give series is absolutely coverget ad so it is coverget. Example 2.4. Test the series! Proof. As the terms a! are positive, we eglect the absolute value sigs, a + a ( + )+! ( + )! ( + )( + ) ( + )! ( +! ) ( + ) e, as As e 2.7 > the series is diverget by the ratio test. This ca be verified by usig the Test for Divergece sice a! 2 ad so a does ot approach 0 as.. The Root Test I the case that the terms of the series ivolve th powers, we have the followig theorem Theorem.. The Root Test i) If lim a L <, the the series a is absolutely coverget ad hece coverget. ii) If lim a L > or lim a, the the series a is diverget. iii) If lim a, the Root Test is icoclusive The third part of the Root Test, tells us that there is o iformatio about the series covergece or divergece. If L i the Ratio test, oe should t use the Root Test, as L will equal oe, ad vice versa.
4 ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS Example.2. Test the series for covergece (. 2+ +2) Proof. Usig the a 2 + + 2 2 + a 2 + + 2 + 2 2 < as, therefore the series coverges by the Root Test. 4. Why we like Absolute Covergece:Rearragemets Fiite sums behave i really ice ways, if we re-arrage the order of terms i a fiite sum, the value of the total sum will ot chage. This is t the case for some ifiite series. By re-arragig a ifiite series, we mea that a ew series is created by chagig the order of the terms. As a example, a possible re-arragemet of a series a + a is a + a + a 2 + a 5 + a 8 + a + a 2 + a 4 +... There is a theorem for absolutely coverget series Theorem 4.. If a is a absolutely coverget series, with sum s, the ay re-arragemet of a has the same sum s. For coditioally coverget series, the situatio is a bit strager. For example, the alteratig harmoic series has the sum 2 + 4 + 5 6 + 7 + l 2 If we multiply the series by 2 we have () 2 4 + 6 8 + 0 2 + 4 + 2 l 2 isertig zeros betwee each term of this series (2) 0 + 2 + 0 4 + 0 + 6 + 0 8 + 0 + 0 + 0 2 + 0 + 4 2 l 2 If we add () to (2) ad use the fact that two coverget series ca be added together, we fid () + 2 + 5 + 7 4 + 2 l 2 While we have oly displayed the first few terms, () is the same as the alteratig harmoic series, with some terms re-arraged so that oe egative term occurs after each pair of positive terms. The problem here is ow the sum is differet. Riema proved that Theorem 4.2. If a is a coditioally coverget series ad r is ay real umber, the there is a re-arragemet of a that has a sum equal to r.
ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS 5 If we treated coditioally coverget ifiite series like fiite sums we could ru ito all sorts of trouble; what would happe if we subtracted the defiitio of the alteratig harmoic series with ()? This is why absolutely coverget series are so desirable.