MATH 572H Sprig 20 Worksheet 7 Topics: absolute ad coditioal covergece; power series. Defiitio. A series b is called absolutely coverget if the series b is coverget. If the series b coverges, while b diverges, the it is called coditioally coverget. Fact. If a series is absolutely coverget, the it is coverget. This statemet may soud a bit tautological ad eve silly, but actually it is ot totally trivial ad requires a proof (which is ot too hard though; see p.467). Defiitio. A series of the form ( ) + a = a a 2 + a a 4 +... where a s are positive umbers, is called alteratig. There is a simple covergece test for alteratig series, which is due to Leibiz : if a sequece of positive umbers {a } is mootoically decreasig ad a = 0, the the alteratig series ( ) + a = a a 2 + a a 4 +... coverges. Notice that this test does ot say aythig about divergece of a alteratig series.. For each of the give series, determie if it is absolutely coverget, coditioally coverget or diverget. a) 6 + 9 2 + + ( )+ +... The series + 6 + 9 + 2 + + +... clearly diverges (as a multiple of the harmoic series). Therefore, the give series caot be absolutely coverget. Let s see if the give series is coditioally coverget. To this ed, observe that it is a alteratig series with a =. The sequece {a } is mootoically decreasig ad a = 0. Therefore, by the alteratig series test, the give series coverges (coditioally). b) 2 2 + 4 4 + + ( ) + +... The series + 2 2 + + 4 4 + + +... coverges by the compariso test. Ideed, o the oe had we kow that the geometric series coverges. O the other had, we have 0 <, =, 2,... Therefore, the give series is absolutely coverget. Gottfried Wilhelm vo Leibiz (646-76) was a promiet Germa mathematicia, scietist ad philosopher.
c) 0 ( ) l d) Sice for ay 0, we have l > ad the harmoic series diverges, the, by the compariso test, the series series is ot absolutely coverget. 0 l 0 diverges. Therefore, the give Let s see if the give series coditioally coverget. To this ed, we observe that this series is alteratig with a = l. The sequece {a } mootoically decreases ad a = 0. Therefore, by the alteratig series test, the series coverges (coditioally). cos 2 +0 0 ( ) l The series cotais both positive ad egative terms, but it s ot alteratig (cos chages its sig i a pretty o-trivial fashio). Thus, we caot use the alteratig series test here. O the other had, we still ca check if the series is absolutely coverget. To this ed, we otice that e) The series 2 0 < our series coverges (absolutely). ( ) + + 2 cos 2 + 0 < 2 + 0 < 2. coverges (as a p-series with p > ). Therefore, by the compariso test, We simply observe that + 2 = 2 0. Thus, the give series diverges by the -th term test. 2. Prove that the series coverges. + ( ) ( + 8 ) ( ) + + 2 4 9 2 +... Will the series coverge if we remove all the paretheses from this formula? Thus, We ca rewrite the formula for the -th term of the give series as follows: 2 2 = 2 ( + ) ( + 8 ) ( ) + + 2 4 9 2 + = 2. 2
This is the p-series with p = 2 >. Hece, it coverges. If we remove the paretheses, the give series will take the followig form: The terms of this series form the sequece + 4 + 8 9 + + 2 2 +...,, 4,, 8,, 5 9 6,... ad we ca see that the it of this sequece is ot goig to be equal to zero (simply because it cotais ifiitely may terms equal to ). Therefore, by the -th term test, a ew series diverges. Moral. I geeral, oe caot use the ordiary algebra rules, whe dealig with ifiite series.. For each of the give series, fid its iterval of covergece. a) ( ) x + By the ratio test, the series will coverge for ay x such that x+ +4 <. x + Let s compute the it o the left: x+ +4 x + = + x = x. + 4 Hece, the series coverges for < x <. Now, let us check the boudary poits. If x =, the ( ) x + = +. Oe ca use the compariso or it compariso test, to show that this series diverges (it has the same divergece rate as the harmoic series). If x =, the ( ) x + = ( ) +. This is a alteratig series, where the absolute values of the terms decrease mootoically ad go to zero. Hece, by the alteratig series test, this series coverges. Fially, we arrive to the coclusio that the iterval of covergece of the give series is (, ].
b) ( ) x By the ratio test, the give power series will coverge for ay x such that c) + + x + x <. To solve this iequality, let s compute the it o the left: + + x + x = x = + x = x. + Hece, the give power series will coverge whe x < or, equivaletly, < x <. Let s check the boudary poits. Whe x =, we have ( ) x =. This is a p-series with p = 2 <. Hece, it diverges. Whe x =, we have ( ) x = ( ). This is a alteratig series, where the absolute values of terms decrease mootoically ad go to zero. Hece, it coverges. Fially, the iterval of covergece of the give series is (, ].! x2 d) x such that We compute By the ratio test, the give power series would coverge as soo as we take + (+)! x2+2! x2 Hece, the series coverges for ay x. 20 x + (+)! x2+2 <! x2 = x = 0 <. + By the ratio test, the series would coverge as soo as we take x such that We compute the it o the left: 20 + x + 20 x <. 20 + x + 20 x = 20 + x = x 20 ( + ) 4 ()
To compute ( + ), let s multiply ad divide the differece + by the cojugate expressio + + : ( + ) = ( + ) + + = + + = 0. Returig to (), we get x 20 ( + ) = x 20 0 = x. Hece, the give series will coverge whe < x <. Let s check the boudary poits. Whe x =, the series becomes ( ) 20. The -th term of this series does t go to zero (it oscillates betwee ad ). Hece, by the -th term test, the series would diverge. Whe x =, the series becomes 20. The -th term of this series does t go to zero (it goes to ). Hece, by the -th term test, the series would diverge. 5