STRATEGY FOR TESTING SERIES We ow have several ways of testig a series for covergece or divergece; the problem is to decide which test to use o which series. I this respect testig series is similar to itegratig fuctios. Agai there are o hard ad fast rules about which test to apply to a give series, but you may fid the followig advice of some use. It is ot wise to apply a list of the tests i a specific order util oe fially works. That would be a waste of time ad effort. Istead, as with itegratio, the mai strategy is to classify the series accordig to its form.. If the series is of the form p, it is a p-series, which we kow to be coverget if p ad diverget if p.. If the series has the form ar or ar, it is a geometric series, which coverges if r ad diverges if r. Some prelimiary algebraic maipulatio may be required to brig the series ito this form. 3. If the series has a form that is similar to a p-series or a geometric series, the oe of the compariso tests should be cosidered. I particular, if a is a ratioal fuctio or algebraic fuctio of (ivolvig roots of polyomials), the the series should be compared with a p-series. (The value of p should be chose as i Sectio 8.3 by keepig oly the highest powers of i the umerator ad deomiator.) The compariso tests apply oly to series with positive terms, but if a has some egative terms, the we ca apply the Compariso Test to a ad test for absolute covergece. 4. If you ca see at a glace that lim l a 0, the the Test for Divergece should be used. 5. If the series is of the form b or b, the the Alteratig Series Test is a obvious possibility. 6. Series that ivolve factorials or other products (icludig a costat raised to the th power) are ofte coveietly tested usig the Ratio Test. Bear i mid that a a l as l for all p-series ad therefore all ratioal or algebraic fuctios of. Thus the Ratio Test should ot be used for such series. 7. If, where x a f f x dx is easily evaluated, the the Itegral Test is effective (assumig the hypotheses of this test are satisfied). I the followig examples we do t work out all the details but simply idicate which tests should be used. EXAMPLE Sice a l 0 as l, we should use the Test for Divergece. EXAMPLE Property of Cegage Learig Not for reproductio s 3 3 3 4 Sice a is a algebraic fuctio of, we compare the give series with a p-series. The compariso series for the Limit Compariso Test is b, where b s 3 3 3 3 3 3 3 3 Thomso Brooks-Cole copyright 007 EXAMPLE 3 e Sice the itegral x dx is easily evaluated, we use the Itegral Test. The Ratio Test xe x also works.
STRATEGY FOR TESTING SERIES EXERCISES EXAMPLE 4 Sice the series is alteratig, we use the Alteratig Series Test. EXAMPLE 5 34 Test the series for covergece or divergece... 3. 4. 3 5. 6. 7. 8. sl 9. k e k 0... l 3. 3 4.! 5. A k 0 6. Click here for aswers. 3! 5 8 3 3 S 3 4 k k k! Sice the series ivolves k!, we use the Ratio Test. EXAMPLE 6 3 Sice the series is closely related to the geometric series 3, we use the Compariso Test. k Click here for solutios. cos k k! k! e 3 5 si 7. 8. 9. l 0. s.. 3. ta 4.! 5. 6. e k l k 7. 8. k 3 k 9. ta 30. s 3. 3. 3 k 4 k k k j 33. si 34. s e s k 5 s 3 5 cos 4 5 j l l sj j 5 Property of Cegage Learig Not for reproductio (s ) Thomso Brooks-Cole copyright 007
STRATEGY FOR TESTING SERIES 3 ANSWERS S Click here for solutios.. D 3. C 5. C 7. D 9. C. C 3. C 5. C 7. D 9. C. C 3. D 5. C 7. C 9. C 3. D 33. C Property of Cegage Learig Not for reproductio Thomso Brooks-Cole copyright 007
4 STRATEGY FOR TESTING SERIES SOLUTIONS. lim a Divergece. + / +/ = = 0, so the series diverges by the Test for +. If a = + ad b = a,the lim b + / +/ =,sotheseries + diverges by the Limit Compariso Test with the harmoic series. 3. + < for all, so + coverges by the Compariso Test with,ap-series that coverges because p =>. 4. Let b = +. The b =0,adb = b3 =,butb >b+ for 3 sice 6 ( ) ( x x + x ) (x )(x +) = = x +x + (x ) x + x (x + x) (x + x) = < 0 for x 3. Thus, (x + x) {b 3} is decreasig ad lim b =0,so ( ) coverges by the Alteratig Series Test. =3 + Hece, the full series ( ) also coverges. + 5. lim a+ a ( 3)+ 3 3(+) ( 3) + 3 3 3 3 3 = 3 <, sotheseries 3 8 ( 3) + is absolutely coverget by the Ratio Test. 3 6. lim a 3 +8 3 / +8 = 3 ( ) 8 <,so 3 coverges by the Root Test. +8 7. Let f(x) = x. The f is positive, cotiuous, ad decreasig o [, ), so we ca apply the Itegral Test. l x [ ] u =lx, Sice x l x dx = u / du =u / + C = l x + C, wefid du = dx/x dx t x l x dx [ t x l x ] t ( l x l t ) l =. Sice the itegral t t diverges, the give series l diverges. = k k! 8. (k +)! = k. Usig the Ratio Test, we get (k +)(k +) k= k= ( lim ak+ a k k+ (k +)(k +) (k +)(k +3) k k + ) =>, so the series diverges. k +3 Or: Use the Test for Divergece. 9. k e k k =. Usig the Ratio Test, we get ek k= k= [ ( ) ] lim ak+ (k +) ek k + e k+ k = k e e = <, so the series coverges. e a k Property of Cegage Learig Not for reproductio Thomso Brooks-Cole copyright 007 0. Let f(x) =x e x3. Thef is cotiuous ad positive o [, ),adf (x) = x( 3x3 ) < 0 for x,sof is e x3 [ decreasig o [, ) as well, ad we ca apply the Itegral Test. x e x3 dx e x3] t t 3 =,sothe 3e itegral coverges, ad hece, the series coverges.
STRATEGY FOR TESTING SERIES 5 Thomso Brooks-Cole copyright 007. b = l the Alteratig Series Test. > 0 for, {b} is decreasig, ad lim b =0, so the give series ( ) + = l coverges by. Let b =.Theb > 0, lim b =0,ad +5 b b + = +5 + + +6 = + 5, which is positive for 5, sothe ( +5)( + + 6) sequece {b } decreases from =5o. Hece, the give series ( ) coverges by the Alteratig +5 Series Test. 3. lim a+ a 3+ ( +) [! 3( +) ] ( +)! 3 + =3 lim =0<, sotheseries ( +) 3 coverges by the Ratio Test.! 4. The series si diverges by the Test for Divergece sice lim si does ot exist. 5. lim a+ a ( +)! 5 8 (3 +) 5 8 (3 +)[3( +)+]! + 3 +5 = 3 <! so the series coverges by the Ratio Test. 5 8 (3 +) =0 6. Usig the Limit Compariso Test with a = + 3 + ad b =,wehave ( a + lim b 3 + ) 3 + 3 + +/ +/ => 0. Sice 3 b is the diverget harmoic series, a is also diverget. 7. lim / = 0 =,so lim ( ) / does ot exist ad the series ( ) / diverges by the Test for Divergece. 8. b = for. {b } is a decreasig sequece of positive umbers ad lim b =0,so = coverges by the Alteratig Series Test. 9. Let f(x) = l x. The f (x) = l x x x 3/ ( ) < 0 whe l x> or x>e,so l is decreasig for >e. l / By l Hospital s Rule, lim / ( ) =0,sotheseries ( ) l coverges by the Alteratig Series Test. 0. lim ak+ k +6 5 k+ k +5 = 5 lim k +6 k +5 = 5 <,sotheseries k +5 coverges by the Ratio a k Test. ( ) (. = 4 Root Test.. 3 + +5 < Property of Cegage Learig Not for reproductio ). lim a 3 + +5 < = 3 for,so with the coverget p-series / (p => ). k= 4 =0<, so the give series is absolutely coverget by the coverges by the Compariso Test 3 + +5
6 STRATEGY FOR TESTING SERIES ( ) 3. Usig the Limit Compariso Test with a =ta ad b =,wehave a ta(/) ta(/x) H sec (/x) ( /x ) lim b / x /x x /x x sec (/x) = => 0. Sice b is the diverget harmoic series, a is also diverget. cos(/) 4. +4 < +4 < ad sice coverges (p => ), Compariso Test. 5. Use the Ratio Test. lim a+ a ( +)! e e (+)!! coverges. e ( +)! e e ++! cos(/) coverges absolutely by the +4 + =0<,so e+ ( ) ( 6. lim a+ a a + + + 5 +/ +/ ) = a 5 + + +/ 5 5 <,so + coverges by the Ratio Test. 5 [ l x 7. dx l x x t x ] t (usig itegratio by parts) H =. So l coverges by the Itegral Test, x k l k ad sice (k +) 3 < k l k = l k k 3 k,thegiveseries k l k 3 coverges by the Compariso Test. k= (k +) { } 8. Sice is a decreasig sequece, e / e / = e for all,ad e coverges (p => ), so e / coverges by the Compariso Test. (Or use the Itegral Test.) 9. 0 < ta < π/ 3/. π/ 3/ = π 3/ which is a coverget p-series (p = 3 3/ ta coverges by the Compariso Test. 3/ > ), so x 30. Let f(x) = x +5. The f(x) is cotiuous ad positive o [, ), ad sice f 5 x (x) = x (x +5) < 0 for x>5, f(x) is evetually decreasig, so we ca use the Alteratig Series Test. lim +5 / +5 =0,sotheseries j ( ) j / j +5 coverges. ( ) 3. lim ak 3 k +4 =[divide by (5/4) k k ( ) k 3 5 k 4k ] lim = sice lim =0ad lim =. (3/4) k + 4 4 Thus, diverges by the Test for Divergece. 3 k +4k k= j= Property of Cegage Learig Not for reproductio Thomso Brooks-Cole copyright 007 3. Note that (l ) l = ( e l l ) l ( ) = e l l l = l l ad l l as,sol l >for sufficietly large. For these we have (l ) l >,so (l ) < l. Sice = coverges (p => ), so does by the Compariso Test. l = (l ) 33. Let a = si(/) ad b = a si(/) si(/). The lim => 0,so coverges by b / limit compariso with the coverget p-series (p =3/ > ). 3/
STRATEGY FOR TESTING SERIES 7 34. Use the Limit Compariso Test with a = a / ad b =/. The lim b / /x H /x l ( /x ) x /x x /x x (/x l ) = l = l > 0. So sice b diverges (harmoic series), so does ( ). Alterate Solutio: = [ratioalize the umerator] ( )/ + ( )/ + ( 3)/ + + / +, ad sice = diverges (harmoic series), so does ( ) by the Compariso Test. Property of Cegage Learig Not for reproductio Thomso Brooks-Cole copyright 007