8.3. Click here for answers. Click here for solutions. THE INTEGRAL AND COMPARISON TESTS. n 3 n 2. 4 n 5 1. sn 1. is convergent or divergent.

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1 SECTION 8. THE INTEGRAL AND COMPARISON TESTS 8. THE INTEGRAL AND COMPARISON TESTS A Click here for aswers. S Click here for solutios.. Use the Itegral Test to determie whether the series is coverget or diverget. Determie whether the series is coverget or diverget s s e l arcta s s cos s si s arcta s s 5 5 s 0 4 Copyright 0, Cegage Learig. All rights reserved.

2 SECTION 8. THE INTEGRAL AND COMPARISON TESTS 8. ANSWERS E Click here for eercises. S Click here for solutios.. Diverget. Coverget 8. Coverges 9. D ive rg e s. Diverget 4. Diverget 0. D ive rg e s. Coverges 5. Coverget 6. Coverget. Coverges. Coverges 7. Diverget 8. Diverget 9. Coverget 0. Coverget. Coverget. Coverget. Coverget 4. Coverget 5. Coverget 6. Coverges 4. D ive rg e s 5. Coverges 6. Coverges 7. D ive rg e s 8. Coverges 9. Coverges 0. Coverges. Coverges. Coverges. Coverges 7. Coverges Copyright 0, Cegage Learig. All rights reserved.

3 SECTION 8. THE INTEGRAL AND COMPARISON TESTS 8. SOLUTIONS E Click here for eercises. Copyright 0, Cegage Learig. All rights reserved = 4. = The fuctio f () = is positive, cotiuous, ad 4 decreasig o [, ), so the Itegral Test applies. = 4 = lim b b 4 = lim [ b 4 l (4 )] b [ = lim l (4b ) l ] = b 4 4 so the improper itegral diverges, ad so does the series. ( ). =5 /.000 is a p-series, p =.000 >, soit coverges.. = 0.99 = ( ) = / 0.99 which diverges sice p =0.99 <. 4. =, which is a p-series, p = / <,soit = = diverges. ( 5. + ) = = + / =,bothof which are coverget p-series because > ad >,so ( + ) coverges by Theorem 8 i = Sectio ( 4) = is a p-series, p =>, soit =5 = coverges. 7. f () = is positive, cotiuous, ad + decreasig o [, ), so applyig the Itegral Test, [ ] t + = lim l ( +) = t = 8. Sice is diverget. + + is cotiuous, positive, ad decreasig o [0, ) we ca apply the Itegral [ ( )] t = lim l + + t [usig the substitutio u = +,so =(u ) du] ([ ( )] ) = lim t l t + ( l) t Now t l ( t + ) ( e ) t =l ad so t + [ ( )] lim t l t + = (usig l Hospital s Rule) t so both the itegral ad the origial series diverge. 9. f () = is positive, cotiuous, ad decreasig o [, ), so applyig the Itegral Test, ( / = + + / ) [ ( ) ] / t = lim l =l t + = coverges. 0. f () =e is cotiuous ad positive o [, ), ad sice f () =e ( ) < 0 for >, f is decreasig as well. Thus, we ca use the Itegral Test: [ e = lim e ] t =0 ( e ) = t e. Sice the itegral coverges, the series coverges.. f () = is positive ad cotiuous o [, ), adsice f () = l < 0 whe >.44, f is l evetually decreasig, so we ca apply the Itegral Itegratig by parts, we get ( = lim t sice lim t coverges. l = l + (l) [ + l ] t t =0by l Hospital s Rule, ad so t ) =. f () = is cotiuous, positive ad 4 + decreasig o [, ), so applyig the Itegral Test, [ ] t arcta 4 + = lim = π t 4 arcta <, so the series coverges.. f () = arcta is cotiuous ad positive o [, ). + f arcta () = ( + ) < 0 for >, sice arcta π > for. Sof is decreasig ad we ca use the Itegral arcta [ = lim + t (arcta )] t = (π/) so the series coverges. (π/4) = π

4 4 SECTION 8. THE INTEGRAL AND COMPARISON TESTS Copyright 0, Cegage Learig. All rights reserved. 4. f () = l is cotiuous ad positive for, ad f () = l < 0 for, sof is decreasig. [ l = lim l t ] t H (by parts) =. Thus, = l = = l coverges by the Itegral 5. f () = is cotiuous ad positive o [, ), + + ad f + () = ( < 0 for, sof is + +) decreasig ad we ca use the Itegral + + = ( +) = lim [arcta ( +)] t t = π arcta so the series coverges as well. + < sice + > for all, ad sice is a coverget p-series (p => ), = = coverges also by the Compariso < 4 ad coverges (geometric with 4 = r = 4 < ) sobythecomparisotest, coverges also. =. is a geometric series with = r = <, ad hece coverges, so also, by the Compariso > ad p = < ) so = = = 4 +5 coverges diverges (p-series with diverges by the Compariso ( +5 ) > =. 4 =0 geometric series ( r = 5 4 > )so the Compariso ( ) 5 is a diverget 4 = diverges by si = ad coverges / / (p = > )so = si = coverges by the Compariso.. ( +) <. = p-series (p => ) so Compariso ( +)( +) < = = = is a coverget = = coverges by the ( +) = ad sice / coverges (p = > ), so does / ( +)( +) by the Compariso 4. Use t he Limit Compar iso Test w ith a = ad b = ( +)( +) a lim = lim b ( +)( +) = lim (+/)(+/) => 0 so sice diverges, so does. ( +)( +) = = ( +) < ad is a coverget geometric = series ( r = < ), so Compariso = coverges by the ( +) +cos 4 sice cos. 4 is a geometric = series with r = < so it coverges, ad so +cos coverges by the Compariso = 5 5 > 5 = 5 (harmoic series) so does ( ) ad sice 5 = 5 +4 < = 5. / p-series (p = > ) so Compariso = = diverges 5 by the Compariso 5 = is a coverget / coverges by the 5 +4

5 SECTION 8. THE INTEGRAL AND COMPARISON TESTS 5 9. arcta 4 < π/ 4 ad π = arcta 4. Let a = ad b =. The a 4 + lim = lim b 4 + => 0. Sice is a = + coverget p-series (p => ), so is by the 4 + Limit Compariso = coverges (p =4> )so 4 coverges by the Compariso 0. Use the Limit Compariso Test with a = 4 ad b = : lim a b coverges (p => ), = lim 4 => 0. Sice = also coverges. 4 = = b. Let a = + ad b =. The a + lim = lim => 0. Sice b = coverget geometric series ( r = < ), coverges by the Limit Compariso = is a +. Use the Limit Compariso Test with a = 0 4 ad b = 4/. a lim b 0/ 7/ = lim = lim 0 4 => 0 so sice b coverges (p = 4 > ), so does = = 0 4. / 4 8 Copyright 0, Cegage Learig. All rights reserved.