Tests for Convergence of Series. a n > 1 n. 0 < a n < 1 n 2. 0 < a n <.

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1 Tests for Covergece of Series ) Use the compari test to cofirm the statemets i the followig eercises.. 4 diverges, 4 3 diverges. Aswer: Let a / 3), for 4. Sice 3 <, we have / 3) > /, a >. The harmoic series 4 diverges, the compari test tells us that the series 4 3 al diverges.. coverges, + Aswer: Let a / + ). Sice + >, we have / + ) < /, 0 < a <. The series coverges, the compari test tells us that the series 3. coverges, e + al Aswer: Let a e /. Sice e <, for,we have e <, 0 < a <. The series coverges, the compari test tells us that the series e al ) Use the compari test to determie whether the series i the followig eercises coverge Aswer: Let a /3 + ). Sice 3 + > 3, we have /3 + ) < /3, 3) ) 0 < a <. 3 Thus we ca compare the series 3 + with the geometric series coverges sice /3 <, the compari test tells us that. 4 +e Aswer: Let a / 4 + e ). Sice 4 + e > 4, we have ) This geometric series al 4 + e < 4, 0 < a < Sice the p-series 4 coverges, the compari test tells us that the series l 4 +e al Aswer: Sice l for, we have / l /, the series diverges y compari with the harmoic series, /.

2 Aswer: Let a / 4 + ). Sice 4 + > 4, we have 4 + < 4, therefore a 4 + < 4, 0 < a <. Sice the p-series coverges, the compari test tells us that the series 5. si 3 + Aswer: We kow that si <, si < 3. Sice the p-series coverges, compari gives that si Aswer: Let a + )/ ). Sice < + + ), we have + > + + ). 4 + coverges al. Therefore, we ca compare the series + with the diverget harmoic series. The compari test tells us that + al diverges. 3) Use the ratio test to decide if the series i the followig eercises coverge or diverge.. )! Aswer: Sice a /)!, replacig y + gives a + / + )!. Thus a + +)! )! )! + )! )! + ) + ))! + ) + ), a + L lim lim + ) + ) 0. Sice L 0, the ratio test tells us that )!.!) )! Aswer: Sice a!) /)!, replacig y + gives a + + )!) / + )!. Thus, a + +)!) +)!!) )! + )!) + )! )!!). However, sice + )! + )! ad + )! + ) + ))!, we have a + + )!) )! + ) + ))!!) + ) + ) + ) + 4 +, a + L lim 4. Sice L <, the ratio test tells us that!) )!

3 3. )!!+)! Aswer: Sice a )!/! + )!), replacig y + gives a + + )!/ + )! + )!). Thus, a + +)! +)!+)! )!!+)! + )! + )! + )!! + )!. )! However, sice + )! + ) + )! ad + )! + ) + ))!, we have a + Sice L >, the ratio test tells us that + ) + ) + ) + ) a + L lim 4. )!!+)! diverges. + ), + 4. r!, r > 0 Aswer: Sice a /r!), replacig y + gives a + /r + + )!). Thus a + r + +)! r! r! r + + )! r + ), a + L lim r lim + 0. Sice L 0, the ratio test tells us that r! coverges for all r > e Aswer: Sice a /e ), replacig y + gives a + / + )e +. Thus a + e ) + )e + + e. Therefore +)e + e Sice L <, the ratio test tells us that a + L lim e <. e Aswer: Sice a / 3 + ), replacig y + gives a + + / + ) 3 + ). Thus a + + +) ) ) 3 +, a + L lim. Sice L > the ratio test tells us that the series diverges. 4) Use the itegral test to decide whether the followig series coverge or diverge.. 3 Aswer: We use the itegral test with f) / 3 to determie whether this series coverges or diverges. We determie whether the correspodig improper itegral d coverges or diverges: 3 d lim d lim 3 3 lim + ). Sice the itegral 3 d coverges, we coclude from the itegral test that the series 3

4 Aswer: We use the itegral test with f) / +) to determie whether this series coverges or diverges. We determie whether the correspodig improper itegral d coverges or diverges: + d lim + d lim + l + ) lim l + ) ) l. Sice the itegral + d diverges, we coclude from the itegral test that the series + diverges. e Aswer : We use the itegral test with f) /e to determie whether this series coverges or diverges. To do we determie whether the correspodig improper itegral d coverges or diverges: e d lim e d lim e e lim e + e ) e. Sice the itegral e d coverges, we coclude from the itegral test that the series e We ca al oserve that this is a geometric series with ratio /e <, ad hece it l ) Aswer: We use the itegral test with f) /l ) ) to determie whether this series coverges or diverges. We determie whether the correspodig improper itegral d coverges or diverges: l ) d lim d lim l ) l ) l lim l + ) l l. Sice the itegral l ) d coverges, we coclude from the itegral test that the series l ) 5) Use the alteratig series test to show that the followig series coverge.. ) Aswer: Let a /. The replacig y + we have a + / +. Sice + >, we have + <, hece a + < a. I additio, lim a 0 ) 0 coverges y the alteratig series test.. ) + Aswer: Let a / + ). The replacig y + gives a + / + 3). Sice + 3 > +, we have 0 < a < + a. We al have lim a ) ++ Therefore, the alteratig series test tells us that the series ) + Aswer: Let a / + + ) / + ). The replacig y + gives a + / + ). Sice + > +, we have + ) < + )

5 0 < a + < a. We al have lim a 0. Therefore, the alteratig series test tells us that the series ) ) e Aswer: Let a /e. The replacig y + we have a + /e +. Sice e + > e, we have e < + e, hece a + < a. I additio, lim a 0 ) e coverges y the alteratig series test. We ca al oserve that the series is geometric with ratio /e ca hece coverges sice <. 6) I the followig eercises determie whether the series is alutely coverget, coditioally coverget, or diverget.. ) Aswer: Both ) alutely coverget.. ) Aswer: The series ) multiple of the harmoic series. Thus ) 3. ) ) + Aswer: Sice ) ad ) are coverget geometric series. Thus ) is coverges y the alteratig series test. However is coditioally coverget. lim + ), diverges ecause it is a the th term a ) + ) does ot ted to zero as. Thus, the series ) + ) is diverget. 4. ) 4 +7 Aswer: The series ) 4 +7 coverges y the alteratig series test. Moreover, the series y compari with the coverget p-series. Thus ) is alutely coverget coverges 5. ) l Aswer: We first check alute covergece y decidig whether / l ) coverges y usig the itegral test. Sice d l lim d l lim ll)) lim ll)) ll))), ad sice this limit does ot eist, l diverges. We ow check coditioal covergece. The origial series is alteratig we check whether a + < a. Cosider a f), where f) / l ). Sice d d l ) l + ) l is egative for >, we kow that a is decreasig for. Thus, for a + Sice / l ) 0 as, we see that ) 6. ) arcta/) + ) l + ) < l a. l is coditioally coverget. Aswer: We first check alute covergece y decidig whether arcta/) Sice arcta is the agle etwee π/ ad π/, we have arcta/) < π/ for all. We compare arcta/) < π/, is a- ad coclude that sice π/) / coverges, arcta/) lutely coverget. Thus ) arcta/)

6 7) I the followig eercises use the limit compari test to determie whether the series coverges or diverges Aswer: We have Sice. + 3 Aswer: We have, y comparig to a 5 + )/3 ) 5 + / 3, a 5 + lim lim c 0. is a diverget harmoic series, the origial series diverges. ), y comparig to 3) a + )/3)) /3) + ) + ), a lim lim + e c 0. ) Sice 3) is a coverget geometric series, the origial series 3. cos ), y comparig to / Aswer: The th term is a cos/) ad we are takig /. We have a cos/) lim lim /. This limit is of the idetermiate form 0/0 we evaluate it usig l Hopital s rule. We have cos/) si/) / ) si/) lim / lim / 3 lim lim / 0 si. The limit compari test applies with c /. The p-series / coverges ecause p >. Therefore cos/)) al Aswer: The th term a / 4 7) ehaves like / 4 for large, we take / 4. We have a / 4 7) 4 lim lim / 4 lim 4 7. The limit compari test applies with c. The p-series / 4 coverges ecause p 4 >. Therefore / 4 7) al Aswer: The th term a )/ 4 ) ehaves like 3 / 4 / for large, we take /. We have a )/ 4 ) lim lim lim / 4. The limit compari test applies with c. The harmoic series / diverges. Thus ) / 4 ) al diverges Aswer: The th term a /3 ) ehaves like /3 for large, we take /3. We have a /3 ) lim lim /3 lim 3 3 lim. 3 The limit compari test applies with c. The geometric series /3 /3) Therefore /3 ) al

7 7. ) Aswer: The th term, a 4, ehaves like /4 ) for large, we take /4 ). We have a /4 ) lim lim /4 lim ) 4 4 lim /). The limit compari test applies with c. The series /4 ) coverges ecause it is a multiple of a p-series with p >. Therefore ) al Aswer: The th term a / + + ) ehaves like /3 ) for large, we take /3 ). We have a / + + ) lim lim /3 ) 3 lim lim + ) + / 3 lim + + / The limit compari test applies with c. The series /3 ) diverges ecause it is a multiple of a p-series with p / <. Therefore / + + ) al diverges. 8) Eplai why the itegral test caot e used to decide if the followig series coverge or diverge... Aswer: The itegral test requires that f), which is ot decreasig. e si Aswer: The itegral test requires that f) e si, which is ot positive, or is it decreasig. 9) Eplai why the compari test caot e used to decide if the followig series coverge or diverge... ) Aswer: The compari test requires that a ) / e positive. It is ot. si Aswer: The compari test requires that a si e positive for all. It is ot. 0) Eplai why the ratio test caot e used to decide if the followig series coverge or diverge.. ) Aswer: With a ), we have a + /a, ad lim a + /a, the test gives o iformatio.

8 . si Aswer: With a si, we have a + /a si + )/ si, which does ot have a limit as, the test does ot apply. ) Eplai why the alteratig series test caot e used to decide if the followig series coverge or diverge.. ) Aswer: The sequece a does ot satisfy either a + < a or lim a 0.. ) ) Aswer: The alteratig series test requires a / which is positive ad satisfies a + < a ut lim a 0. ) JAMBALAYA!!! Determie if the followig series coverge or diverge.. 8! Aswer: We use the ratio test with a 8!. Replacig y + gives a )! ad Thus a + 8+ / + )! 8 /! 8! + )! 8 +. a + 8 L lim lim + 0. Sice L <, the ratio test tells us that 8!. 3 Aswer: We use the ratio test with a 3. Replacig y + gives a + +)+ 3 + ad Thus a + + )+ )/3 + /3 + ). 3 a + + ) + /) L lim lim lim Sice L <, the ratio test tells us that e Aswer: The first few terms of the series may e writte + e + e + e 3 + ; this is a geometric series with a ad e /e. Sice <, the geometric series coverges to S e e e. 4. ta ) Aswer: We compare the series with the coverget series /. From the graph of ta, we see that ta < for 0, ta/) < for all. Thus ) ta <, the series coverges, sice / Alteratively, we try the itegral test. Sice the terms i the series are positive ad decreasig, we ca use the itegral test. We calculate the correspodig itegral usig the sustitutio w /: ) ta d lim ta ) d lim l cos ) lim l cos Sice the limit eists, the itegral coverges, the series ta /) )) ) lcos ) lcos ).

9 Aswer: We use the limit compari test with a Because a ehaves like 5 we take /. We have a 5 + ) lim lim By the limit compari test with c 5/) sice diverges, 6. ) al diverges. Aswer: Let a / 3. The replacig y + gives a + / 3 + ). Sice 3 + ) > 3, as, we have a + < a. I additio, lim a 0 the alteratig series test tells us that the series ) 3 7. si Aswer: Sice 0 si for all, we may e ale to compare with /. We have 0 si / / for all. So si / coverges y compari with the coverget series / ). Therefore si / ) al coverges, sice alute covergece implies covergece l Aswer: Sice 3 l 3 l, our series ehaves like the series / l. More precisely, for all, we have 0 l 3 l 3 l, 3 l diverges y compari with the diverget series. 9. +) 3 + Aswer: Let a + )/ 3 +. Sice ), we have a + ) a grows without oud as, therefore the series +) 3 + diverges.