Math 2414 Homework Set 7 Solutions 10 Points
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1 Mah Homework Se 7 Soluios 0 Pois #. ( ps) Firs verify ha we ca use he iegral es. The erms are clearly posiive (he epoeial is always posiive ad + is posiive if >, which i is i his case). For decreasig we ll eed a lile Calc I. ( ) ( ) f e f e I ll leave i o you o verify ha he derivaive is egaive if > ad so he fucio ad hece he series erms are decreasig. We ca he apply he iegral es ad his requires iegraio by pars. u dv e d du d v e ( ) e d ( ) e + e d ( ) e e e ( ) ( ) e d lim e d lim e lim e + e e The iegral coverges ad so he series also coverges by he Iegral Tes. #. ( ps) Because he leadig coefficies of he polyomials are boh posiive we ca see ha he series erms will eveually be posiive. Also from all he mius ad plus sigs here s o way he Compariso Tes ca be used here. However, by lookig a he larges powers of i looks like his series should behave like so we ca use his wih he Limi Compariso Tes c lim lim 0 < < So, by he Limi Compariso Tes boh series will have he same covergece ad we kow ha is a coverge series so our origial series mus coverge. #5. ( ps) The erms of his series are all posiive ad i looks like i should behave like which we kow is coverge ad so we ca guess ha his series is coverge so we ll wa o fid a larger series ha we kow is coverge. + si ( ) + si ( ) + 0 b/c 0 si So, we kow is a coverge series ad is erms are larger ha he origial series erms so by he Compariso Tes he origial series coverges.
2 Mah Homework Se 7 Soluios 0 Pois #8. ( ps) We ll firs eed o do a quick rewrie of he series o ge io proper form. ( ) ( ) ( + 7) ( ) ( + 7) ( ) ( ) Noe ha ( ) oly chages he sig ad so ca be i eiher he umeraor or deomiaor wihou chagig he acual value. Now, b > 0 lim b lim The limi is fairly clearly zero ad if we icrease he oly hig ha happes is he deomiaor ges larger ad so he b mus ge smaller. Therefore, he b decrease ad so by he Aleraig Series Tes he series mus coverge. #. ( ps) This looks like a Raio Tes so, L lim lim lim < So, by he Raio Tes he series coverges. ( 9 + )( ) No Graded #. Wih his series we do really eed o use he iegral es if we do a quick rewrie So, because p < we kow by he p-series es ha his series diverges. #. Firs verify ha we ca use he iegral es. The erms are clearly posiive (he deomiaor is always posiive ad he umeraor is posiive if 0, which i is i his case). For decreasig we ll eed a lile Calc I. ( ) f ( ) f ( ) We ca see ha for >. he derivaive is egaive ad so he fucio ad hece he series erms are decreasig eveually. We ca he apply he iegral es. Here s he iegral es work, ( ) d lim lim d l + 8 lim l + 8 l The iegral diverges ad so he series also diverges by he Iegral Tes.
3 Mah Homework Se 7 Soluios 0 Pois #6. All he erms i his series are posiive ad oice ha e < ad so i looks like his series should behave like series ha we kow is coverge. which we kow is diverge ad so we ca guess diverge ad we ll wa smaller + b/c > + > e e > b/c e < So, we kow is a diverge series ad is erms are larger ha he origial series erms so by he Compariso Tes he origial series diverges. #7. We ca see ha, b > 0 lim b lim Fially, if we icrease he oly hig ha happes is he deomiaor ges larger ad so he b mus ge smaller. Therefore, he b decrease ad so by he Aleraig Series Tes he series mus coverge. #9. I his case we have, l ( ) l b lim b lim lim Usig L Hospials Rule we ca see ha he limi is zero. For he decreasig we ll eed a lile Calc I. l ( ) l ( ) f ( ) f ( ) 5 5 So, we ca see ha he derivaive will be egaive provided > e.59 ad so he fucio ad hece b will be eveually decreasig. Therefore, by he Aleraig Series Tes he series coverges. #0. This looks like a Raio Tes so, lim ( + 5)( + )( + ) + 5!! !! L lim! +!! +! lim >
4 Mah Homework Se 7 Soluios 0 Pois So, by he Raio Tes he series diverges. #. This looks like a Roo Tes so, L lim 5 + lim 5 + lim 0 < + So, by he Roo Tes he series coverges. #. This looks like a Raio Tes so, L lim lim lim > So, by he Roo Tes he series diverges. #. This looks like a Raio Tes so, + ( ) L lim lim So, he Raio Tes wo work. This is really se up o do he Roo Tes however if we did we d also ge L so eiher will work i his case. This looks like a aleraig series however we ca also see ha he erms wo go o zero i he limi so le s jus hi his wih a Divergece Tes. ( ) lim lim ( ) lim lim ( ) does' eis So, by he Divergece Tes his series diverges. #5. lim + + 0! So, by he Divergece Tes his series diverges. #6. This looks like a iegral es problem. The erms are clearly posiive ad, f ( ) e f ( ) e ( ) e We ca see ha he derivaive will be egaive for > ad so he series erms will be decreasig as well for ha rage. I ll leave he iegraio deails o you o verify. ( ) e d lim d lim lim e e + e + e e The iegral is coverge ad so by he Iegral Tes he series is also coverge. Noe ha you could also have doe he Raio Tes o his oe ad ha probably would have bee a lile quicker. #7. This looks like he Raio Tes will be easies.
5 Mah Homework Se 7 Soluios 0 Pois ( ) L lim lim lim 6 > So, by he Raio Tes he series diverges. #8. This looks like a compariso es ad from larges powers of i looks like i should behave like 6 ad 6 0 will diverge by he p-series es so le s guess diverge ad look for a larger series ha we kow diverges. csc ( ) csc ( ) 0 si csc < 6 Now, as oed above diverges ad is erms are smaller ha he erms i he origial series so by 6 0 he Compariso Tes his series diverges. #9. This looks like a aleraig series wih, b lim b lim Also, icreasig oly icreases he deomiaor i he b ad so he b are a decreasig sequece so by he Aleraig Series Tes he series coverges.
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