MAC 3 ) Note that 6 3 + INFINITE SERIES PROBLEMS-SOLUTIONS 6 3 + <, for all, 3 3, ad coverges by the p-series test. Therefore, coverges by the Compariso Test. 6 3 + ) 3) +3 8 +3 3 8 64 ad ( 8 ) commo ratio r 64. Sice r <, +3 8 is a geometric series with coverges by the Geometric Series Test. Therefore, coverges (o-zero costat multipliers do ot affect covergece or divergece). ( + ) 4 The Ratio Test would be a good cadidate for a covergece test because the th term! cotais a factorial expressio. u + u ( + ) 4 + ( + )! + ( + ) 4 ( + )!! u ρ lim + 4 + lim u + Therefore, ( ) 4 + ( ) 4 lim ( ) 3 ( + ) 4 ( ) ( )! 4 + ( + ) 4 + ( ) ( )! ( + )! 4 + + + 4 + 4 3 + 3 + 3 + 4 lim 4 lim 3 0 coverges by the ratio test because ρ <.! ( ) ( ) 3 + 4 + + 4) 4 3 3 >,, ad diverges by the p-series test (p < ). / / 4 We coclude that by the Compariso Test, diverges. 4 3 ) 9 7 3 3 9 diverges. 7 3 3 9 7 3 3 > 7 3 3 > 3 ad diverges. Therefore, by the Compariso Test,
! 6) This is a series for the Ratio Test (see #3 for the reasoig). ( )! u + u ( + ) + ( + )! ( + )! ( + ) + ( + )! ( )!! ( + )!! + ( )! + + 4 + 6 + + lim 4 + + + 4 + 6 + + 4 e ad 4 e < ( )( + ) + ( + ) ( + ) u ρ lim + lim u + + 4 + 6 + + Therefore,! coverges by the Ratio Test. ( )! ( ) 7) + 6 + This is a alteratig series. Try the Alteratig Series test (remider: the other covergece tests such as the Compariso Test ad Ratio Test apply oly to positive term series.) Now chec if the two coditios for the Alteratig Series Test hold: ) u + < u, for all? u + ( + ) + 6( + ) + + 8 + 8 < + 6 + u, ad ) lim u lim + 6 + 0. Sice both statemets are true, the series coverges by the Alteratig Series Test. + 8) π e. This is a geometric series with commo ratio r π e ad r π e >. Thus, π e 0 diverges by the Geometric Series Test. 0 + 9) +. Note that for all >, + 4 + 4 4 + 4 < 6 4 + 4, 6 4 + 4 6 3 + 4 < 6 6 3 ad 6 is a + coverget p-series. Therefore, by the Compariso Test coverges. 4 + 4
0) 4 + cos 4 + cos 4 + ad is a coverget p-series. Therefore, from the compariso test, we coclude that 4 + cos coverges. ) e. Write the th term as ad use the Ratio Test. e u + + u e e + e + e + ρ lim coverges by the Ratio Test. u + u e lim + e ( ) e <. Sice ρ <, the series ) ( ) 3 l. This is ot a geometric series, the Ratio Test will be icoclusive (verify that), ad the Compariso will be impractical because it will be difficult to fid a appropriate series to compare to. I a case such as this, perhaps the itegral test could be applied. Lettig f ( x) checig to see if f is positive, o-icreasig, ad cotiuous o, hypotheses of the Itegral Test), we have ( ) 3 dx x l x 0 + l u du lim 3 c l ul x ( ) ( l ) < c l u 3 du lim c u c l lim c c + l The improper itegral coverges. Thus, by the Itegral Test, the series ( ), ad the 3 x( l x) ) (those three coditios are the lim c c + l ( ) coverges. l ( ) 3 3) Observe that lim u 9 lim 9 0. Therefore, the series diverges by the th term 9 Divergece Test. 4) cosπ cosπ if is eve ad cos π if is odd. Therefore, + cosπ is the alteratig + series ( ). Note that the absolute value of the th term is + ( ) + + ad the series + ca be show to coverge usig the itegral test or direct compariso with the coverget p-series.
cosπ The series of the absolute value terms coverges. Therefore, the series is + absolutely coverget. ) ( ) ( ) 4 4 ( ) ( ). The series of the absolute value terms is ad / ( ) is a diverget p-series (p < ). Thus, ( ) diverges which meas that 4 / 4 / is ot absolutely coverget. We ow apply the alteratig series test to ) u + < u : ( ) < 4 4 + Yes ) lim u 0 : lim 4 0 ( ). 4 ( ) Therefore, coverges by the Alteratig Series Test. We showed that the series of absolute 4 ( ) value terms diverges but the series itself is coverget. We the coclude that the series is 4 coditioally coverget. 6) 3 Divergece Test. y lim x x x Noticig that the limit of the th term a idetermiate form of type, try that th term l y x l liml y lim x l l x lim x x lim y e limu 0 Yes lim x x x Sice the limit of the th term is ot zero, the series Test. 3 diverges by the th term Divergece 7) ta Try the Itegral Test. First verify the hypotheses of the test; that is, show that + ( ) ta x f x is a positive, cotiuous, ad o-icreasig fuctio o the iterval, + x ).
ta π / x dx u du + x u π / 4 π / π / 4 uta x du + x dx Also, sice ta x > 0 o ( 0,), absolute value terms is coverget ad so 3π 3 <. Sice the improper itegral coverges, so does the series. ta ta, for all positive itegers. Thus, the series of + + ta is absolutely coverget. + 8) 4 + 6 9 4 + 6 4 9 + 3. 9 4 9 ad 3 are each coverget geometric series (why?). We ca the say that 4 + 6 coverges (See Theorem 9.4.3) ad sice the series is 9 a positive term series, it is cocluded that the series coverges absolutely. Absolute covergece ad covergece have the same meaig for positive term series. 9) ( ) Try usig the Ratio Test for Absolute Covergece. u + u ( ) + + ( ) ( ) + + ( ) ( ) + + ρ lim u + u lim. Sice ρ >, the + series diverges. 0). The geometric series has a commo ratio of r. r < the series coverges. The sum of the series is ) S ST term r 3 3 + 0 3 8 + 3 9 3 3 0 6 8 9 6 3 r 8 < the series coverges. 9 The sum is S ST term r 8 7 8 9 8 7 9 8 7 9 8 3
( ) ) 0.9999 S ST term r ( ) r 0.9999 0.9999 < the series coverges. The sum is ( ) ( ( ) 0.9999 ) 0.9999 0.9999 + 9999 0000 9999 0000 9999 0000 3) sec 4 r sec 4 the series diverges 9999 99990000 9998000 99990000 ( ) ( ) 4) log log / log r < The series coverges. 0 first term The sum is S r Decide whether the give statemet is true or false. Explai your reasoig. ) If a ad b TRUE: Suppose that { s } are coverget, the so is a a ad b ( ) + b are both coverget. The the sequeces of th partial sums ad { t } both coverge. If their limits are L ad M, the the sequece s + t { } coverges ad its limit is L + M (by Thm 9..3). But s + t ( a + b ). Therefore, ( a + b ) is coverget. 6) If ( a + b ) is coverget, the so are a ad FALSE: A couterexamples is which is coverget but a a ad b ad b. b both diverge. { } is the sequece of th partial sums of. ( a + b ) 0
7) If lim a 0, the a is coverget. FALSE: Couterexamples iclude all p-series where 0 < p. What is true is that if a is coverget, the lim a 0, which is the th term test for divergece. The statemet give i the problem is the coverse of the th term test. The coverse statemet q p of q p may or may ot be true. 8) Oe ca prove that FALSE: tha ad coverges by applyig the Compariso Test usig the series 0. 0. >. is larger tha 0. ad sice. coverges, a larger series could still coverge of diverge. However, for all >, 0. < 0. 0. coverges. Thus, by the Compariso Test, is coverget. 0. 9) is a coverget series for all x i the domai of sec x. sec x FALSE The rage of sec x is (,, ). If x π, where is a iteger, the sec x ± sec x ad diverges. sec x 30) The Ratio Test ca be used to prove the p-series test. FALSE: If a, the p a ρ lim + ( + ) p lim lim a ( + ) lim p ( + ) p lim + p lim +. If p + 0 ρ, the ratio test i icoclusive. I that case, aother test must be tried. I fact, the p-series test is a direct cosequece of the Itegral Test.