Review for Test 3 Math 1552, Integral Calculus Sections 8.8,

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1 Review for Test 3 Math 55, Itegral Calculus Sectios 8.8, Termiology review: complete the followig statemets. (a) A geometric series has the geeral form k=0 rk.theseriescovergeswhe r is less tha oe ad diverges whe r is greater tha or equal to oe. (b) A p-series has the geeral form k= k p. The series coverges whe p is greater tha oe ad diverges whe p is less tha or equal to oe. To show these results, we ca use the itegral test. (c) The harmoic series diverges ad telescopig series coverge. (d) If you wat to show a series coverges, compare it to a larger series that also coverges. If you wat to show a series diverges, compare it to a smaller series that also diverges. (e) If the direct compariso test does ot have the correct iequality, you ca istead use the it compariso test. I this test, if the it is a fiite, positive umber (ot equal to 0) the both series coverge or both series diverge. (f) I the ratio ad root tests, the series will coverge if the it is less tha ad diverge if the it is greater tha. If the it equals, the the test is INCONCLUSIVE. (g) If a =0,thewhatdowekowabouttheseries k a k? absolutely NOTHING! (h) A itegral is improper if either oe or both its of itegratio are ifiite, or the fuctio has a vertical asymptote o the iterval a, b]. (i) A sequece is a ifiite list of terms. Asequece{a } coverges if: the it of the terms exists ad is fiite as.

2 (j) The smallest value that is greater tha or equal to every term i a sequece is called the least upper boud (l.u.b.). The largest value that is less tha or equal to every term i the sequece is called the greatest lower boud (g.l.b). Ifbothofthesevaluesarefiite, the we say the sequece is bouded. (k) A sequece is called mootoic if the terms are icreasig, mootoicallyicreasig, decreasig, ormootoicallydecreasig. If a sequece is both mootoic ad bouded, the we kow it must coverge.. Sum the series Solutio: 4 k 7 k = = =7 =7 ( 4 k 7 k ) 6 k 7 k 7 3. Fid the sum of the series ( ) k k= 4 k 7 k. 7 k 7 k 7 ( ) k 7 ] 7 7 (k )(k +3). ] = Solutio: Usigpartialfractiosothetelescopigseries,weseethat: (k )(k +3) = /4 k /4 k +3, so the series becomes: (k )(k +3) = 4 k ] k +3 k= k= = ( 4 5 )+( 3 7 )+( 5 9 )+( 7 ] )+... = + ] =

3 4. Determie whether the followig series coverge or diverge. Justify your aswers usig the tests we discussed i class. (a) k= e k (+4e k ) 3. Solutio: Use the Itegral Test. We first check the coditios to apply this test. Sice k, e k > 0sothefuctioispositiveadcotiuous. Lettigf(x) = fid the derivative: f (x) = ex ( 8.8e x ) ( + 4e x ) 4.. e x (+4e x ) 3.,weca Note that f (x) < 0whex, so the fuctio is decreasig. Now evaluate the itegral: e x b dx = ( + 4e x ) 3. b = 4 b = 4 b x=b = 8.8 b e x ( + 4e x ) 3. x= u du (u 3. =+4ex ) ] b.( + 4e x ). ( + 4e b ). ( + 4e). = 8.8 (0 ( + 4e). )= 8.8( + 4e).. Sice the itegral coverges, the series also coverges. (b) ( k 5 k ) k Solutio: UsetheRootTest: ( 5 ) ] / ( = 5 = e ) 5. ] Sice the it is e 5 <, the series coverges by the Root Test. (c) k+ k k= k! Solutio: UsetheRatioTest: a + a ( +) + = ( +)!! + ( +)! = ( +)! ( +) = =0.

4 Sice the it is 0, which is less tha, the series coverges by the ratio test. (d) k= k Solutio: Recalltheformula: i= i = (+),sowecarewritethisseriesas: k= k = k(k +). The series is telescopig, so it coverges ad we ca fid its sum usig partial fractios: k= k(k +) = k= k= k ] =. k + Alterately, we ca compare to the series k= k.sicek +>k,weseethat k+ < k ad thus k(k+) < k. Sice k= k coverges (p-series with p => ), k= k = k= k coverges, so the series k= k = k= k(k+) also coverges by the Basic Compariso Test.

5 5. For each sequece, determie: (i) the l.u.b. ad g.l.b.; (ii) whether the sequece is mootoic; { (iii) whether the sequece coverges or diverges, ad the it if it is coverget. ( ) } 3 (a) + Solutio: First,let sfidtheit: so the sequece coverges. ( ) ( )] 3 3 = ( ) + + = ( ) 3 e = e 6, Sice the sequece is coverget, it is bouded. Writig out a few terms, we ca see the sequece is decreasig (take the derivative to cofirm). Therefore, it is mootoic, ad l.u.b.= (the first term), g.l.b.= (it). 7 e 6 { } (b) cos(π) 4 Solutio: Notethatsice cos(π), we see that: 4 cos(π) 4 4. Sice 4 = 4 =0,bytheSadwichTheorem,wealsohave cos(π) 4 =0,sothesequececoverges. Sice the terms alterate, the sequece is bouded, but it is ot mootoic. We fid that l.u.b.= 6 (first positive term) ad g.l.b.= (first egative term). 4 { } (c) ( ) + +4 Solutio: Note that + +4 =,sowhe is odd, ( ) = adtheterms approach, but whe is eve, ( ) =+adthetermsapproach. Astheits for differet values of are ot equal, there is o it, ad thus the sequece diverges. Sice we have its for values where is positive ad egative, the sequece is bouded, ad those its are the bouds. Thus, l.u.b.= ad g.l.b=-. The sequece alterates sigs, so it is ot mootoic.

6 6. Determie if the improper itergral coverges or diverges. If it coverges, evaluate the itegral. (a) x dx. (x ) 3/ Solutio: Letu = x, the du =xdx, sowehave: Thus, the itegral coverges. x=b b x= u 3/ du = b x b = b b + ] 3 = 3. (b) Solutio: Usigpartialfractios: so the itegral diverges. 0 0 dx x 5x +6. dx c x 5x +6 = c 0 x 3 ] dx x = l x 3 c x c 0 = l c 3 c c l 3 ] =,