Practice Test Problems for Test IV, with Solutios Dr. Holmes May, 2008 The exam will cover sectios 8.2 (revisited) to 8.8. The Taylor remaider formula from 8.9 will ot be o this test. The fact that sums, products, itegrals, atiderivatives of Taylor series are also Taylor series is i 8.9 ad you are expected to kow it for this test (though I do t promise to ask about it; there is plety to do ad the test will ivolve a lot of compressio). REVISED: o 8.9 o this test at all. This practice test cotais problems similar i spirit to those that will be o your test. I m ot worryig about how log it is, though (ad I will as much as I ca cotrol that o the real test). Be aware that actual problems o the test may cover more tha oe of these issues; otice how log the practice documet is! For example, the Ratio ad Root tests might appear oly i tests for radius of covergece of power series.
. 8.2 (geometric series) (a) Evaluate the series or explai why it diverges: =0 2 3 = 2 3 + 2 9 + 2 27... this is a geometric series with first term a = 2, (ot 2! Notice that the first value of is 0, ot!) ad commo ratio 3 r = < (it coverges sice r < ). The sum is 2 = 3. 3 3 (b) (this might also be viewed as 8.7) Write a geometric series which 4 coverges to for suitable values of x. Explai what the rage 4 x 2 of values of x is for which it coverges. Hit: start by dividig the top ad bottom of the expressio by 4; the it will be clearer how to apply the formula! 4 = which is 4 x 2 x2 4 a r for this has first term, commo ratio x2 4. x2 for a =, r =, so a geometric series 4 So the series is + x2 4 + ( x2 4 )2 + ( x2 4 )3... = + x2 4 + x4 4 2 + x6 4 3... or i=0 x 2 4. It coverges if the commo ratio is less tha, that is if x2 4 <, that is x 2 < 4, that is 2 < x < 2. 2
2. 8.3 (a) Use the Itegral Test to determie whether = 2 coverges or diverges. Do ot use aythig else (yes, its a p-series, ad the Itegral Test is how we showed the results about p-series i the first place: do t say aythig about p-series, just do the Itegral Test. ) Be sure to state the facts about the fuctio x 2 which esure that the Itegral Test applies! The fuctio is positive, decreasig, ad approaches zero as x x 2 approaches ifiity, so the Itegral Test does apply. This meas that the give series coverges if ad oly if the improper itegral coverges. I ask you to actually evaluate it: this is lim b [2 x] b or lim b [2 b 2] = : sice the improper itegral diverges, the series also diverges. (b) How may terms of the series = must be added to esure that the partial sum is withi 0 6 of the exact value of the series? The upper boud o the error i N = 2 2 as a estimate is 3 N x 2 dx
. Evaluate this ad you will fid that it is lim b [ ( )] =. b N N < iff N > 0 6 : to get error less tha oe oe millioth, you N 0 6 eed to add more tha oe millio terms. This series does ot coverge very fast! 4
3. 8.4 For each series, determie whether it coverges or diverges by compariso with a series whose covergece or divergece behavior is kow (you should quickly idicate the status of the kow series as geometric, p-series with p > or p : you must also state whether the kow series is coverget or diverget). State the iequalities you use for compariso (ot as iequalities betwee whole series whose covergece behavior you do t kow yet!). Of course you eed to give a fial aswer: does the give series coverge or diverge? You might wat to use the limit compariso test o oe or more of these. (a) (b) e < e = +e 2 e 2 e The series =0 e + e 2 =0 coverges because it is a geometric series with r = <. So the e origial series coverges by compariso because it is term by term less tha this coverget geometric series. =0 e 2 3 + 2 2 should be very close to = for large so we wat to compare this with the diverget series (p ). Ufortuately, = 3 + 3 2 <, so direct compariso doest work. Limit compariso 3 + does, though: lim 2 3 + ( 2 ) = lim 3 + = lim 3 3 + = 5
so our series coverges if ad oly if = coverges: we kow that the secod series diverges, so the first does as well by the limit compariso test. 6
4. 8.5 Use the Ratio Test or the Root Test to settle the covergece behavior of each of these series. Show all work!!! O oe of them the Ratio ad Root Tests will be icoclusive: say so, ad explai whether the series coverges or diverges (usig some other fact about it). If you use special limits, please idicate explicitly which special limits you used. (a) (b) (c) = Ratio Test or Root Test works. I ll use the Root Test: L = lim = lim e = < so the series coverges e e by the Root Test. Use the Ratio Test. lim + +2 + = e + ( + ) 2 = lim ( + 2) = The Ratio Test gives o iformatio sice L =. But this is easily see to diverge, sice lim the series diverges by the th term test. =! 3 + = 0, so Use the Ratio Test (the presece of a factorial meas we are ot likely to be able to use the Root Test). L = lim (+)! (+) 3! 3 ( + )! 3 = lim!( + ) = lim ( + ) 3 3 ( + ) = 3 7
Sice L diverges to ifiity, this series diverges by the Ratio Test. Also, this shows that the terms are strictly icreasig after a while, so it must diverge by the th term test as well. 8
5. 8.5b Give two series, both of which have covergece behavior which caot be decided by the Ratio Test or Root Test (L = ), oe of which coverges ad oe of which does ot coverge. (There are very familiar examples of this!) Show the work verifyig that the Ratio Test is icoclusive for each series ad idicate why oe coverges ad the other does ot. the obvious examples are ad = =. Both of 2 these have L = with either the ratio or root test; both are p-series, the first oe divergig ad the secod oe covergig. 9
6. 8.6 Two series are give. Oe satisfies the coditios of the Alteratig Series Test ad oe does ot. Explai i each case. Particularly i the case of the oe which does satisfy the coditios, you eed to spell out the three thigs that are true (o the other oe you eed oly poit out what fails). That the three coditios hold is obvious (i the part where AST works): you do ot eed to give verificatios of them, just spell out clearly what they are. For the oe which does satisfy the coditios of the AST, determie whether it coverges absolutely ad explai why it does or does ot. You do ot eed to say aythig about whether the other oe coverges or ot. Further, how may terms of the series that does satisfy AST do you eed to add to approximate the exact value of the series withi.0? (a) (b) The coditios = ( ) + + 0 (positive), (decreasig) ad lim = 0 (coverget to 0) o the terms which are alterately added ad subtracted are all obviously true, so this coverges by the AST. It coverges oly coditioally because = ( ) + = = is a diverget p-series (p = ). The error i the sum of the 2 first N terms of a alteratig series is bouded by the (N +)-th term: N+ <.0 is true iff N + > 00 that is N + > 0000 or N > 9999. You eed to add at least 0000 terms to get error less tha.0. This series has rotte covergece behavior.... = + si() ( ) 3 0
This oe does ot satisfy the coditios of the test, as the terms si() which are alterately added ad subtracted are ot all positive. No further work is requested by the directios, though this 3 ca be show to coverge absolutely.
7. 8.7 Use the Ratio Test or Root Test to determie the radius of covergece ad iterval of covergece of each of the followig power series. Remember that you do eed to check separately the behavior of the edpoits. Remember whe usig the Ratio or Root Test that you are checkig for absolute covergece. (a) (b) = + x ( ) We use the Root Test to test for absolute covergece. L = lim x = lim x = x So the series will coverge absolutely if x < ad diverge if x >, by the Root Test. At x = ad x = we eed to check separately because the Root Test is icoclusive (L = ). coverges because it is a alteratig series. x = : = ( ) + x = : + = ( ) = = ( ) diverges because it is the result of multiplyig the diverget harmoic series by. So the radius of covergece is ad the iterval is < x. = Seeig a factorial, we thik the Ratio Test is more likely to help. x 2! L = lim x 2(+) (+)! x 2! x 2(+)! = lim x 2 ( + )! = lim x 2 + = 0 < Sice L = 0 < o matter what x is, this series coverges for all values of x. The radius of covergece is ad the iterval of ocvergece is (, ). 2
(c) = 2 x 2 Here we use the Root Test. L = lim 2 x 2 2x = lim ( ) = 2x 2 so L < whe 2x < that is x < that is < x <. For 2 2 2 these values of x the series coverges absolutely. If L = 2x > the the series diverges. We oly eed to check x = ad x =. 2 2 If x = we get the series 2 = 2 which is a coverget p-series ad if x = 2 we get ( ) = 2 which coverges absolutely (drop the mius sigs ad we have the p-series agai). So the radius of covergece is 2 ad the iterval is 2 x 2. 3
8. 8.7b Determie the first five terms of the power series for l( + x) by writig dow a series for (this is a 8.2 problem!) the itegratig +x term by term. What is the radius ad iterval of covergece for this series (do t forget to check edpoits!) +x is would be the value of a geometric series with a =, r = x, that x + x 2 x 3 + x 4 x 5... Itegratig this term by term gives x x2 2 + x3 3 x4 4... = = ( ) + x Applyig the Ratio or Root Test shows that the radius of covergece is (Im ot repeatig this, it is very much like oe of the problems above). We eed to check for covergece at x = ad x = : at x = it coverges coditioally by AST, while at x = it diverges because it is the harmoic series times mius oe. 4
9. 8.8 Compute the first five terms of the Maclauri series of (+x) 2. Show all work. f(x) = ( + x) 2 f (x) = ( 2)( + x) 3 f (x) = ( 2)( 3)( + x) 4 f (x) = ( 2)( 3)( 4)( + x) 5 f (x) = ( 2)( 3)( 4)( 5)( + x) 6 so f(0) = ( + 0) 2 = f (0) = ( 2)( + 0) 3 = 2 f (0) = ( 2)( 3)( + 0) 4 = 6 f (0) = ( 2)( 3)( 4)( + 0) 5 = 24 f (0) = ( 2)( 3)( 4)( 5)( + 0) 6 = 20 so the series is or equivaletly 2x + 6 2! x2 24 3! x3 + 20 4! x4... 2x + 3x 2 4x 3 + 5x 3... The obvious patter does cotiue to hold. Write dow the power series for ad multiply it by itself. Check (+x) that you get the same first few terms of a series for this way as (+x) 2 you got by computig the Taylor series. I m omittig this part: I didt put it o the test ad while it is easy to show i hadwritte form, it is very hard to typeset the solutio i LaTeX. Of course you get the same series as i the first part! 5