Exam-3 Solutions, Math n + 2. Solution: First note that this series is a telescoping series. Let S n be the nth partial sum of the series.
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1 Exam-3 Solutios, Math Fid the sum of the followig series. [ l( + + ] l( Solutio: First ote that this series is a telescopig series. Let S be the th partial sum of the series. S l 3 l 3 4 S l 3 l l 3 4 l 4 5 l 3 l 4 5 S 3 l 3 l l 3 4 l l 4 5 l 5 6 l 3 l 5 6 S l 3 l l 3 4 l l l( + l( l( This gives S l 3 l(+ +3. Now, lim S l l( + lim l( + l 3 lim l 3 lim x l 3 lim x l 3 0 l l(x + x + 3 x + [where x R, i.e. x is a real umber] [Usig L Hospital] This gives us [ l( + + ] l( + l
2 . Use the compariso test or limit compariso test to determie which of the followig series are coverget: (I si ( + (II (III Solutio: The aswer is oly (II ad (III coverge. We first cosider si (+. Usig si ( +, we get si (+. The series diverges by p-test. Now compariso test implies that si (+ diverges. Next, we ivestigate ++. It looks like we wat to use compariso test But the problem is that it is ot that clear to come up with a icer series you wat to compare with. Therefore we go for limit compariso test. Let a We eed b which is simpler. We fid b by lookig at the domiatig factors i a which are i the umerator ad 4 i the deomiator. Hece, we take b 4. Now, a + + lim lim b lim lim /4 / 4 lim + / + / + / + / 4 We kow that the series coverges by p-test. Limit Compariso Test implies that ++ coverges too Fially, ote that. The series is a geometric series with commo ratio ad hece coverges. Now by compariso test implies coverges.
3 3. Cosider the followig series (I ( (II l( (III 3 + (! Which of the followig statemets is true? (a Oly (I ad (III coverge. (b Oly(III coverges. (c Oly (I ad (II coverge. (dall three coverge. (e All three diverge. Solutio: Here the correct aswer is oly (I ad (III coverge. Let b. The b > 0, b is decreasig ad lim b 0. Hece by the alteratig series test the series ( coverges. Note that for all, because l( for all > 0. I other l( l( words, all of the terms i the sequece{ l( } are greater tha. This meas that lim 0. By divergece test, l( diverges. l( For 3 + we will use ratio test. Let a (! 3+, the a (! Usig the (+! fact that ( +! ( +! we get a + a 3. Now, lim which is + less tha. Applyig ratio test we see that 3 + (! is coverget. 4. Cosider the followig series (I ( +! e (II ( +. + Which of the followig statemets is true? (a They both diverge. (b They both coverge. (c (I coverges ad (II diverges. (d (I diverges ad (II coverges. (e Decidig whether these series coverge or diverge is beyod the scope of the methods taught i this course.
4 Solutio: The correct aswer is they both diverge. For (+! e we will use ratio test. Let a (+! e, the a + (+! (+ e +. Now, a + a ( +! ( + e e + ( +! + e ( + which gives lim a + (+! e ( Let a diverges. + + a which is greater tha. Now by ratio test the series. The a + + which is bigger tha. Now by root test. This gives lim + + a ( + + diverges. 5. Which oe of the followig series coverges coditioally? ( (a + (b (c cos ( 3 ( ( (d (e 5 ( e e + Solutio: Correct aswer is ( + coverges coditioally. The sum coverges by the Alteratig Series Test sice b + > 0, b b, for all ad lim diverges by the p-series test with p However, Note that a, b, ad e are absolutely coverget ad d is diverget. +
5 6. Fid a power series represetatio for the fuctio x ( x 3 i the iterval (,. (Hit: Differetiatio of power series may help. (a x 3 (b ( 3x 3 (c x x 3+ (d 3 + (e x 3 Solutio: Write f(x x ( x 3 d dx [ We ote that ] x 3 We use the well-kow power series Pluggig i x 3 we obtai: 3x ( x 3 3f(x. x x for < x <. x x 3 for < x <. 3 We differetiate this power series i order to compute a power series for f(x. 3f(x d [ ] dx x 3 d x 3 dx (3x 3.
6 So, f(x x 3 i the iterval (,. 7. Use you kowledge of a well kow power series to calculate the limit cos (x + x 4 lim x 0 x 8 (a (b 8! (c (d (e The limit does ot exist Solutio: We kow the Taylor (Maclauri series expasio for cos(x aroud x 0 is give by cos(x. Substitutig i x we obtai cos(x ( x (! ( x 4 (! Now we ca evaluate the limit as follows: cos (x + x 4 lim x 0 x 8 x4! + x8 4! x + 6! ( x4 + x8 x + x6! 4! 6! 8! lim x 0 x 8 + x 4 ( x 4 + x8 x + x6 4! 6! 8! + x 4 lim x 0 x 8 lim x 0 x4 x8 + 6! 8!
7 8. Which of the followig is the third Taylor polyomial of the fuctio ( x f(x si cetered at a π? (x π (a 4(! ( x π (b 4(! (x π3 (c(x π 3! (d x 4(! (e(x π (x π3 (x π Solutio: The aswer is. 4(! Let P 3 (x ( deote the third Taylor polyomial cetered at a π of the fuctio x f(x si. The, P 3 (x f (0 (π + f (π! (x π + f (π! (x π + f (π (x π 3 3! Hece, to fid P 3 (x we eed to fid the values: f (0 (π f(π, f (π, f (π, f (π. ( x ( π f (0 (x f(x si f (0 (π si [ ( x ] f ( x (x si cos f (π ( π cos 0 [ ( x ] f (x cos ( x 4 si f (π ( π 4 si 4 [ f (x ( x ] 4 si ( x 8 cos f (π ( π 8 cos 0
8 Hece, P 3 (x f (0 (π + f (π! (x π + f (π! + 0! (x π + 4! (x π + 0 (x π3 3! (x π 4(! (x π + f (π (x π 3 3! 9. Compute the radius of covergece, R, of the followig power series (ar (br 5 (cr (dr / (er x Solutio: The aswer is R. We use the ratio test: First we compute a + a So, x + + (+ x a + a x x ( + x x ( + x ( + L lim a + a lim x ( + x To get covergece, the Ratio Test says that we eed L < ad so: Hece, the radius of covergece R. L x < x <
9 0. Which of the followig gives a power series represetatio of the fuctio ( x (a! ( x (b! ( x + (c (! ( x + (d! x (e (! f(x e x ( x Solutio: The aswer is.! Ideed, we kow that e x x x ad so if we plug i! we obtai: e x ( x! ( x.! (l. Cosider the series (. Fill i the followig blaks ad be sure to show 3 your work. I each case idicate which test you are usig ad show how it is applied. Is the series absolutely coverget? (YES or NO Solutio: Aswer is No. Note that we kow that l which gives By the Compariso Test, 3 (l (l ( (l. For Recall that the series diverges. (l diverges as well. 3
10 Is the series coverget? (YES or NO Solutio: This series is a Alteratig series. There is a possibility that we ca use Alteratig Series Test. Let b (l. The b > 0. Now, (l lim (l x lim x x lim l x x x lim 0 x x [Where x is a real umber.] [Usig L Hospital] [Usig L Hospital] If we ca show b is decreasig the we ca use the Alteratig Series Test. We will (l x use Calculus I to see if b is decreasig. Let f(x. The f l x (l x (x x x l x( l x. For large x we have l x < 0. This shows that f (x < 0 for large x x ad hece f(x is decreasig for large x. (l Now by Alteratig Series Test, ( coverges. 3. (a Give the Taylor series expasio for the atiderivative F (x cos ( x dx about 0 (McLauri Series where F (0 0. Hit: Use your kowledge of a well kow series. Solutio: We kow that the Taylor series expasio for cos(x aroud x 0 is cos(x ( x, which has radius of covergece R. Pluggig i (! x we obtai cos( x ( x which is valid for all x 0. Fially, we (! compute the idefiite itegral F (x ( x dx (! ( (! ( (! x dx x C
11 Pluggig i x 0 we obtai So, C 0, ad F (0 ( (! F (x C 0 + C C. ( (! x + +. (b Use part (a to fid a expressio for the defiite itegral as a sum of a ifiite series. 0 cos( x dx Solutio: By the Fudametal Theorem of Calculus we kow 0 cos( x dx F ( F (0 ( (! ( (! + (c Use the alteratig series estimatio theorem to estimate the value of the above defiite itegral so that the error of estimatio is less tha. 00 (you may write your aswer as a sum of fractios. Solutio: The series i part (b is of the form ( b with b (!(+. We check that this series satisfies the coditios for the Alteratig Series Estimatio Theorem i b + ((+!(+ (!(+ b holds for all 0, ii lim b lim (!( + 0. Thus R S S b +. We eed to fid the value of which makes
12 b + < 00. We compute: b0 b!( 4 b 4!(3 7 b 3 6!( < 00 So E S S b 3 <. So S 00 gives approximatio of the itegral which is withi of the actual value. Fially, we compute our estimate for the itegral, 00 0 cos( xdx S b 0 b + b Fid the radius of covergece ad iterval of covergece of the followig power series: ( (x Solutio: We use the ratio test: First we compute a + a ( + (x ( (x+3 5 (x + 3 (x So, L lim a + a lim (x a + a 5 (x + 3 (x x lim + To get covergece, the Ratio Test says that we eed L < ad so: x L x < x + 3 < 5
13 Hece, the radius of covergece is R 5. Now, to get the iterval of covergece we have that it cotais (a R, a + R where a 3, R 5. That is, I.O.C cotais the iterval ( 8,. What is left to do is to check the ed poits to see if they belog to the iterval of covergece. x 8 : ( ( ( ( 5 5 ( 5 5 So, for x 8 the series diverges sice it is a p series with p. Therefore, x 8 is NOT i the iterval of covergece. x : ( ( ( 5 5 ( So, for x the series coverges by the Alteratig Series Test. Therefore, x IS i the iterval of covergece. Hece, the iterval of covergece is ( 8, ].
14 Formula Sheet si x + cos x + ta x sec x si x ( cos x cos x ( + cos x si x si x cos x si x cos y ( si(x y + si(x + y si x si y ( cos(x y cos(x + y cos x cos y ( cos(x y + cos(x + y sec θ l sec θ + ta θ + C
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