MTH 26 TEST April, 20 (PLEASE PRINT YOUR NAME!!) Name:. (6 poits each) Evaluate lim! a for the give sequece fa g. (a) a = 2 2 5 2 5 (b) a = 2 7 2. (6 poits) Fid the sum of the telescopig series p p 2. (a) Write the Nth partial sum S N of the series i the telescopig form ad the simplify it to a closed formula. S N =
(b) Sum the series by dig the limit of S N. lim S N = N!. (6 poits each) Determie whether the followig geometric series coverges or diverges. If coverges, d the sum of the series. (a) P (b) P (2 ) =0. (6 poits each) Use the Test for Divergece to determie whether the give series diverges. (a) 5 75 2
(b) P 2 2 5 2 5 5. (56 poits) Use the Itegral Test to determie whether the series diverges. 2 (a) Applyig the useful Property of First Derivative to the correspodig fuctio f (x) = x > to verify that forms a decreasig sequece. 2 coverges or x2 x for (b) Evaluate the improper itegral correspodig to the series, that is, Z x 2 x dx. (c) I view of the Itegral Test, determie whether the series 2 coverges or diverges.
6. (0 poits) Use the Alteratig Series Test for Covergece to determie whether the alteratig P series ( ) 2 coverges or diverges. (You must verify that a = 2 is decreasig i.) 7. (0 poits) Determie whether the series cos() 2 is absolutely coverget, coditioally coverget, or diverget. (Hit: jcos ()j for all ad the use the Compariso Test.)
. (0 poits) Use the Ratio Test to decide whether the series! coverges absolutely or di- verges. [Hit: You must calculate L a! a ad use the well-kow fact that e.]! 9. (0 poits) Use the Root Test to decide whether the series or diverges. (You must calculate L! ja j =.) ( ) 20 coverges absolutely 5
KEY TO WORK-SHEET -. (a) Dividig every term by 2 gives lim a!! 2 2 5 2 5 (b) Decompose a ito two fractios:! 2 2 5 5 2 lim a 2!! 7! 7 2 7! 2. (a) The Nth partial sum ca be writte S N = = NX p p 2 p p = p p N = p p5 = 2 0 0 0 = 2 5 0 0 ) 5 : 7 p5 p6 p p : N (b) The the give series coverges ad has sum p 2, sice lim S N p p = p N! N! N p = 2 : 2 = 0 0 = 0: 7 p p N 2 N. (a) Note that =0 = 0 Sice jrj = = <, the geometric series P coverges ad has the sum ( ) = 7. Cosequetly, has the sum =0 = 7 = =0 = 7 : (b) Sice 2 = (2 ) =, the series ca be writte =0 The geometric series P =0 =0 2 = =0 = =0 coverges for jrj = = < ad has the sum = 2 = = 7 : 6
whereas the geometric series P =0 =0 = coverges for jrj = = < ad has the sum 2 Hece, the give series coverges ad has the sum = = 2 : =0 2 = 7 2 = 5 :. (a) Sice lim a 5!! 7 5 5! 7 5 it follows from the Test for Divergece that the series (b) Sice lim a!! 2 2 5 2 5! it follows from the Test for Divergece that the series = 5 7 0 = 5 6= 0; 7 5 75 diverges. 2 2 5 5 2 5. (a) Cosider the correspodig fuctio f (x) = x x 2. The which implies that fa g = = 2 0 0 0 0 0 0 = 2 6= 0; 5 2 2 5 2 5 diverges. f 0 (x) = x2 () (x) (2x) (x 2 ) 2 = x2 (x 2 ) 2 < 0 for x > o 2 (b) Let u = x 2. The du = 2xdx ad Z x x 2 dx = Z 2 Now we evaluate the improper itegral Z is a decreasig sequece. x x 2 2xdx = 2 x x 2 dx = Z lim N! = 2 lim N! (c) I view of the Itegral Test, the series 6. Cosider First we show that f (x) = x x 2 Z u du = 2 l juj C = 2 l x 2 C: x x 2 dx N! 2 l x 2 l N 2 l (2) = : 2 diverges. a = f () = 2 : is decreasig for x >. By the Quotiet Rule for Di eretiatio, f 0 (x) = x2 () (x) (2x) (x 2 ) 2 = x2 (x 2 ) 2 < 0 for x > N 7
It follows that f is decreasig for x > ad, thus, fa g is a decreasig sequece for 2 (that is, a 2 a a : : :). Next, we see that lim a!! 2! 2 = 0 0 = 0: Therefore, it follows from the Alteratig Series Test that the alteratig series coverges. 7. For all, we have 0 cos() 2 2. Sice the series 2 is a coverget geometric series for jrj = 2 <. I view of the Compariso Test for Covergece, it follows that the series cos() 2 coverges absolutely.. Set a =! for all. The lim a a!!!! () 2 ()!! ( ) ( ) 2!!! Hece, it follows from the Ratio Test that the series P! ( )! ( ) ( )!! lim!! diverges. = e = e > : 9. Sice lim ja j =!! ( ) 20 = =! 20! = 20 = 20 = > ; () it follows from the Root Test that the series P ( ) 20 diverges.
5. (a) Cosider the correspodig fuctio f (x) = x2 x. The which implies that fa g = f 0 (x) = x (2x) x 2 x 2 (x ) 2 = x (x ) 2 < 0 for x > 2 (b) Let u = x. The du = x 2 dx ad Z x 2 x dx = Z Now we evaluate the improper itegral Z o is a decreasig sequece. x x2 dx = Z u du = l juj C = l x C: x 2 x dx = Z lim N! = lim N! x 2 x dx N! l x l N l (2) = : N (c) I view of the Itegral Test, the series 2 diverges. 9
F (Partial Sum of Series) The Nth partial sum S N of the series a is give by The series NX a = a a 2 a a N : a is said to coverge to L if lim S N = L ad diverge if lim S N does ot exist. N! N! F (Geometric Series) If jrj <, the the series ar coverges to a r. That is, ar = a ar ar 2 ar ar = a r for jrj < : If jrj, the the series ar diverges F (Basic Properties of Series) Suppose that a coverges to A ad b coverges to B. The (a) If (a b ) coverges to A B ad (a b ) coverges to A B; (b) If (a ) coverges to A, where is ay real costat. F (Test for Divergece) If lim! u 6= 0, the the series u diverges. F (Itegral Test) Let f be a positive, cotiuous, decreasig fuctio o the iterval [; ), ad de ed a = f () for all. Z (a) If f (x) dx coverges, the a coverges. Z (b) If f (x) dx diverges, the a diverges. F (p-series) The p-series p coverges if p > ad diverges if p. F (Erichmet: Approximatig the Value of a I ite Series) Let f be a positive, cotiuous, decreasig fuctio o the iterval [; ). If the i ite series f () coverges, the MX f () Z M f (x) dx 0
uderestimates the series f () by a amout that is most f (M). Let 0 a b for all. If the series b co- F (Basic Compariso Test for Covergece) verges, the the series a also coverges. F (Basic Compariso Test for Divergece) Let 0 b a for all su cietly large. If the series b diverges, the the series a also diverges. F (Limit Compariso Test) is a ( ite) positive umber, the F (Alteratig Series Test) or coverges if Let a ad a coverges if ad oly if A series of the form a b be series of positive terms. If lim! b b coverges. ( ) a = a a 2 a a ( ) a ( ) a = a a 2 a a ( ) a (a) a a 2 a ad (b) lim! a = 0. F (Absolute Covergece) terms. If the series u coverges absolutely. Let exists ad u be a series, possibly cotaiig both positive ad egative ju j of absolute values coverges, the we say that the series F (Covergece of Absolute Values Implies Covergece Test) does u. If ju j coverges, the so F (Basic Properties of Absolute Covergece) Suppose that u coverges absolutely to A ad v coverges absolutely to B. The (a) If (u v ) coverges absolutely to A B ad (u v ) coverges absolutely to A B;
(b) If (u ) coverges absolutely to A, where is ay real costat. F (Coditioal Covergece) say that the series F (Ratio Test) If a series u coverges coditioally. (a) If L <, the Let u coverges but the series u be a series. Suppose that u coverges absolutely. (b) If L >, or if L =, the u diverges. (c) If L =, the test gives o iformatio. ju j lim = L:! ju j ju j diverges, the we F (Root Test) Let u be a series. Suppose that (a) If L <, the u coverges absolutely. (b) If L >, or if L =, the u diverges. (c) If L =, the test gives o iformatio. lim ju j = = L:! F (L Hôpital s Rule) Suppose that fuctios f ad g are di eretiable o a ope iterval (a; b) cotaiig c, except possibly at c itself. If g (x) 6= 0 for x 6= c, ad if f (x) =g (x) has the idetermiate form 0=0 or = at x = c, the f (x) lim! g (x) f 0 (x)! g 0 (x) provided f 0 (x) =g 0 (x) has a limit or becomes i ite as x approaches c. F (Usig the Derivative to Tell Whe the Fuctio Is Icreasig or Decreasig) If f 0 (x) > 0 for each x i a iterval I, the f is icreasig o I. If f 0 (x) < 0 for each x i a iterval I, the f is decreasig o I. F (Some Useful Formulas) a b c d Z a 2 x 2 dx = x a arcta C a = ad bc lim! x=x = 2