Math 115 HW #4 Solutions

Transcription

1 Math 5 HW #4 Solutios From Does the series coverge or diverge? ( ) = Aswer: This is a alteratig series, so we eed to check that the terms satisfy the hypotheses of the Alteratig Series Test. To see that the terms are decreasig i absolute value (i.e. that b + < b ), defie x f(x) = x The x x f 2 x (x) = x2 x 2 = 2(x3 + 2) 3x 3 2x 2 = x x 3 2x 2 x So log as x > 0, the term o the right is egative, so we see that f is a decreasig fuctio. Therefore, the terms of the sequece are decreasig i absolute value. To see that the terms are goig to zero, we eed to show that = 0. I the lefthad side, multiply both umerator ad deomiator by. This yields = Sice the umerator is costat ad the deomiator goes to ifiity as, this it is equal to zero. Therefore, we ca apply the Alteratig Series Test, which says that the series coverges. 2. Does the series coverge or diverge? ( ) = e/ Aswer: Agai, we wat to use the Alteratig Series Test, so we eed to cofirm that the terms are decreasig i absolute value ad goig to zero. To see that the terms are decreasig, we wat to show that e + + < e ;

2 i other words, we wat to show that + e e + <. But certaily + e < e ad + >, so the above iequality is true, which implies that the first iequality is true ad that the terms are decreasig i absolute value. To see that the terms are goig to zero, we just examie the it sice []e =. e = e = 0, Therefore, by the Alteratig Series Test, the series coverges. 6. Does the series coverge or diverge? = si(π/2)! Aswer: This series does coverge, but, we ll see that the slightly less complicated series coverges. ( ) = To use the Alteratig Series Test, we eed to show that the terms are decreasig i absolute value, so we eed to show that ( + )! <!. Multiplyig both sides by ( + )! shows that this iequality is equivalet to the iequality < ( + )!!! = + = +, which is certaily true. Therefore, the origial iequality is true ad the terms are decreasig i absolute value. Also, sice 0 <! < ad 0.! = 0 Therefore, the Alteratig Series Test tells us that the series coverges. 24. Show that the series = ( ) 5 2

3 is coverget. How may terms of the series do we eed to add i order to approximate the sum of the series to withi 0.000? Aswer: To see that the terms are decreasig i absolute value we eed to show that ( + )5 + < 5. Multiplyig by ( + )5 +, we see that this iequality is equivalet to < ( + )5+ 5 = 5 +, which is clearly true sice + >. Hece, the terms are decreasig i absolute value. To see that the it of the terms is zero, just take 5 = 0, sice the umerator is costat ad the deomiator. Therefore, by the Alteratig Series Test, the series coverges. If we wat the error to be less tha 0.000, the we eed to fid such that R = s s is bouded by R However, we kow that, for alteratig series, the remaider is bouded as R b + = Therefore, if we ca fid such that (+)5 + as desired. R ( + )5 +. < 0.000, we will have that < 0.000, ( + )5+ But clearly, if = 4, the ( + )5 + = = 5, 625 > 0, 000, meaig that ( + )5 + < = 0, 000, 0, 000 so we eed oly add the first 4 terms of the series to approximate the sum withi For what values of p is the series coverget? ( ) = Aswer: Whe p < 0, the terms are blowig up, so the series diverges. Whe p = 0, we have ( ), which also diverges. I claim that whe p > 0 the series coverges. 3 p

4 To see this, I ll use the Alteratig Series Test, which requires that the terms decrease i absolute value ad that they go to zero. To see that they decrease i absolute value, I eed to show that ( + ) p < p. Multiplyig both sides by ( + ) p, this is equivalet to the iquality < ( + )p p = ( ) + p, which is certaily true sice + >. To see that the terms go to zero, cosider the it This it is certaily zero sice the umerator is costat ad the deomiator is goig to (because p > 0). Therefore, the Alteratig Series Test tells us that ( ) p coverges for p > 0. From Is the series = p. si 4 4 absolutely coverget, coditioally coverget, or diverget? Aswer: Usig the fact that si x for ay x, we kow that si 4 si 4 4 = 4 4. Sice the series coverges (sice it s a geometric series), we kow, by the Compariso Test, that = = 4 si 4 4 coverges. Thus, the series si 4 = 4 coverges absolutely. 8. Is the series =! absolutely coverget, coditioally coverget, or diverget? 4

5 Aswer: Usig the Ratio Test, (+)! = (+) +! ( + )! ( + ) +! = ( + ) ( + ) +. Cacelig a factor of + from both umerator ad deomiator yields ( + ). Dividig umerator ad deomiator by gives ( + ) = ( + ) = e ( ) sice + = e. Therefore, sice e coverges absolutely. <, the Ratio Test says that the series 22. Is the series =2 ( ) absolutely coverget, coditioally coverget, or diverget? Aswer: Usig the Root Test, ( ) 2 5 ( ) + = 2 5 ( ) 2 5 = Sice + = 2, the above it is equal to 32, which is certaily >. Therefore, by the Root Test, the series diverges. 24. Is the series =2 (l ) absolutely coverget, coditioally coverget, or diverget? Aswer: Usig the Root Test, (l ) = l = 0 sice = ad l =. Therefore, the Root Test says that the series coverges absolutely. 30. The series a is defied by the equatios a = a + = 2 + cos a. 5

6 Determie whether a coverges or diverges. Aswer: Usig the Ratio Test, a + a = Sice cos, we kow that so the above it is bouded above by 2+cos a a = 2 + cos 2 + cos 3, 3 = cos. (*) O the other had, sice the it i (*) has to be o-egative, we see that it is actually equal to 0. Hece, sice 0 <, the Ratio Test says that the series coverges absolutely. 6