REVIEW 1, MATH n=1 is convergent. (b) Determine whether a n is convergent.
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1 REVIEW, MATH 00. Let a = +. a) Determie whether the sequece a ) is coverget. b) Determie whether a is coverget.. Determie whether the series is coverget or diverget. If it is coverget, fid its sum. a) ) ; b) ) 6) ; c) 5 ; 5 ; e) i) π ; j) l + ; k) + ; f) = ; g) = ) ; h) + ).. If the th partial sum of the series 4. Determie whether the series is coverget: a) + 6 ; a is s = +, fid a ad + + ; b) + si + 4 ; c) = ; + ; e)! ; f) For what values of p does the series = 6. Determie whether the series is coverget. a) ; b) ; c) l l + l4 l5 + ; l ) ; e) ) si π ; f) l) p coverge? a. ) ; g) 5 ) cos π 7. If the statemet is true, prove it; if it is false, provide a couter-example: a) If b ) is a positive sequece ad lim b = 0, the b) If b ) is a decreasig sequece the ) b is coverget. c) If a ) ad b ) are sequeces such that a ad If a ) is a sequece such that a is coverget the e) If a is coverget the 8. Determie whether the series is coverget: ) a is also coverget. ) b is coverget. b are coverget the a is coverget. a b is coverget as well.
2 a) g) ; b) 0) ; c)! )! ; e) ) + ; ) + 4 ; ) + ) + 4 ; f) si4) 4 ) ; i) cosπ). 9. Determie whether the series is absolutely coverget, coditioally coverget, or diverget. a) + ; b) ) = l ; c) ) = l) l ; ) + cos ; e) +. = 0. Fid the Maclauri series for f x) usig the defiitio. Fid the associated radius of covergece. a) f x) = l + x); b) f x) = cosx; c) f x) = 5x ; f x) =. Fid the Taylor series for f x) cetered at the give value a: + x). a) f x) = + x + x, a = ; b) f x) = x, a = ; c) f x) = cosx, a = π. f x) = x, a = 4; e) f x) = x, a =.
3 Solutios. Let a = +. a) Determie whether the sequece a ) is coverget. b) Determie whether a is coverget. Solutio. The sequece is coverget ad its limit is lim a = lim + =. The series is diverget by divergece test, because the limit of the sequece is ot 0.. Determie whether the series is coverget or diverget. If it is coverget, fid its sum. a) ) ; b) ) 6) ; c) 5 ; 5 ; e) i) Solutio. π ; j) l + ; k) + ; f) = ; g) = ) ; h) + ). + 6 ; a) The series is diverget because of the divergece test: lim = + 0. b) The series is diverget because of the divergece test: lim ) does ot exist. c) The series is coverget ad we will use the summatio =0 r = r for r,). The we have 5 5 = 5. The series is diverget because of the divergece test: lim 6 5) does ot exist. π e) We have + = π ). π The last series is diverget because lim = + 0. f) = = = ) + ) = = ). We will ow fid the -th partial sum: s = = + + g) This was doe i class, ad is similar to the previous problem. h) + 6 = = + = + =. i) Hit: Use the represetatio: = + ) + ) = + +. k= ) = k ) = k +. Therefore lim s = ad the sum of the series is. j) Let s = k= l k+ k = l +l +l + +l + = l l+l l+l l4+ +l l+) = l+). Thus lim s = lim l + )) = hece the series is diverget. k) Sice lim + ) = the divergece test implies that the series + ) is diverget.. If the th partial sum of the series a is s = +, fid a ad Solutio. Sice s = a + + a ad s = a + + a we obtai that We also have a = lim s =. a = s s = + ) + ) ) = = + ) + ). a.
4 4. Determie whether the series is coverget: a) Solutio. + + ; b) + si + 4 ; c) = ; + ; e)! ; f) +. a) We will use the limit compariso test. Let a = ++ ad b = a. The lim =. Sice the series b we coclude that the series is coverget. + + b) We will use the limit compariso test. Let a = +si +4, ad b = =. The the series b is coverget ad lim a b = lim +si +4 Usig the squeeze theorem ad the iequality si to the limit compariso test the series a is coverget. si. + si = lim + 4 = + lim 4 + lim is coverget si we obtai that lim = 0. Thus lim a = ad accordig b c) Let a = ad b =. The series a is diverget. From lim = we coclude that the series b a is diverget accordig to the limit compariso test. Let a =. Sice lim a + + a = lim +4 = <, thus the series is coverget accordig to the ratio test. + e) Hit: Use ratio test. f) Let a = + ad let b =. The series well. 5. For what values of p does the series Solutio. Let us take f x) = = l) p coverge? a is coverget, ad lim =, hece the origial series is coverget as b xlx) p. The fuctio f is positive ad decreasig sice f x) = x lx) p xlx)p ) = lx)p + plx) p x lx) p. For sufficietly large x the umerator is positive regardless of p. Thus f x) 0, ad the fuctio is decreasig. Sice f is positive + ad decreasig the itegral test implies that the series = l) p ad the itegral f x)dx are equicoverget. + We ca evaluate the itegral f x)dx usig the substitutio u = lx. The we have du = x dx. The itegral becomes: f x)dx = xlx) p dx = l u p du = u p du. l The ati-derivative of u p is p+ u p+ for p, ad the ati-derivative of u is lu. Thus for p, the itegral is diverget ad for p > the itegral is coverget. 6. Determie whether the series is coverget. 4
5 a) ; b) ; c) l l + l4 l5 + ; Solutio. l ) ; e) ) si π ; f) a) The geeral term of the series satisfies a = ) the series ) b) The give series is equal to 4 is diverget. + ) ) ; g) 5 ) cos π is decreasig ad coverges to 0. Thus, accordig to the alteratig series the series +. Sice lim =, we ca use the divergece test to coclude that Let a = + 6. The sequece a ) is a sequece of positive terms that ) c) ; ; e) ; f) These series are coverget accordig to the alteratig series test. Prove this!) f) Sice > for 5, we coclude that the series is diverget accordig to the divergece test. 5) g) Sice lim cos π = cos0 =, accordig to the divergece test the origial series is diverget. 7. If the statemet is true, prove it; if it is false, provide a couter-example: a) If b ) is a positive sequece ad lim b = 0, the b) If b ) is a decreasig sequece the ) b is coverget. c) If a ) ad b ) are sequeces such that a ad If a ) is a sequece such that a is coverget the e) If Solutio. a is coverget the ) a is also coverget. ) b is coverget. b are coverget the a is coverget. a) This statemet is false. A couter-example is the sequece {, if is eve b = if is odd. The b ) The series has all terms positive, lim b = 0, ad ) b = is diverget while the series b) The statemet is false. A couter-example is b =. ) + + ) 5 + = is coverget. + 6 a b is coverget as well. 4. is coverget. Therefore, their differece is diverget. 4 5
6 c) The statemet is false. A couter-example is a = b = ). Each of the series a ad to the alteratig series test, while their product is diverget because a b = ad a b = b is coverget accordig The statemet is false. A couter-example is a = ). The argumet is the same as i the previous problem. e) The statemet is false. A couter-example is a = ). The while ) a = is diverget. 8. Determie whether the series is coverget: a) g) ; b) 0) ) + ; c)! 4 ; )! ; e) ) + ) + 4 ; f) ) ) + ; ; i) si4) 4 cosπ).. a is coverget accordig to the alteratig series test, Solutio. a) Let a = +). Sice lim a + a = lim + + ) + ) = lim + = lim = lim lim + + =. Sice < accordig to ratio test, the origial series is coverget. 0) b) Let a = 0). Sice lim a +! a = lim + +)! 0! = lim 0) +! 0) + )! = lim 0 = 0. Sice 0 < accordig to + 0) ratio test, the series is coverget. Moreover, it is absolutely coverget.! c) For this series we use the alteratig series test. We take a = 4. It is easy to prove that a ) is decreasig ad coverget to 0. Thus the series ) + 4 is coverget. We use the ratio test for this series. Let us take a = )!. The we have lim a + a = lim )! + ))! = lim )! )! + ) + ) = lim Sice 0 < the ratio test ca be applied to coclude that the series is coverget. + ) + ) = 0. e) I this problem we use the alteratig series test. Let a = The sequece a cotais positive terms. We will ow prove that the sequece is decreasig. It suffices to prove that a + < a. The iequality is equivalet to: + + ) + ) ) < This iequality ca be trasformed ito a equivalet oe by takig the reciprocals of both sides: Sice ) ) + + ) > = = +, the last iequality ca be rewritte as: ) + >
7 Cacelig from both sides of the equatio, we obtai: + + ) + >. It is sufficiet to prove that > + +, which is equivalet to + >. The last iequality is true sice + + = ) + > 0. We ca ow use the alteratig series test to prove that + is coverget. + 4 f) We use the compariso test for this problem. Let a = si4) 4, ad let b = 4. The we have a < b. Sice the series si4) b is coverget, the series a is coverget as well. We have proved that 4 is absolutely coverget, hece it is coverget. g) We will prove that this series is absolutely coverget. I order to do this we eed to prove that ad let d + = = is coverget. We + will use the limit compariso test to attai this goal. Let c = /. The we have lim = ad the d series d = is coverget p-series with p = > ). Accordig to the limit compariso test we have that + is coverget. Cosequetly, the series ) is coverget as well. + h) This problem is very similar to a) ad the ratio test ca be applied to establish the absolute covergece of the give series. i) This problem is very similar to f). We eed to use the compariso test to prove the absolute covergece of the series from the questio. 9. Determie whether the series is absolutely coverget, coditioally coverget, or diverget. a) Solutio. + ; b) ) = l ; c) ) = l) l ; ) + cos ; e) +. = a) Usig the ratio test, oe ca prove that the series is diverget. Ideed, if a = lim a + a = lim + ) Alteratively, this problem ca be solved usig the root test. = lim + ) + = 7 lim b) The series is absolutely coverget ad this ca be proved usig the root test. Ideed, lim a = lim l) = lim l = 0. Sice 0 < the root test implies that the series is coverget. +. The we have + ) c + ) = +. c) The series is absolutely coverget ad we will prove this usig the compariso test. Let a = ). The l) l for > e e. Sice the series a = l) l = e ll)l ) = e l ll) = e lll ) = ll <, is coverget the origial series is coverget as well. This problem ca be solved usig the root test. The aswer is: absolutely coverget. e) This series is absolutely coverget ad this ca be proved usig the limit compariso test. 7
8 0. Fid the Maclauri series for f x) usig the defiitio. Fid the associated radius of covergece. a) f x) = l + x); b) f x) = cosx; c) f x) = 5x ; f x) = Solutio. The Maclauri series is obtaied usig the formula Mx) = f 0) + f 0) x + f 0)! + x). x + f ) 0)! x +. a) f x) = + x, f x) = + x), f ) x) = + x). The geeral term is: f k) x) = )k k )! + x) k. We ow have f 0) = 0, f 0) =, f 0) =, f k) 0) = ) k k )!. Thus the Maclauri series is: b) See Wikipedia. Mx) = x x + x 4 x4 + 5 x5 + = k= ) k xk k. c) We first express f x) i the followig form: f x) = e l5x ) = e 5xl = e x 5l. We ca ow easily calculate f x), f x), f k) x), ad apply the formula. This is a routie exercise.. Fid the Taylor series for f x) cetered at the give value a: a) f x) = + x + x, a = ; b) f x) = x, a = ; c) f x) = cosx, a = π. f x) = x, a = 4; e) f x) = x, a =. 8
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