Math 2260 Exam #3 Practice Problem Solutions

1 Math 2260 exam #3 Practice Problem Solutions1. Does the following series converge or diverge? Explain your answer. n=02n3n+ :Since 3n+n3>3nfor alln 1, it follows that2n3n+n3<2n3n=(23) , n=02n3n+n3< n=0(23)n=11 23= , the given series Does the following series converge or diverge? Explain your answer. n= :Use the Ratio Test:limn n+13n+1n3n= limn n+ 13n+1 3nn= limn 13 n+ 1n= <1, the Ratio Test implies that this series Does the following series converge or diverge? Explain your answer. n=12nsin(1n).Answer:Notice that the terms of this series are not going to zero:limn 2nsin(1n)= limx 2xsin(1x)= limx sin(1x)12x= limx cos(1x) 2x2 2(2x)2= limx 2 cos(1x)x2 4x2 2= limx 4 cos(1x)= 4where I went from the second to the third lines using L H opital s Rule. Since the limit of the terms isequal to 4, not zero, the series must Does the following series converge or diverge?

2 If it converges, find the sum. If it diverges, explain why. n=12n+ :Re-writing slightly, the given series is equal to n=1(2n4n+3n4n)= n=12n4n+ n= both of these series are convergent geometric series, I know the original series converges, so itremains only to determine the sum. Notice that n=12n4n=24+416+864+..=24(1 +24+416+..)= n=124(24)n 1=2/41 2/4=1/21/2= , n=13n4n=34+916+2764+..=34(1 +34+916+..)= n=134(34)n 1=3/41 3/4=3/41/4= , n=12n+ 3n4n= n=12n4n+ n=13n4n= 1 + 3 = Find the interval of convergence of the power series n=1(2x 5) :We use the Ratio Test on the series of absolute values to first determine the radius ofconvergence:limn (2x 5)n+1(n+1)23n+1 (2x 5)nn23n = limn |2x 5|n+1(n+ 1)23n+1 n3n|2x 5|n= limn |2x 5|3 n2(n+ 1)2=|2x 5| , the given series converges absolutely when|2x 5|3<1, meaning when|2x 5|< we check the endpoints.

3 When 2x 5 = 3, the series becomes n=13nn23n= n=11n2,which , when 2x 5 = 3, then series becomes n=1( 3)nn23n= n=1( 1)n3nn23n= n=1( 1)nn2,which also , the series converges for allxso that 3 2x 5 3,which is the interval [1,4].26. Use the first two non-zero terms of an appropriate Taylor series to approximate 10sin(x2) the error of your approximation ( the difference between your answer and the actual valueof the integral).Answer:First, recall that the Taylor series centered atx= 0 for sin(x) issin(x) =x x33!+x55! x77!+..Therefore, the Taylor series centered atx= 0 for sin(x2) issin(x2) = (x2) (x2)33!+(x2)55! (x2)77!+..=x2 x63!+x105! x147!+..Hence, 10sin(x2)dx= 10(x2 x63!+x105! ..)dx=[x33 x77 3!+x1111 5! ..]10=13 142+111 5! ..Therefore, we can approximate this number by13 142=1342,and we know the error is no bigger than111 5!=111 120=11320,so in particular the estimate1342is accurate to 3 decimal places.

4 (It would be totally fine in an examsituation to leave your answer as111 5!.)7. Find the radius of convergence of the Taylor series n=2( 1)n 1 +nn2(x 2) :Use the Ratio Test on the series of absolute values:limn ( 1)n+1 1+(1+n)(n+1)2(x 2)n+1 ( 1)n 1+nn2(x 2)n = limn 2 +n|x 2|n+1(n+ 1)2 n2 1 +n|x 2|n= limn |x 2| 2 +n 1 +n n2(n+ 1)2=|x 2|.Therefore, the series converges absolutely when|x 2|<1, so the radius of convergence is equal to Write the second-degree Taylor polynomial forf(x) = xcentered atc= 25. Use this polynomial toapproximate 26 and estimate the error of this :I can write down the Taylor series centered atx= 25 for the functionf(x) = xby firstcomputing the derivatives off:f (x) =12x 1/2=12 xf (x) =12 12x 3/2= 14x3/2f (x) = 14 32x 5/2=38x5/2 Therefore,f(25) = 25 = 5f (25) =12 25=110f (25) = 14 253/2= 14 125= 1500f (25) =38 255/2=38 3125=325,000 Hence, the Taylor series centered atx= 25 for xis x= 5 +(x 25)10 (x 25)21000+3(x 25)3150,000.

5 Therefore, 26 5 +(26 25)10 (26 25)21000= 5 +110 11000= = ,with an error no more than3(26 25)3150,000=3150,000=150,000= (In fact, 26 )9. Does the series n=1n nn2converge or diverge. Be sure to give a complete :Since limn n n= 1 as we discussed in class, a limit comparison to the series 1n2is anatural:limn n nn21n2= limn n nn2 n21= limn n n= , since the series 1n2converges, the Limit Comparison Test implies that the given seriesconverges as Does the following series converge or diverge? Explain your answer. n=1n!(n+ 1)!(3n)!.Answer:Use the Ratio Test:limn (n+1)!(n+2)!(3(n+1))!n!(n+1)!(3n)!= limn (n+ 1)!(n+ 2)!(3n+ 3)! (3n)!n!(n+ 1)!= limn (n+ 1)(n+ 2)(3n+ 3)(3n+ 2)(3n+ 1)= 0,since the numerator is a polynomial of degree 2 but the denominator is a polynomial of degree , since 0<1 the Ratio Test implies that the series Does the sequence(arctan(n2n2+ 1)) n=1converge or diverge?

6 If it converges, find the limit; if it diverges, explain :First, notice thatlimn n2n2+ 1= , the term inside the arctangent is going to 1, solimn arctan(n2n2+ 1)= arctan(1) = Does the series n=21n2 nconverge or diverge? Explain your :For largen, then2should dominate the n, so let s do a limit comparison to the convergentseries 1n2 n1n2= limn 1n2 n n21= limn n2n2 n= limn 11 1n3/2= , since 1n2converges, so does 1n2 Does the series n=1n!n5converge or diverge? Explain your :Use the Ratio Test:limn (n+1)!(n+1)5n!n5= limn (n+ 1)!(n+ 1)5 n5n!= limn (n+ 1)n5(n+ 1)5= since the expressionn5(n+1)5is going to 1 but (n+ 1) is going to .Therefore, the Ratio Test implies that the series Does the series n=13nn3converge or diverge? Explain your :Use the Ratio Test:limn 3n+1(n+1)33nn3= limn 3n+1(n+ 1)3 n33n= limn 3 (n+ 1)3n3= 3>1, the Ratio Test implies that the series Does the series n=02n+ 3(n2+ 3n+ 6)2converge or diverge?

7 Explain your :Fornvery large, the denominator will be dominated by the termn4, so do a limit comparisonto the convergent series nn4:limn 2n+3(n2+3n+6)2nn4= limn 2n+ 3(n2+ 3n+ 6)2 n4n= limn 2n+ 3n n4(n2+ 3n+ 6)2= 2 1 = , since the limit is finite and the series nn4=1n3converges, the Limit Comparison Testimplies that the given series converges as For which values ofxdoes the series n=0(x 4)n5nconverge? What is the sum of the series when it converges?Answer:First, use the Ratio Test on the series of absolute values:limn (x 4)n+15n+1 (x 4)n5n = limn |x 4|n+15n+1 5n|x 4|n=|x 4|5,so the given series converges absolutely whenever|x 4|5<1, meaning when|x 4|<5 (from this wesee that the radius of convergence of the series is 5).Now check the endpoints. Whenx 4 = 5, the series becomes n=05n5n= n=01,which , whenx 4 = 5, the series becomes n=0( 5)n5n= n=0( 1)n5n5n= n=0( 1)n,which also , the series converges for 5< x 4<5,which is to say, on the interval ( 1,9).

8 When the series does converge, it is just the geometric series n=0(x 45)n=11 x 45=15 (x 4)5=59 x,so when it converges the series converges to the functionf(x) =59 Does the series n=1( 1)n(n2+n3)n4+ 1converge absolutely, converge conditionally, or diverge? Explain your :The series of absolute values n=1 ( 1)n(n2+n3)n4+ 1 = n=1n2+n3n4+ 1should behave similarly to n3n4, so do a limit comparison to this series:limn n2+n3n4n3n4= limn n2+n3n4 n4n3= limn n2+n3n3= , since n3n4= 1ndiverges, the Limit Comparison Test says that the series of absolutevalues diverges as , the given series satisfies the hypotheses of the Alternating Series Test, so it converges. There-fore, since the series converges but the series of absolute values diverges, we conclude that the seriesconverges Does the series n=1(2n)!2n(n!)2converge or diverge? Explain your :Use the Ratio Test:limn (2(n+1))!

9 2n+1((n+1)!)2(2n)!2n(n!)2= limn (2n+ 2)!2n+1((n+ 1)!)2 2n(n!)1(2n)!= limn (2n+ 2)(2n+ 1)2 (n!)2((n+ 1)n!)2= limn (2n+ 2)(2n+ 1)2 1(n+ 1) (2n+ 2)(2n+ 1) = 4n2+ 6n+ 2 and since (n+ 1)2=n2+ 2n+ 1, the above limit is equal tolimn 4n2+ 6n+ 22(n2+ 2n+ 1)=42= 2>1, the Ratio Test implies that the given series Does the series n=2( 1)nnln(n)converge absolutely, converge conditionally, or diverge? Explain your :The series of absolute values n=2 ( 1)nnln(n) = n=21nln(n)diverges (see HW #12, Problem 3 for a proof of this). However, the series satisfies the hypotheses ofthe Alternating Series Test and hence converges, so we see that it converges For what values ofpdoes the series n=01(n2+ 1)pconverge? Explain your :For largen, the +1 will barely contribute, so do a limit comparison with the series 1(n2)p:limn 1(n2+1)p1(n2)p= limn 1(n2+ 1)p (n2)p1= limn (n2n2+ 1)p= 1p= 1,since limn n2n2+1= , the given series and the series 1(n2)p= 1n2pwill either both converge or both 1n2pconverges for 2p >1 and diverges otherwise, we see that the given series converges when2p >1, which is to say whenp > What is the interval of convergence of the following power series?

10 Explain your answer. n=13n(x 2) :Start by applying the Ratio Test to the series of absolute values:limn 3n+1(x 2)n+1(n+1)2 3n(x 2)nn2 = limn 3n+1|x 2|n+1(n+ 1)2 n23n|x 2|n= limn 3|x 2| n2(n+ 1)2= 3|x 2|.Therefore, the ratio test implies that the given series converges absolutely when 3|x 2|<1, meaningwhen|x 2|< , check the endpoints. Whenx 2 =13, the series becomes n=13n(13)nn2= n=11n2,which , whenx 2 = 13, the series becomes n=13n( 13)nn2= n=1( 1)nn2,which also , the series converges for 13 x 2 13,meaning the interval of convergence is[53,73].22. What are the first four nonzero terms of the Taylor series centered atx= 0 for the functionf(x) =xe3x?Answer:Sinceex= 1 +x+x22!+x33!+..,we see thate3x= 1 + (3x) +(3x)22!+(3x)33!+..= 1 + 3x+9x22!+27x33!+..Therefore,xe3x=x(1 + 3x+9x22!+27x33!+..)=x+ 3x2+9x32!+27x43!+..23. Which of the following gives the value of 1/20cos(x2)dxcorrect to within ( within 1/10,000)?