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.

2 Since the limit of the terms isequal to 4, not zero, the series must Does the following series converge or diverge? 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!

4 +(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. (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|.

5 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.

6 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)!

7 (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? 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?

8 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?

9 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).

10 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).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.