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Tests for Convergence of Series 1) Use the comparison test ...

Tests for Convergence of Series1) Use the comparison test to confirm the statements in the following n=41ndiverges, so n=41n : Letan= 1/(n 3), forn 4. Sincen 3< n, we have 1/(n 3)>1/n, soan> harmonic Series n=41ndiverges, so the comparison test tells us that the Series n=41n 3also n=11n2converges, so n=11n2+ : Letan= 1/(n2+ 2). Sincen2+ 2> n2, we have 1/(n2+ 2)<1/n2, so0< an< Series n=11n2converges, so the comparison test tells us that the Series n=11n2+2also n=11n2converges, so n=1e : Letan=e n/n2. Sincee n<1, forn 1,we havee nn2<1n2, so0< an< Series n=11n2converges, so the comparison test tells us that the Series n=1e nn2also ) Use the comparison test to determine whether the Series in the following exercises n=113n+1 Answer: Letan= 1/(3n+ 1).

1 n=0 2n 3+1 diverges. 4) Use the integral test to decide whether the following series converge or diverge. 1. X1 n=1 1 n3 Answer: We use the integral test with f(x) = 1=x3 to determine whether this series converges or diverges. We determine whether the corresponding improper integral Z 1 1 1 x3

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Transcription of Tests for Convergence of Series 1) Use the comparison test ...

1 Tests for Convergence of Series1) Use the comparison test to confirm the statements in the following n=41ndiverges, so n=41n : Letan= 1/(n 3), forn 4. Sincen 3< n, we have 1/(n 3)>1/n, soan> harmonic Series n=41ndiverges, so the comparison test tells us that the Series n=41n 3also n=11n2converges, so n=11n2+ : Letan= 1/(n2+ 2). Sincen2+ 2> n2, we have 1/(n2+ 2)<1/n2, so0< an< Series n=11n2converges, so the comparison test tells us that the Series n=11n2+2also n=11n2converges, so n=1e : Letan=e n/n2. Sincee n<1, forn 1,we havee nn2<1n2, so0< an< Series n=11n2converges, so the comparison test tells us that the Series n=1e nn2also ) Use the comparison test to determine whether the Series in the following exercises n=113n+1 Answer: Letan= 1/(3n+ 1).

2 Since 3n+ 1>3n, we have 1/(3n+ 1)<1/3n=(13)n, so0< an<(13) we can compare the Series n=113n+1with the geometric Series n=1(13) geometric seriesconverges since|1/3|<1, so the comparison test tells us that n=113n+1also n=11n4+enAnswer: Letan= 1/(n4+en). Sincen4+en> n4, we have1n4+en<1n4,so0< an< thep- Series n=11n4converges, the comparison test tells us that the Series n=11n4+enalso n=21lnnAnswer: Since lnn nforn 2, we have 1/lnn 1/n, so the Series diverges by comparison with theharmonic Series , 1 n=1n2n4+1 Answer: Letan=n2/(n4+ 1).

3 Sincen4+ 1> n4, we have1n4+1<1n4, soan=n2n4+ 1<n2n4=1n2,therefore0< an< thep- Series n=11n2converges, the comparison test tells us that the Series n=1n2n4+1converges n=1nsin2nn3+1 Answer: We know that|sinn|<1, sonsin2nn3+ 1 nn3+ 1<nn3= thep- Series n=11n2converges, comparison gives that n=1nsin2nn3+ n=12n+1n2n 1 Answer: Letan= (2n+ 1)/(n2n 1). Sincen2n 1< n2n+n=n(2n+ 1), we have2n+ 1n2n 1>2n+ 1n(2n+ 1)= , we can compare the Series n=12n+1n2n 1with the divergent harmonic Series n= comparisontest tells us that n=12n+1n2n 1also ) Use the ratio test to decide if the Series in the following exercises converge or n=11(2n)!

4 Answer: Sincean= 1/(2n)!, replacingnbyn+ 1 givesan+1= 1/(2n+ 2)!. Thus|an+1||an|=1(2n+2)!1(2n)!=(2n)!(2n+ 2)!=(2n)!(2n+ 2)(2n+ 1)(2n)!=1(2n+ 2)(2n+ 1),soL= limn |an+1||an|= limn 1(2n+ 2)(2n+ 1)= 0, the ratio test tells us that n=11(2n)! n=1(n!)2(2n)!Answer: Sincean= (n!)2/(2n)!, replacingnbyn+ 1 givesan+1= ((n+ 1)!)2/(2n+ 2)!. Thus,|an+1||an|=((n+1)!)2(2n+2)!(n!)2(2n )!=((n+ 1)!)2(2n+ 2)! (2n)!(n!) , since (n+ 1)! = (n+ 1)n! and (2n+ 2)! = (2n+ 2)(2n+ 1)(2n)!, we have|an+1||an|=(n+ 1)2(n!)2(2n)!(2n+ 2)(2n+ 1)(2n)!(n!)2=(n+ 1)2(2n+ 2)(2n+ 1)=n+ 14n+ 2,soL= limn |an+1||an|= <1, the ratio test tells us that n=1(n!)

5 2(2n)! n=1(2n)!n!(n+1)!Answer: Sincean= (2n)!/(n!(n+ 1)!), replacingnbyn+ 1 givesan+1= (2n+ 2)!/((n+ 1)!(n+ 2)!). Thus,|an+1||an|=(2n+2)!(n+1)!(n+2)!(2n)! n!(n+1)!=(2n+ 2)!(n+ 1)!(n+ 2)! n!(n+ 1)!(2n)!.However, since (n+ 2)! = (n+ 2)(n+ 1)n! and (2n+ 2)! = (2n+ 2)(2n+ 1)(2n)!, we have|an+1||an|=(2n+ 2)(2n+ 1)(n+ 2)(n+ 1)=2(2n+ 1)n+ 2,soL= limn |an+1||an|= >1, the ratio test tells us that n=1(2n)!n!(n+1)! n=11rnn!,r >0 Answer: Sincean= 1/(rnn!), replacingnbyn+ 1 givesan+1= 1/(rn+1(n+ 1)!). Thus|an+1||an|=1rn+1(n+1)!1rnn!=rnn!rn+1 (n+ 1)!

6 =1r(n+ 1),soL= limn |an+1||an|=1rlimn 1n+ 1= 0, the ratio test tells us that n=11rnn!converges for allr > n=11nenAnswer: Sincean= 1/(nen), replacingnbyn+ 1 givesan+1= 1/(n+ 1)en+1. Thus|an+1||an|=1(n+1)en+11nen=nen(n+ 1)en+1=(nn+ 1) limn |an+1||an|=1e< <1, the ratio test tells us that n= n=02nn3+1 Answer: Sincean= 2n/(n3+ 1), replacingnbyn+ 1 givesan+1= 2n+1/((n+ 1)3+ 1). Thus|an+1||an|=2n+1(n+1)3+12nn3+1=2n+1(n+ 1)3+ 1 n3+ 12n= 2n3+ 1(n+ 1)3+ 1,soL= limn |an+1||an|= >1 the ratio test tells us that the Series n=02nn3+ ) Use the integral test to decide whether the following Series converge or n=11n3 Answer: We use the integral test withf(x) = 1/x3to determine whether this Series converges or determine whether the corresponding improper integral 11x3dxconverges or diverges.

7 11x3dx= limb b11x3dx= limb 12x2 b1= limb ( 12b2+12)= the integral 11x3dxconverges, we conclude from the integral test that the Series n= n=1nn2+ 1 Answer: We use the integral test withf(x) =x/(x2+1) to determine whether this Series converges or determine whether the corresponding improper integral 1xx2+ 1dxconverges or diverges: 1xx2+ 1dx= limb b1xx2+ 1dx= limb 12ln(x2+ 1) b1= limb (12ln(b2+ 1) 12ln 2)= .Since the integral 1xx2+ 1dxdiverges, we conclude from the integral test that the Series n=1nn2+ n=11enAnswer : We use the integral test withf(x) = 1/exto determine whether this Series converges or do so we determine whether the corresponding improper integral 11exdxconverges or diverges.

8 11exdx= limb b1e xdx= limb e x b1= limb ( e b+e 1)=e the integral 11exdxconverges, we conclude from the integral test that the Series n= can also observe that this is a geometric Series with ratiox= 1/e <1, and hence it n=21n(lnn)2 Answer: We use the integral test withf(x) = 1/(x(lnx)2) to determine whether this Series converges ordiverges. We determine whether the corresponding improper integral 21x(lnx)2dxconverges or diverges: 21x(lnx)2dx= limb b21x(lnx)2dx= limb 1lnx b2= limb ( 1lnb+1ln 2)=1ln the integral 21x(lnx)2dxconverges, we conclude from the integral test that the Series n=21n(lnn) ) Use the alternating Series test to show that the following Series n=1( 1)n 1 nAnswer: Letan= 1/ n.

9 Then replacingnbyn+ 1 we havean+1= 1/ n+ 1. Since n+ 1> n, we have1 n+1<1 n, hencean+1< an. In addition, limn an= 0 so n=0( 1)n nconverges by the alternating n=1( 1)n 12n+1 Answer: Letan= 1/(2n+ 1). Then replacingnbyn+ 1 givesan+1= 1/(2n+ 3). Since 2n+ 3>2n+ 1, wehave0< an+1=12n+ 3<12n+ 1= also have limn an= 0. Therefore, the alternating Series test tells us that the Series n=1( 1)n 12n+ n=1( 1)n 1n2+2n+1 Answer: Letan= 1/(n2+ 2n+ 1) = 1/(n+ 1)2. Then replacingnbyn+ 1 givesan+1= 1/(n+ 2)2. Sincen+ 2> n+ 1, we have1(n+ 2)2<1(n+ 1)2so0< an+1< also have limn an= 0.

10 Therefore, the alternating Series test tells us that the Series n=1( 1)n 1n2+2n+ n=1( 1)n 1enAnswer: Letan= 1/en. Then replacingnbyn+ 1 we havean+1= 1/en+1. Sinceen+1> en, we have1en+1<1en, hencean+1< an. In addition, limn an= 0 so n=1( 1)nenconverges by the alternating seriestest. We can also observe that the Series is geometric with ratiox= 1/ecan hence converges since|x|< ) In the following exercises determine whether the Series is absolutely convergent, conditionallyconvergent, or ( 1)n2nAnswer: Both ( 1)n2n= ( 12)nand 12n= (12)nare convergent geometric Series .


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