› ~apilking › math10560 › work › old exams › practice … · question 1question 2question...

Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 10 Question 11 Question 12

Question 1

1. Find∞Xn=1

3 · 5n−1.

I This is a geometric series of forP∞

n=1 arn−1 = a + ar + . . . .

n=122n

3·5n−1 = 43

+ 1615

+ . . .

I a = 43

and r = 1615

I ThereforeP∞

n=122n

3·5n−1 = a1−r

= 4/31−(4/5)

= 4/31/5

= 203.

I AlternativelyP∞

n=122n

3·5n−1 =P∞

n=14·4n−1

3·5n−1 =43

P∞n=1

´n−1= 4

1− 45

”= 20

Annette Pilkington Solutions PE3

Question 1

1. Find∞Xn=1

3 · 5n−1.

n=1 arn−1 = a + ar + . . . .

n=122n

3·5n−1 = 43

+ 1615

+ . . .

I a = 43

and r = 1615

I ThereforeP∞

n=122n

3·5n−1 = a1−r

= 4/31−(4/5)

= 4/31/5

= 203.

I AlternativelyP∞

n=122n

3·5n−1 =P∞

n=14·4n−1

3·5n−1 =43

P∞n=1

´n−1= 4

1− 45

”= 20

Question 1

1. Find∞Xn=1

3 · 5n−1.

n=1 arn−1 = a + ar + . . . .

n=122n

3·5n−1 = 43

+ 1615

+ . . .

I a = 43

and r = 1615

I ThereforeP∞

n=122n

3·5n−1 = a1−r

= 4/31−(4/5)

= 4/31/5

= 203.

I AlternativelyP∞

n=122n

3·5n−1 =P∞

n=14·4n−1

3·5n−1 =43

P∞n=1

´n−1= 4

1− 45

”= 20

Question 1

1. Find∞Xn=1

3 · 5n−1.

n=1 arn−1 = a + ar + . . . .

n=122n

3·5n−1 = 43

+ 1615

+ . . .

I a = 43

and r = 1615

I ThereforeP∞

n=122n

3·5n−1 = a1−r

= 4/31−(4/5)

= 4/31/5

= 203.

I AlternativelyP∞

n=122n

3·5n−1 =P∞

n=14·4n−1

3·5n−1 =43

P∞n=1

´n−1= 4

1− 45

”= 20

Question 1

1. Find∞Xn=1

3 · 5n−1.

n=1 arn−1 = a + ar + . . . .

n=122n

3·5n−1 = 43

+ 1615

+ . . .

I a = 43

and r = 1615

I ThereforeP∞

n=122n

3·5n−1 = a1−r

= 4/31−(4/5)

= 4/31/5

= 203.

I AlternativelyP∞

n=122n

3·5n−1 =P∞

n=14·4n−1

3·5n−1 =43

P∞n=1

´n−1= 4

1− 45

”= 20

Question 1

1. Find∞Xn=1

3 · 5n−1.

n=1 arn−1 = a + ar + . . . .

n=122n

3·5n−1 = 43

+ 1615

+ . . .

I a = 43

and r = 1615

I ThereforeP∞

n=122n

3·5n−1 = a1−r

= 4/31−(4/5)

= 4/31/5

= 203.

I AlternativelyP∞

n=122n

3·5n−1 =P∞

n=14·4n−1

3·5n−1 =43

P∞n=1

´n−1= 4

1− 45

”= 20

Question 2

2. The seriesP∞

n=2(−1)n+1√

a. does not converge absolutely but does converge conditionally.b. converges absolutely.c. diverges because the terms alternate.

d. diverges because limn→∞(−1)n+1√

n6= 0.

e. diverges even though limn→∞(−1)n+1√

I This is an alternating series of typeP

(−1)n+1bn, with bn > 0.

I We apply the alternating series test: (ii) limn→∞ bn = limn→∞ 1/√

I (i) The sequence {bn}∞n=2 is decreasing since√

n + 1 >√

n and thusbn+1 = 1/

√n + 1 < 1/

√n = bn for all n ≥ 2.

I Therefore the seriesP

(−1)n+1bn converges, ruling out 3 of the answers.

I The series∞Xn=2

| (−1)n+1

√n| =

∞Xn=2

diverges since it’s a p series and

p = 12< 1. Therefore the series does not converge absolutely.

I Therefore The seriesP∞

n=2(−1)n+1√

ndoes not converge absolutely but does

converge conditionally.

Question 2

2. The seriesP∞

n=2(−1)n+1√

n6= 0.

(−1)n+1bn, with bn > 0.

n + 1 >√

n and thusbn+1 = 1/

√n + 1 < 1/

I The series∞Xn=2

| (−1)n+1

√n| =

∞Xn=2

n=2(−1)n+1√

Question 2

2. The seriesP∞

n=2(−1)n+1√

n6= 0.

(−1)n+1bn, with bn > 0.

n + 1 >√

n and thusbn+1 = 1/

√n + 1 < 1/

I The series∞Xn=2

| (−1)n+1

√n| =

∞Xn=2

n=2(−1)n+1√

Question 2

2. The seriesP∞

n=2(−1)n+1√

n6= 0.

(−1)n+1bn, with bn > 0.

n + 1 >√

n and thusbn+1 = 1/

√n + 1 < 1/

I The series∞Xn=2

| (−1)n+1

√n| =

∞Xn=2

n=2(−1)n+1√

Question 2

2. The seriesP∞

n=2(−1)n+1√

n6= 0.

(−1)n+1bn, with bn > 0.

n + 1 >√

n and thusbn+1 = 1/

√n + 1 < 1/

I The series∞Xn=2

| (−1)n+1

√n| =

∞Xn=2

n=2(−1)n+1√

Question 2

2. The seriesP∞

n=2(−1)n+1√

n6= 0.

(−1)n+1bn, with bn > 0.

n + 1 >√

n and thusbn+1 = 1/

√n + 1 < 1/

I The series∞Xn=2

| (−1)n+1

√n| =

∞Xn=2

n=2(−1)n+1√

Question 2

2. The seriesP∞

n=2(−1)n+1√

n6= 0.

(−1)n+1bn, with bn > 0.

n + 1 >√

n and thusbn+1 = 1/

√n + 1 < 1/

I The series∞Xn=2

| (−1)n+1

√n| =

∞Xn=2

n=2(−1)n+1√

Question 3

3. Use Comparison Tests to determine which one of the following series isdivergent.

I (a)∞Xn=1

n32 + 1

converges by comparison with∞Xn=1

, a p-series with

I (b)∞Xn=1

n2 + 8converges by comparison with

∞Xn=1

n2, a p-series with

p = 2 > 1.

I (c)∞Xn=1

n2 − 1

n3 + 100diverges by limit comparison with

∞Xn=1

n, a p-series with

p = 1.

I (d)∞Xn=1

converges since it is a geometric series with |r | =5

I (e)∞Xn=1

geometric series with |r | =1

Question 3

I (a)∞Xn=1

n32 + 1

, a p-series with

I (b)∞Xn=1

∞Xn=1

n2, a p-series with

p = 2 > 1.

I (c)∞Xn=1

n2 − 1

∞Xn=1

n, a p-series with

p = 1.

I (d)∞Xn=1

I (e)∞Xn=1

Question 3

I (a)∞Xn=1

n32 + 1

, a p-series with

I (b)∞Xn=1

∞Xn=1

n2, a p-series with

p = 2 > 1.

I (c)∞Xn=1

n2 − 1

∞Xn=1

n, a p-series with

p = 1.

I (d)∞Xn=1

I (e)∞Xn=1

Question 3

I (a)∞Xn=1

n32 + 1

, a p-series with

I (b)∞Xn=1

∞Xn=1

n2, a p-series with

p = 2 > 1.

I (c)∞Xn=1

n2 − 1

∞Xn=1

n, a p-series with

p = 1.

I (d)∞Xn=1

I (e)∞Xn=1

Question 3

I (a)∞Xn=1

n32 + 1

, a p-series with

I (b)∞Xn=1

∞Xn=1

n2, a p-series with

p = 2 > 1.

I (c)∞Xn=1

n2 − 1

∞Xn=1

n, a p-series with

p = 1.

I (d)∞Xn=1

I (e)∞Xn=1

Question 3

I (a)∞Xn=1

n32 + 1

, a p-series with

I (b)∞Xn=1

∞Xn=1

n2, a p-series with

p = 2 > 1.

I (c)∞Xn=1

n2 − 1

∞Xn=1

n, a p-series with

p = 1.

I (d)∞Xn=1

I (e)∞Xn=1

Question 4

4. Consider the following series

(I )∞Xn=1

“2n2 + 7

n2 + 1

(II )∞Xn=2

n − 1(III )

∞Xn=1

Which of the following statements is true?They all diverge.(I ) converges, (II ) diverges, and (III ) diverges.They all converge.(I ) converges, (II ) diverges, and (III ) converges.(I ) diverges, (II ) diverges, and (III ) converges.

I For (I), we apply the n th root test. limn→∞np|an| = limn→∞

2n2+7n2+1

= limn→∞2+7/n2

1+1/n2 = 2 > 1. Therefore the series diverges.

n=221/n

n−1diverges by direct comparison with the series

, since21/n

n−1> 1

nfor all n.

I For III, we apply the ratio test, limn→∞|an+1||an| = limn→∞

(n+1)!

‹n!en

= limn→∞n+1e

=∞ > 1. Therefore the series diverges.

I Therefore they all diverge.

Question 4

(I )∞Xn=1

“2n2 + 7

n2 + 1

(II )∞Xn=2

n − 1(III )

∞Xn=1

2n2+7n2+1

= limn→∞2+7/n2

n=221/n

, since21/n

n−1> 1

nfor all n.

(n+1)!

‹n!en

= limn→∞n+1e

Question 4

(I )∞Xn=1

“2n2 + 7

n2 + 1

(II )∞Xn=2

n − 1(III )

∞Xn=1

2n2+7n2+1

= limn→∞2+7/n2

n=221/n

, since21/n

n−1> 1

nfor all n.

(n+1)!

‹n!en

= limn→∞n+1e

Question 4

(I )∞Xn=1

“2n2 + 7

n2 + 1

(II )∞Xn=2

n − 1(III )

∞Xn=1

2n2+7n2+1

= limn→∞2+7/n2

n=221/n

, since21/n

n−1> 1

nfor all n.

(n+1)!

‹n!en

= limn→∞n+1e

Question 4

(I )∞Xn=1

“2n2 + 7

n2 + 1

(II )∞Xn=2

n − 1(III )

∞Xn=1

2n2+7n2+1

= limn→∞2+7/n2

n=221/n

, since21/n

n−1> 1

nfor all n.

(n+1)!

‹n!en

= limn→∞n+1e

Question 5

4. Which series below is the MacLaurin series (Taylor series centered at 0) for

1 + x?

a.∞Xn=0

(−1)n xn+2 b.∞Xn=0

x2n+2 c.∞Xn=0

n + 2d.∞Xn=2

(−1)nx2n−2

e.∞Xn=0

(−1)nx2n

1+x= x2

1−(−x)= x2P∞

n=0(−x)n =P∞

n=0(−1)nxn+2, for |x | < 1.

Question 5

4. Which series below is the MacLaurin series (Taylor series centered at 0) for

1 + x?

a.∞Xn=0

(−1)n xn+2 b.∞Xn=0

x2n+2 c.∞Xn=0

n + 2d.∞Xn=2

(−1)nx2n−2

e.∞Xn=0

(−1)nx2n

1+x= x2

1−(−x)= x2P∞

n=0(−x)n =P∞

n=0(−1)nxn+2, for |x | < 1.

Question 6

6. Which series below is a power series for cos(√

∞Xn=0

(−1)nxn

∞Xn=0

(−1)n√xn

∞Xn=0

(−1)nx2n− 12

∞Xn=0

(−1)nxn

(2n + 1)!∞Xn=0

(−1)nxn

n2 + 1

I cos x =∞Xn=0

(−1)n x2n

(2n)!= 1− x2

4!− x6

6!+ . . .

I Thereforecos(√

x) =P∞

n=0(−1)n (√

(2n)!=P∞

n=0(−1)n xn

(2n)!= 1− x

2!+ x2

4!− · · · ..

Question 6

∞Xn=0

(−1)nxn

∞Xn=0

(−1)n√xn

∞Xn=0

(−1)nx2n− 12

∞Xn=0

(−1)nxn

(2n + 1)!∞Xn=0

(−1)nxn

n2 + 1

I cos x =∞Xn=0

(−1)n x2n

(2n)!= 1− x2

4!− x6

6!+ . . .

I Thereforecos(√

x) =P∞

n=0(−1)n (√

(2n)!=P∞

n=0(−1)n xn

(2n)!= 1− x

2!+ x2

4!− · · · ..

Question 6

∞Xn=0

(−1)nxn

∞Xn=0

(−1)n√xn

∞Xn=0

(−1)nx2n− 12

∞Xn=0

(−1)nxn

(2n + 1)!∞Xn=0

(−1)nxn

n2 + 1

I cos x =∞Xn=0

(−1)n x2n

(2n)!= 1− x2

4!− x6

6!+ . . .

I Thereforecos(√

x) =P∞

n=0(−1)n (√

(2n)!=P∞

n=0(−1)n xn

(2n)!= 1− x

2!+ x2

4!− · · · ..

Question 7

7. Calculate

limx→0

sin(x3)− x3

I We have sin x =∞Xn=0

(−1)n x2n+1

(2n + 1)!= x − x3

5!− · · ·

I therefore

sin(x3) =∞Xn=0

(−1)n x6n+3

(2n + 1)!= x3 − x9

5!− · · · ,

limx→0

sin(x3)− x3

x9= lim

(x3 − x9

3!+ x15

5!− · · · )− x3

= limx→0

− x9

3!+ x15

5!− · · ·

x9= −1

Question 7

7. Calculate

limx→0

sin(x3)− x3

(−1)n x2n+1

(2n + 1)!= x − x3

5!− · · ·

I therefore

sin(x3) =∞Xn=0

(−1)n x6n+3

(2n + 1)!= x3 − x9

5!− · · · ,

limx→0

sin(x3)− x3

x9= lim

(x3 − x9

3!+ x15

5!− · · · )− x3

= limx→0

− x9

3!+ x15

5!− · · ·

x9= −1

Question 7

7. Calculate

limx→0

sin(x3)− x3

(−1)n x2n+1

(2n + 1)!= x − x3

5!− · · ·

I therefore

sin(x3) =∞Xn=0

(−1)n x6n+3

(2n + 1)!= x3 − x9

5!− · · · ,

limx→0

sin(x3)− x3

x9= lim

(x3 − x9

3!+ x15

5!− · · · )− x3

= limx→0

− x9

3!+ x15

5!− · · ·

x9= −1

Question 7

7. Calculate

limx→0

sin(x3)− x3

(−1)n x2n+1

(2n + 1)!= x − x3

5!− · · ·

I therefore

sin(x3) =∞Xn=0

(−1)n x6n+3

(2n + 1)!= x3 − x9

5!− · · · ,

limx→0

sin(x3)− x3

x9= lim

(x3 − x9

3!+ x15

5!− · · · )− x3

= limx→0

− x9

3!+ x15

5!− · · ·

x9= −1

Question 8

8. Which of the following is a graph of the parametric curve defined by

x = sin (t) y = cos (3t)

for 0 ≤ t ≤ 2π?

I We check the values of x and y at some specific values of t:

I At t = 0, (x , y) = (0, 1), At t = π6

, (x , y) = ( 12, 0).

I At t = π4

, (x , y) = ( 1√2,− 1√

2), At t = π

3, (x , y) = (

2,−1).

I At t = π2

, (x , y) = (1, 0).

I Joining the dots, we see that the answer has to be III.

Question 8

for 0 ≤ t ≤ 2π?

I At t = 0, (x , y) = (0, 1), At t = π6

, (x , y) = ( 12, 0).

I At t = π4

, (x , y) = ( 1√2,− 1√

2), At t = π

3, (x , y) = (

2,−1).

I At t = π2

, (x , y) = (1, 0).

Question 8

for 0 ≤ t ≤ 2π?

I At t = 0, (x , y) = (0, 1), At t = π6

, (x , y) = ( 12, 0).

I At t = π4

, (x , y) = ( 1√2,− 1√

2), At t = π

3, (x , y) = (

2,−1).

I At t = π2

, (x , y) = (1, 0).

Question 8

for 0 ≤ t ≤ 2π?

I At t = 0, (x , y) = (0, 1), At t = π6

, (x , y) = ( 12, 0).

I At t = π4

, (x , y) = ( 1√2,− 1√

2), At t = π

3, (x , y) = (

2,−1).

I At t = π2

, (x , y) = (1, 0).

Question 8

for 0 ≤ t ≤ 2π?

I At t = 0, (x , y) = (0, 1), At t = π6

, (x , y) = ( 12, 0).

I At t = π4

, (x , y) = ( 1√2,− 1√

2), At t = π

3, (x , y) = (

2,−1).

I At t = π2

, (x , y) = (1, 0).

Question 8

for 0 ≤ t ≤ 2π?

I At t = 0, (x , y) = (0, 1), At t = π6

, (x , y) = ( 12, 0).

I At t = π4

, (x , y) = ( 1√2,− 1√

2), At t = π

3, (x , y) = (

2,−1).

I At t = π2

, (x , y) = (1, 0).

Question 9

9. Does the series∞Xn=1

converge or diverge? Show your reasoning and state clearly any theorems ortests you are using.Solution 1:

I Let an =(n!)n

n2n= (

I . Since

limn→∞

n2= lim

n→∞

n − 1

n· (n − 2)! =∞,

we have

limn→∞

an = limn→∞

´n=∞.

I Hence limn→∞

an 6= 0. By the Divergence Test, the series is divergent.

Solution 2:

I Another possibility is to use the Root Test:

limn→∞

np|an| = lim

n→∞

n2=∞.

I Since the limit is > 1, the series diverges.

Question 9

I Let an =(n!)n

n2n= (

I . Since

limn→∞

n2= lim

n→∞

n − 1

n· (n − 2)! =∞,

we have

limn→∞

an = limn→∞

´n=∞.

I Hence limn→∞

Solution 2:

limn→∞

np|an| = lim

n→∞

n2=∞.

Question 9

I Let an =(n!)n

n2n= (

I . Since

limn→∞

n2= lim

n→∞

n − 1

n· (n − 2)! =∞,

we have

limn→∞

an = limn→∞

´n=∞.

I Hence limn→∞

Solution 2:

limn→∞

np|an| = lim

n→∞

n2=∞.

Question 9

I Let an =(n!)n

n2n= (

I . Since

limn→∞

n2= lim

n→∞

n − 1

n· (n − 2)! =∞,

we have

limn→∞

an = limn→∞

´n=∞.

I Hence limn→∞

Solution 2:

limn→∞

np|an| = lim

n→∞

n2=∞.

Question 9

I Let an =(n!)n

n2n= (

I . Since

limn→∞

n2= lim

n→∞

n − 1

n· (n − 2)! =∞,

we have

limn→∞

an = limn→∞

´n=∞.

I Hence limn→∞

Solution 2:

limn→∞

np|an| = lim

n→∞

n2=∞.

Question 9

I Let an =(n!)n

n2n= (

I . Since

limn→∞

n2= lim

n→∞

n − 1

n· (n − 2)! =∞,

we have

limn→∞

an = limn→∞

´n=∞.

I Hence limn→∞

Solution 2:

limn→∞

np|an| = lim

n→∞

n2=∞.

I Since the limit is > 1, the series diverges.Annette Pilkington Solutions PE3

Question 10

10. Use the Integral Test to discuss whether the series∞Xn=1

(ln n)2

nconverges.

I Let f (x) =(ln x)2

x. It is continuous and positive when x > 1.

I Note f ′(x) =2x ln x· 1

x−(ln x)2

x2 = 2 ln x−(ln x)2

x2 = ln xx2 (2− ln x), and

2− ln x < 0, i.e., 2 < ln x , if x > e2. Then f ′(x) < 0 and hence f (x) isdecreasing for x > e2. Therefore, we can use the Integral Test.

I We compute the indefinite integralR (ln x)2

xdx using the substitution

u = ln x , gettingR (ln x)2

Ru2du = u3

3+ C = (ln x)3

3+ C .

I SoR∞

1(ln x)2

xdx = limt→∞

1(ln x)2

xdx = limt→∞

(ln x)3

˛̨̨t1

= limt→∞(ln t)3

=∞. Hence the improper integralR∞

1(ln x)2

xdx is divergent.

I By the Integral Test, the series∞Xn=1

(ln n)2

ndiverges. (Note: It is easier to

prove this series diverges using the Comparison Test, comparing to theharmonic series. But we were not at liberty to use this test.)

Question 10

(ln n)2

nconverges.

x−(ln x)2

x2 = 2 ln x−(ln x)2

Ru2du = u3

3+ C = (ln x)3

3+ C .

I SoR∞

1(ln x)2

xdx = limt→∞

1(ln x)2

xdx = limt→∞

(ln x)3

˛̨̨t1

= limt→∞(ln t)3

1(ln x)2

xdx is divergent.

(ln n)2

Question 10

(ln n)2

nconverges.

x−(ln x)2

x2 = 2 ln x−(ln x)2

Ru2du = u3

3+ C = (ln x)3

3+ C .

I SoR∞

1(ln x)2

xdx = limt→∞

1(ln x)2

xdx = limt→∞

(ln x)3

˛̨̨t1

= limt→∞(ln t)3

1(ln x)2

xdx is divergent.

(ln n)2

Question 10

(ln n)2

nconverges.

x−(ln x)2

x2 = 2 ln x−(ln x)2

Ru2du = u3

3+ C = (ln x)3

3+ C .

I SoR∞

1(ln x)2

xdx = limt→∞

1(ln x)2

xdx = limt→∞

(ln x)3

˛̨̨t1

= limt→∞(ln t)3

1(ln x)2

xdx is divergent.

(ln n)2

Question 10

(ln n)2

nconverges.

x−(ln x)2

x2 = 2 ln x−(ln x)2

Ru2du = u3

3+ C = (ln x)3

3+ C .

I SoR∞

1(ln x)2

xdx = limt→∞

1(ln x)2

xdx = limt→∞

(ln x)3

˛̨̨t1

= limt→∞(ln t)3

1(ln x)2

xdx is divergent.

(ln n)2

Question 10

(ln n)2

nconverges.

x−(ln x)2

x2 = 2 ln x−(ln x)2

Ru2du = u3

3+ C = (ln x)3

3+ C .

I SoR∞

1(ln x)2

xdx = limt→∞

1(ln x)2

xdx = limt→∞

(ln x)3

˛̨̨t1

= limt→∞(ln t)3

1(ln x)2

xdx is divergent.

(ln n)2

Question 11

111. Find the radius of convergence and interval of convergence of the powerseries

P∞n=1

(−1)n√n

(x − 3)n

I Set an =(−1)n

(x − 3)n. Using the Ratio Text,

limn→∞

|an+1||an|

= limn→∞

√n√

n + 1|x − 3| = |x − 3|.

I Hence, the radius of convergence is 1, and the series converges absolutelyfor |x − 3| < 1, or 2 < x < 4.

I For the end points, when x = 2, the series is∞Xn=1

(−1)n(−1)n

=∞Xn=1

which is divergent since it is a p-series with p = 12< 1.

I when x = 4, the series is∞Xn=1

(−1)n(1)n

which is convergent since it’s an

alternating series, and bn = 1√n

is decreasing and limn→∞

= 0. (See the

solution to Problem #3 for details.)

I Hence, the interval of convergence is 2 < x ≤ 4.

Question 11

P∞n=1

(−1)n√n

(x − 3)n

I Set an =(−1)n

limn→∞

|an+1||an|

= limn→∞

√n√

n + 1|x − 3| = |x − 3|.

(−1)n(−1)n

=∞Xn=1

(−1)n(1)n

= 0. (See the

Question 11

P∞n=1

(−1)n√n

(x − 3)n

I Set an =(−1)n

limn→∞

|an+1||an|

= limn→∞

√n√

n + 1|x − 3| = |x − 3|.

(−1)n(−1)n

=∞Xn=1

(−1)n(1)n

= 0. (See the

Question 11

P∞n=1

(−1)n√n

(x − 3)n

I Set an =(−1)n

limn→∞

|an+1||an|

= limn→∞

√n√

n + 1|x − 3| = |x − 3|.

(−1)n(−1)n

=∞Xn=1

(−1)n(1)n

= 0. (See the

Question 11

P∞n=1

(−1)n√n

(x − 3)n

I Set an =(−1)n

limn→∞

|an+1||an|

= limn→∞

√n√

n + 1|x − 3| = |x − 3|.

(−1)n(−1)n

=∞Xn=1

(−1)n(1)n

= 0. (See the

Question 11

P∞n=1

(−1)n√n

(x − 3)n

I Set an =(−1)n

limn→∞

|an+1||an|

= limn→∞

√n√

n + 1|x − 3| = |x − 3|.

(−1)n(−1)n

=∞Xn=1

(−1)n(1)n

= 0. (See the

Question 12

12. (a) Show thatP∞

n=0(−1)nx2n = 11+x2 provided that |x | < 1.

(b) FindP∞

n=0(−1)n

(2n+1)(√

3)2n+1 .

I (a) When |x | < 1 , we have |x2| < 1. Therefore1

1+x2 = 11−(−x2)

n=0(−x2)n =P∞

n=0(−1)nx2n.

I (b) Integrate both the left and right hand sides of (a) to getR P∞n=0(−1)nx2ndx =

1+x2 dx

I ThereforeP∞

R(−1)nx2ndx =

1+x2 dx

I ThereforeP∞

n=0(−1)n x2n+1

2n+1= arctan x + C .

I Letting x = 0, we have C = 0.

I Hence, we haveP∞

n=0(−1)n x2n+1

2n+1= arctan x . Let x =

I We getP∞

n=0(−1)n

(2n+1)(√

3)2n+1 = arctan 1√3

= π6.

Question 12

(b) FindP∞

n=0(−1)n

(2n+1)(√

3)2n+1 .

1+x2 = 11−(−x2)

n=0(−x2)n =P∞

n=0(−1)nx2n.

1+x2 dx

I ThereforeP∞

R(−1)nx2ndx =

1+x2 dx

I ThereforeP∞

n=0(−1)n x2n+1

I We getP∞

n=0(−1)n

(2n+1)(√

3)2n+1 = arctan 1√3

= π6.

Question 12

(b) FindP∞

n=0(−1)n

(2n+1)(√

3)2n+1 .

1+x2 = 11−(−x2)

n=0(−x2)n =P∞

n=0(−1)nx2n.

1+x2 dx

I ThereforeP∞

R(−1)nx2ndx =

1+x2 dx

I ThereforeP∞

n=0(−1)n x2n+1

I We getP∞

n=0(−1)n

(2n+1)(√

3)2n+1 = arctan 1√3

= π6.

Question 12

(b) FindP∞

n=0(−1)n

(2n+1)(√

3)2n+1 .

1+x2 = 11−(−x2)

n=0(−x2)n =P∞

n=0(−1)nx2n.

1+x2 dx

I ThereforeP∞

R(−1)nx2ndx =

1+x2 dx

I ThereforeP∞

n=0(−1)n x2n+1

I We getP∞

n=0(−1)n

(2n+1)(√

3)2n+1 = arctan 1√3

= π6.

Question 12

(b) FindP∞

n=0(−1)n

(2n+1)(√

3)2n+1 .

1+x2 = 11−(−x2)

n=0(−x2)n =P∞

n=0(−1)nx2n.

1+x2 dx

I ThereforeP∞

R(−1)nx2ndx =

1+x2 dx

I ThereforeP∞

n=0(−1)n x2n+1

I We getP∞

n=0(−1)n

(2n+1)(√

3)2n+1 = arctan 1√3

= π6.

Question 12

(b) FindP∞

n=0(−1)n

(2n+1)(√

3)2n+1 .

1+x2 = 11−(−x2)

n=0(−x2)n =P∞

n=0(−1)nx2n.

1+x2 dx

I ThereforeP∞

R(−1)nx2ndx =

1+x2 dx

I ThereforeP∞

n=0(−1)n x2n+1

I We getP∞

n=0(−1)n

(2n+1)(√

3)2n+1 = arctan 1√3

= π6.

Question 12

(b) FindP∞

n=0(−1)n

(2n+1)(√

3)2n+1 .

1+x2 = 11−(−x2)

n=0(−x2)n =P∞

n=0(−1)nx2n.

1+x2 dx

I ThereforeP∞

R(−1)nx2ndx =

1+x2 dx

I ThereforeP∞

n=0(−1)n x2n+1

I We getP∞

n=0(−1)n

(2n+1)(√

3)2n+1 = arctan 1√3

= π6.

Question 12

(b) FindP∞

n=0(−1)n

(2n+1)(√

3)2n+1 .

1+x2 = 11−(−x2)

n=0(−x2)n =P∞

n=0(−1)nx2n.

1+x2 dx

I ThereforeP∞

R(−1)nx2ndx =

1+x2 dx

I ThereforeP∞

n=0(−1)n x2n+1

I We getP∞

n=0(−1)n

(2n+1)(√

3)2n+1 = arctan 1√3

= π6.

› ~apilking › math10560 › work › old exams › practice … · question 1question 2question...

Documents