Limit - Calculus Two - Solved Exam, Exams for Calculus. Biju Patnaik University of Technology, Rourkela
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naghm23 February 2013

Limit - Calculus Two - Solved Exam, Exams for Calculus. Biju Patnaik University of Technology, Rourkela

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This is the Solved Exam of Calculus Two which includes Volume Obtained, Area, Firstquadrant, Evaluate, Integrals Necessary, Region, Disc, Shell Method, Washer Method etc. Key important points are: Limit, First Four Terms...
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Solution: APPM 1360 Exam 3 Spring 2011

1. (18 pts) Answer the following questions, justify your answers.

(a) (6 pts) Evaluate the limit using series: lim x→0

cos(x2)− 1 x4

(b) (6 pts) Find the first four terms of the Maclaurin series of f(x) = x2√ 3 + x

, simplify your answer.

(c) (6 pts) Find the Maclaurin series of f(x) = x3 sin(3x), give your answer in ∑

-notation.

Solution:

(a) Since cos(x) =

∞∑ n=0

(−1)nx2n

(2n)! , we have cos(x2) =

∞∑ n=0

(−1)nx4n

(2n)! = 1− x

4

2! +

x8

4! − x

12

6! + · · · , so

lim x→0

cos(x2)− 1 x4

= lim x→0

( 1− x42! +

x8

4! − x12

6! + · · · ) − 1

x4 = lim

x→0 − 1

2! +

x4

4! − x

8

6! + · · · = −1

2

(b) NOTE: Problem should have said “Find the first four nonzero terms of ...”

Now, here f(x) = x2√

3 (1 + x/3)−1/2 and

(1 + x/3)−1/2 = ∞∑ n=0

( −12 n

) (x/3)n =

( −12 0

) +

( −12 1

) x

3 +

( −12 2

)(x 3

)2 +

( −12 3

)(x 3

)3 + · · ·

= 1− 1 2 · x

3 +

(−1/2)(−1/2− 1) 2!

· x 2

9 − (−1/2)(−1/2− 1)(−1/2− 2)

3! · x

3

27 + · · ·

= 1− x 6

+ x2

24 − 5x

3

432 + · · ·

and so the first four nonzero terms of the Maclaurin series are,

x2√ 3

(1 + x/3)−1/2 ≈ x 2

√ 3 − x

3

6 √

3 +

x4

24 √

3 − 5x

5

432 √

3

(c) Since sin(x) = ∞∑ n=0

(−1)nx2n+1

(2n + 1)! , we have

x3 sin(3x) = x3 ∞∑ n=0

(−1)n(3x)2n+1

(2n + 1)! =

∞∑ n=0

(−1)n32n+1x2n+4

(2n + 1)!

2. (12 pts) Show all work and justify your answers. Consider the series

∞∑ n=1

(−1)n(x + 2)n

n2 ,

(a) For what values of x does this series converge conditionally?

(b) For what values of x does this series converge absolutely?

(c) For what values of x does this series diverge?

Solution:

(a) We do the Ratio Test,

lim n→∞

∣∣∣∣(x + 2)n+1(n + 1)2 · n2(x + 2)n ∣∣∣∣ = |x + 2| limn→∞ n2n2 + 2n + 1 L′H= |x + 2|

and so |x + 2| < 1 which implies −3 < x < −1.

Check endpoints:

(∗) If x = −3, we have ∞∑ n=1

(−1)n(−1)n

n2

anbn=(ab)n︷︸︸︷ =

∞∑ n=1

(1)n

n2 =

∞∑ n=1

1

n2 which is a (absolutely) con-

vergent p-series.

(∗) And if x = −1, we have ∞∑ n=1

(−1)n

n2 , which is absolutely convergent since

∞∑ n=1

∣∣∣∣(−1)nn2 ∣∣∣∣ = ∞∑

n=1

1

n2 .

So the power series is NOT conditionally convergent for any values of x.

(b) The power series is absolutely convergent for −3 ≤ x ≤ −1 (see work from part (a).)

(c) The power series is divergent for (−∞,−3) ∪ (−1,+∞).

3. (15 pts) Do the following series converge or diverge? Justify your answer, name any test that you use.

(a) ∞∑ n=2

ln(n)

ln(3n) (b)

∞∑ n=1

√ 4n

n2 − 5 (c)

∞∑ n=1

ln(5n)

7n

Solution:

(a) Note that lim n→+∞

ln(n)

ln(3n)

L′H = lim

n→+∞

1/n

3/3n = 1 6= 0, so the series diverges by the Divergence Test.

(b) Note that for large n we have

√ 4n

n2 − 5 ≈ √ n

n2 =

1

n3/2 and

lim n→∞

√ 4n

n2 − 5 · n

3/2

1 = lim

n→∞

2n2

n2 − 5 L′H = 2.

Now, since 0 < 2 < ∞ and ∞∑ n=3

1

n3/2 is a convergent p-series we have that

∞∑ n=3

√ 4n

n2 − 5 converges and

so

∞∑ n=1

√ 4n

n2 − 5 converges by Limit Comparison Test.

(c) Note that for n ≥ 1 we have ln(5n) > ln(e) = 1, so ∞∑ n=1

ln(5n)

7n >

∞∑ n=1

1

7n , now note

∞∑ n=1

1

7n is

divergent since it is a non-zero constant multiple of the harmonic series, which is a divergent p-series.

And so ∞∑ n=1

ln(5n)

7n diverges by Direct Comparison Test.

4. (18 pts) Do the following series converge or diverge? If the series converges, does it converge condi- tionally or absolutely? Justify your answer, name any test that you use.

(a) ∞∑ n=1

(−1)n 1 4n− 5

(b) ∞∑ n=1

(−1)n 4 n

n! (c)

∞∑ n=1

(−1)n n! 7n

Solution:

(a) Note that this is only an alternating series for n > 1. Now note ∞∑ n=1

∣∣∣∣ (−1)n4n− 5 ∣∣∣∣ = ∞∑

n=1

1

4n− 5 ,

which is divergent by Limit Comparison Test with

∞∑ n=1

1

n . Now, since

∞∑ n=2

(−1)n

4n− 5 is an alternating

series with bn = 1

4n− 5 , which satisfies bn+1 ≤ bn and lim

n→∞ bn = 0, it is convergent and therefore

∞∑ n=1

(−1)n

4n− 5 is also convergent, and so

∞∑ n=1

(−1)n

4n− 5 is conditionally convergent.

(b) Note that,

lim n→∞

∣∣∣∣an+1an ∣∣∣∣ = limn→∞ 4n+1(n + 1)! · n!4n = limn→∞ 4(n + 1) = 0 < 1

and so the series is absolutely convergent by the Ratio Test.

(c) Note that lim n→∞

(−1)n n! 7n

= lim n→∞

(−1)n (n− 1)! 7

6= 0, and so the series diverges by Divergence Test.

5. (17 pts) Answer the following questions with as much detail as possible.

(a) (6 pts) What function does the series ∞∑ n=0

3nx3n+2

n! represent?

(b) (6 pts) What number does the series ∞∑ n=0

(−1)n

n! converge to?

(c) (5 pts) If we use the first four terms of the series to approximate the number in part (b), what is the error bound? Is this approximation an underestimate or an overestimate? Justify your answer.

Solution:

(a) Note that

∞∑ n=0

3nx3n+2

n! =

∞∑ n=0

3nx3nx2

n! = x2

∞∑ n=0

(3x3)n

n! = x2e3x

3 since

∞∑ n=0

xn

n! = ex for all x.

(b) Since ∞∑ n=0

xn

n! = ex for all x, it follows that

∞∑ n=0

(−1)n

n! = e−1.

(c) If we approximate e−1 ≈ 1−1+ 1 2! − 1

3! , then by the Alternating Series Error Estimation Theorem

we have |error| ≤ 1 4!

= 1

24 , and since the first unused term was positive, we have an underestimate.

6. (20 pts–4pts ea.) Answer “Always True” or “False” . You do NOT need to justify your answer.

(a) If the sequence {an}∞n=1 converges to zero, then the series ∞∑ n=1

an will converge.

(b) If you show that 0 ≤ bn ≤ an for all n, and the series ∞∑ n=1

bn converges then

∞∑ n=1

an converges.

(c) If ∞∑ n=1

an and ∞∑ n=1

bn are both divergent series, then ∞∑ n=1

an + bn will also be divergent.

(d) If ∞∑ n=1

bn7 n converges then

∞∑ n=1

bn(−5)n also converges.

(e) If 0 ≤ an ≤ bn for all n, and the sequence {bn}∞n=1 converges then the sequence {an} ∞ n=1 also

converges.

Solution:

(a) F (b) F (c) F (d) A.T. (e) F

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