University of Toronto

MAT137Y1 – Calculus!

Test 4 – 23 March 2018

Sample solutions

1

1. [5 points] Calculate ∫ ∞0

1

ex + e−xdx

Hint: A substitution may help.

Your answer:π

4

Solution. We have ∫ ∞0

1

ex + e−xdx = lim

a→∞

∫ a

0

1

ex + e−xdx.

We can rewrite the integral as ∫ a

0

1

ex + e−xdx =

∫ a

0

ex

e2x + 1dx.

This suggests making the substitution t = ex. Then dt = exdx and the integral becomes∫ a

0

ex

e2x + 1dx =

∫ ea

1

dt

t2 + 1= arctan t

∣∣∣ea1

= arctan(ea)− arctan(1) = arctan(ea)− π

4.

It then follows that∫ ∞0

1

ex + e−xdx = lim

a→∞

(arctan(ea)− π

4

)=π

2− π

4=π

4.

2

2. [5 points] Let R be the bounded region in the first quadrant bounded by the curves y = x andy = x2. Compute the volume of the solid of revolution obtained by rotating R around the liney = 1.

Your answer:π

5

Solution. First of all, observe that x = x2 if and only if x = 0 or x = 1. We want to revolvethe shaded region

0.2 0.4 0.6 0.8 1.0

0.2

0.4

0.6

0.8

1.0

about the line y = 1. At a point x, note that the cross-section is an annulus of inner radius1− x and outer radius 1− x2. Therefore by the washer method the volume is given by

V =

∫ 1

0

π[(1− x2)2 − (1− x)2]dx = π

∫ 1

0

(x4 − 3x2 + 2x)dx =π

5.

Alternatively, we can slice this solid using y as a variable (cylindrical shells). We get:

V =

∫ 1

0

2π(1− y) [√y − y] dy = 2π

∫ 1

0

[y1/2 − y − y3/2 + y2

]dy =

π

5

3

3. [3 points] Give the equation of a function g that satisfies all the following properties at once:

• ∀a > 1, limx→∞

g(x)

xa= 0

• ∀a ≤ 1, limx→∞

g(x)

xa=∞

Your answer: g(x) = x lnx

Solution. We can choose for instance g(x) = x lnx.

• When a > 1, we have a− 1 > 0 and then

limx→∞

x lnx

xa= lim

x→∞

lnx

xa−1= 0

from the Big Theorem.

• When a < 1, we have 1− a > 0 and then

limx→∞

x lnx

xa= lim

x→∞x1−a lnx =∞

• Finally, when a = 1, we have

limx→∞

x lnx

xa= lim

x→∞lnx = ∞

NOTE: It makes no sense to choose a function g that depends on a. The question asks for onefunction g that satisfies all the properties at once.

4

4. [9 points] We want to compute the sum of the series∞∑n=1

1

4n2 − 1

(a) [2 points] Compute the first three partial sums.

Your answer:1

3,

2

5,

3

7

Solution. We have

s1 =1

4− 1=

1

3,

s2 =1

3+

1

15=

2

5,

s3 =2

5+

1

35=

3

7.

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(b) [5 points] Obtain and prove a formula for the value of the k-th partial sum.

Suggestion: One option is to make a conjecture looking at the first three partial sums,then prove it by induction.

Proof #1. We prove by induction that

sk =k

2k + 1, k = 1, 2, . . .

• Base case: s1 = 1/3.

• Induction step: Let k ≥ 1 and assume that

sk =k

2k + 1, k ≥ 1.

I want to prove that sk+1 =k + 1

2(k + 1) + 1Then

sk+1 = sk +1

4(k + 1)2 − 1=

k

2k + 1+

1

(2k + 1)(2k + 3)=

k(2k + 3) + 1

(2k + 1)(2k + 3)

=2k2 + 3k + 1

(2k + 1)(2k + 3)=

(k + 1)(2k + 1)

(2k + 1)(2k + 3)=

k + 1

2k + 3=

k + 1

2(k + 1) + 1,

as required.

Proof #2. This series is telescopic! We can prove that sk =k

2k + 1for all k ≥ 1 directly.

sk =k∑

n=1

1

4n2 − 1=

k∑n=1

1

2

[1

2n− 1− 1

2n+ 1

]=

1

2

[k∑

n=1

1

2n− 1

]− 1

2

[k∑

n=1

1

2n+ 1

]

=1

2

[1 +

1

3+

1

5+ . . .+

1

2k − 1

]− 1

2

[1

3+

1

5+ . . .+

1

2k − 1+

1

2k + 1

]=

1

2

[1− 1

2k + 1

]=

k

2k + 1

6

(c) [2 points] Use the definition of series to compute the value of∞∑n=1

1

4n2 − 1.

Solution. We have

∞∑n=1

1

4n2 − 1= lim

n→∞sn = lim

n→∞

n

2n+ 1=

1

2.

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5. [8 points] Determine whether each of the following series is convergent or divergent. Circleyour final answer and justify it.

Note: You will only get points for a correct answer with a correct justification. A wrongjustification or no justification will get you 0 points.

(a) [2 points]∞∑n=1

√n6 + 2n+ 1

n5 + 3n3 + 2Circle one: Convergent Divergent

Solution. We compare the given series with the series∞∑n=1

1

n2. Observe that

limn→∞

√n6 + 2n+ 1

n5 + 3n3 + 21

n2

= limn→∞

n2

(√n6 + 2n+ 1

n5 + 3n3 + 2

)= lim

n→∞

√1 + 2

n5 + 1n6

1 + 3n2 + 2

n5

= 1.

Since the series∞∑n=1

1

n2converges we conclude that the given series also converges by the

Limit Comparison Test.

(b) [2 points]∞∑n=1

(−1)n

arctannCircle one: Convergent Divergent

Solution. The limit

limn→∞

(−1)n

arctann

does not exist and the series diverges by the Divergence Test (or the “Necessary Conditiontest”). More specifically,

limn→∞

1

arctann=

2

π.

Even more specifically, for n = 2k

limk→∞

(−1)2k

arctan(2k)= lim

k→∞

1

arctan(2k)=

2

π,

whereas for n = 2k + 1

limk→∞

(−1)2k+1

arctan(2k + 1)= lim

k→∞

−1

arctan(2k + 1)= − 2

π,

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(c) [2 points]∞∑n=1

3n (n!)2

(2n)!Circle one: Convergent Divergent

Solution. We apply the Ratio Test

limn→∞

3n+1((n+ 1)!)2

(2(n+ 1))!

3n(n!)2

(2n)!

= limn→∞

(3n+1(n+ 1)!(n+ 1)!

(2n+ 2)!

)((2n)!

3nn!n!

)

= limn→∞

3(n+ 1)2

(2n+ 2)(2n+ 1)=

3

4< 1.

So the given series converges by the Ratio test.

(d) [2 points]∞∑n=2

1

n (lnn)2Circle one: Convergent Divergent

Solution. Here we apply the Integral Test. Consider the function

f(x) =1

x(lnx)2, x ≥ 2.

Then f(n) = an. Clearly f is continuous. Since x ≥ 2, we have lnx > 0, so f ispositive. The function x(lnx)2 is increasing and so f is decreasing. So the hypotheses ofthe integral test are satisfied. Therefore the series

∞∑n=2

1

n (lnn)2

is convergent if and only if the improper integral∫ ∞2

1

x (lnx)2dx

is convergent.

But we can prove the improper integral is convergent (for the proof, see Q1a of PS9), sothe series is convergent as well.

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6. [5 points] Here is a new definition. We say that the sequence

{an}∞n=1 = a1, a2, a3, a4, . . .

is eventually bounded above when ∃m ∈ N such that the sequence

{an}∞n=m = am, am+1, am+2, am+3, . . .

is bounded above.

Is the theorem below true or false?

Circle one: True False

Theorem. IF a sequence is eventually bounded above,THEN it is bounded above.

If true, prove it. If false, give a counterexample.

Proof.

• Let {an}∞n=1 be a sequence that is eventually bounded above, i.e. there exists m ∈ N suchthat the sequence

{an}∞n=m = am, am+1, am+2, am+3, . . .

is bounded above. Since the sequence {an}∞n=m is bounded above there exists a numberM such that

an ≤M for all n ≥ m.

• Now takeM̃ = max{M,a1, . . . , am−1}.

We will show that the sequence {an}∞n=1 is bounded above by M̃ .

• Let n ≥ 1. We need to prove that an ≤ M̃ .

– If n < m, then an ∈ {a1, . . . , am−1}, so an ≤ M̃ by definition of M̃ .

– On the other hand, for n ≥ m, we have an ≤M ≤ M̃

In either case we have proven that an ≤ M̃ as needed. This completes the proof.

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