MAT 137Y: Calculus with proofs
Test 4 - Part B - Sample solutions
QUESTION 1
Q1 - The problem
For which values of a, b > 0 are the following integrals convergent?
(a)
∫ ∞
1
1
xa(1 + xb)
dx
(b)
∫ 1
0
1
xa(1 + xb)
dx
(c)
∫ ∞
0
1
xa(1 + xb)
dx
Q1 - Important notes
• The behaviour of powers of x is different as x→∞ and as x→ 0+:
lim
x→∞
x + 2x2
3x + 5x5
=
2
5
, but lim
x→0+
x + 2x2
3x + 5x5
=
1
3
In other words, in the sum xa + xa+b ...
– ... the “dominant” term is xa+b as x→∞
– ... the “dominant” term is xa as x→ 0+
• It is not enough to prove the integral is convergent for some values of a and b.
You also need to prove that it isn’t convergent for the rest of the values.
Q1 - Solution
(a)
∫ ∞
1
1
xa(1 + xb)
dx is convergent iff a + b > 1.
Proof: The integral is improper at ∞. I want to compare it with
∫ ∞
1
1
xa+b
dx.
lim
x→∞
1
xa(1 + xb)
1
xa+b
= lim
x→∞
xa+b
xa + xa+b
= lim
x→∞
1
1
xb
+ 1
= 1
By LCT,
∫ ∞
1
1
xa(1 + xb)
dx is convergent iff
∫ ∞
1
1
xa+b
dx is convergent.
We know the second integral is convergent iff a + b > 1.
(b)
∫ 1
0
1
xa(1 + xb)
dx is convergent iff a < 1.
Proof: The integral is improper at 0+. I want to compare it with
∫ 1
0
1
xa
dx.
lim
x→0+
1
xa(1 + xb)
1
xa
= lim
x→0+
1
1 + xb
= 1
By LCT,
∫ 1
0
1
xa(1 + xb)
dx is convergent iff
∫ 1
0
1
xa
dx is convergent.
We know the second integral is convergent iff a < 1.
(c)
∫ ∞
0
1
xa(1 + xb)
dx is convergent iff (a + b > 1 AND a < 1 )
We can also write the condition as 1− b < a < 1.
Proof: ∫ ∞
0
1
xa(1 + xb)
dx =
∫ 1
0
1
xa(1 + xb)
dx +
∫ ∞
1
1
xa(1 + xb)
dx
The first integral is convergent if and only if the other two integrals are convergent independently.
Then I use the result from Questions 1a and 1b.
QUESTION 2
Q2 - The problem
Prove that IF a sequence is divergent to ∞, THEN it is bounded below.
Suggestion: You know that every convergent sequence is bounded. Revisit the proof.
Q2 - Important notes
This proof is extremely similar to the one in Video 11.5. You only need to modify a few things. You
won’t get any credit for copying down the proof of Video 11.5. You only get credit for understanding
what needs to be modified and writing it correctly. This means you could write a proof that is partially
right and still get 0 points.
In particular, if you write something like...
“The sequence is divergent to ∞ so
∀M ∈ R, ∃n0 ∈ N, ∀n ∈ N, n ≥ n0 =⇒ an ≥M
“Take A = min{a0, a1, a2, . . . , an0−1,M}”
... then your proof is incorrect. You need to fix a value of M first. Otherwise your variable M
is quantified – it is a dummy variable, and it does not mean anything. Moreover, the value of n0
depends on M . The above set does not make sense (or is not finite) unless we fix one single value of
M and as a consequence one single value of n0.
Q2 - Solution
As per the suggestion, watch the proof in Video 11.5. This one is very similar.
• Let {an}∞n=0 be a sequence. Assume it is divergent to ∞. I want to prove that it is bounded
below. In other words, I want to prove that
∃A ∈ R, ∀n ∈ N, A ≤ an
• By assumption {an}∞n=0 is divergent to∞. Using 42 as the bound in the definition of “divergent
to ∞”, I know ∃n0 ∈ N such that
∀n ∈ N, n ≥ n0 =⇒ an ≥ 42 (1)
Notice that I chose 42 at random. I could have chosen any other number as the bound. The
point is to fix one.
• Next I choose
A = min{a0, a1, a2, . . . , an0−1, 42} (2)
I will now prove that this value of A satisfies
∀n ∈ N, A ≤ an.
• Let n ∈ N. There are two cases to consider:
– If n < n0, then A ≤ an by the way we defined A in (2)
– If n ≥ n0, then an ≥ 42 from (1). Therefore A ≤ 42 ≤ an
Either way, I have concluded that A ≤ an, which is what I wanted to show.





















































































































































































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