Name:
Id #:
Math 153 (26) - Midterm Test 2
Spring Quarter 2016
Thursday, November 10, 2016 - 09:00 am to 09:50 am
Instructions:
Prob.
Points Score
possible
1
14
2
18
3
8
TOTAL
40
• Read each problem carefully.
• Write legibly.
• Show all your work on these sheets. Feel free to
use the opposite side.
• This exam has 6 pages, and 3 problems. Please
make sure that all pages are included.
• You may not use books, notes, calculators, etc.
Cite theorems from class or from the texts as appropriate.
• Proofs should be presented clearly (in the style
used in lectures) and explained using complete
English sentences.
Good luck!
Math 153 (26) - Midterm Test 2
Spring Quarter 2016
Page 2 of 6
Question 1. (Total of 14 points)
a) (6 points) Show that 5 divides 8n − 3n for for any n ∈ Z+ .
Solution. New question. We use induction on n. For n = 1 note that 81 −31 = 5
which is clearly divisible by 5 and so the result holds in this case. Assume, by way
of induction hypothesis, that 5 divides 8n − 3n for some n ∈ Z+ . Now consider
8n+1 − 3n+1 = 8(8n − 3n ) + 5 · 3n .
By the induction hypothesis, 8n − 3n = 5k for some k ∈ Z and so
8n+1 − 3n+1 = 5(8k + 3n )
is divisible by 5. Hence, by the principle of mathematical induction, the result
holds for all k ∈ Z+ .
b) (8 points) Consider the recursively defined sequence
a1 = 4
√
an+1 = 2an + 4 for all n ∈ Z+
Show this sequence converges and find the value of the limit.
Solution. Similar to homework. We use induction to show that the sequence
√
an+1 ≥ an and an ≤ 8 for all n ∈ Z+ . When n = 1 we have a2 = 2 · 4 + 4 >
4 = a1 and a1 = 4 ≤ 8 and so the result holds in this case.
Assume, by way of induction hypothesis, that an+1 ≥ an and an ≤ 8 for some
n ∈ Z+ . Now consider
p
√
an+2 = 2an+1 + 4 ≥ 2an + 4 = an+1 ,
where the above inequality is by the induction hypothesis. On the other hand,
√
√
an+1 = 2an + 4 ≤ 16 + 4 = 8 by the induction hypothesis. Hence by the
principle of mathematical induction, the result holds for all n ∈ Z+ .
By the monotone convergence theorem, the sequence (an )∞
n=1 must converge to
some limit ` ≥ 4. Note that
√
√
` = lim an+1 = lim 2an + 4 = 2` + 4
n→∞
n→∞
by the continuity of the square root function. Hence,
√
2` = ` − 4
and squaring both sides yields 2` = `2 −8`+16. Thus, ` is a root of the polynomial
(x − 2)(x − 8) and so ` = 2 or ` = 8. Since ` ≥ 4, we conclude that ` = 8.
Math 153 (26) - Midterm Test 2
Spring Quarter 2016
Page 3 of 6
Question 2. (Total of 18 points)
a) (2 points) Let (an )∞
n=1 be a sequence of real numbers. State what it means for the
P∞
series k=1 ak to converge.
Solution. Bookwork. The series
P
sums nk=1 ak converges.
P∞
k=1
ak converges if the sequence of partial
b) (8 points) Let (an )∞
n=1 be a non-negative, non-increasing sequence and suppose
∞
∞
X
X
k
that
2 a2k converges. Prove that the series
ak also converges.
k=0
k=1
Solution. Bookwork. Since
∞
X
2k a2k converges, by the boundedness theorem
k=0
there exists some M > 0 such that
n
X
2k a2k ≤ M
for all n ∈ Z+ .
k=0
Observe that, since the sequence is non-increasing,
2n+1
X−1
ak = a1 + (a2 + a3 ) + (a4 + a5 + a6 + a7 ) + · · · + (a2n + · · · + a2n+1 −1 )
k=1
≤ a1 + (a2 + a2 ) + (a4 + a4 + a4 + a4 ) + · · · + (a2n + · · · + a2n )
n
X
n
= a1 + 2a2 + 4a4 + · · · + 2 a2n =
2k a2k ≤ M.
k=0
Hence, again since the sequence is non-increasing,
n
X
ak ≤
k=1
2n+1
X−1
k=1
and so the sequence of partial sums for
edness theorem
∞
X
k=1
∞
X
k=1
ak converges.
ak ≤ M
ak is bounded. Hence, by the bound-
Math 153 (26) - Midterm Test 2
Spring Quarter 2016
Page 4 of 6
c) (8 points) Determine whether the following series converge or diverge.
i)
∞
X
√ k
√
k+2− k
k=1
ii)
∞
X
k!
2k2
k=1
You may use without proof the fact k/2k → 0 as k → ∞.
Solution. Similar to homework.
√ k
√
i) Let ak :=
k + 2 − k for k ≥ 1 and consider the kth roots
1/k
ak
=
√
k+2−
√
√ √
√
√
( k + 2 − k)( k + 2 + k)
2
√
√ .
k=
=√
√
k+2+ k
k+2+ k
Hence,
1/k
0 ≤ ak
1
≤√
k
1/k
and so, by the squeeze theorem, ak → 0 as k → ∞. Thus, by the root test
the series converges.
k!
ii) Let ak := k2 for k ≥ 1 and consider the ratios
2
2
k+1
ak+1
(k + 1)! 2k
= 2k+1 .
= (k+1)2
ak
k!
2
2
Hence,
0≤
and so, by the squeeze theorem,
the series converges.
ak+1
k
≤ k
ak
2
ak+1
→ 0 as k → ∞. Thus, by the ratio test
ak
Math 153 (26) - Midterm Test 2
Spring Quarter 2016
Page 5 of 6
Question 3. (Total of 8 points)
1. (5 points) Let a, b ∈ R3 with a, b 6= 0. Show that
kprojb ak ≤ kak.
Draw a picture and give a simple geometric interpretation of this result.
Solution. Similar to homework / examples in class. Recall that
projb a = a ·
and so
b b
kbk kbk
b b = a · b .
kprojb ak = a ·
kbk kbk
kbk
By the Cauchy-Schwarz inequlity,
a · b ≤ kak b = kak,
kbk
kbk
as required.
Geometrically, the identity tells us that the length of the projection of a onto b
is at most the length of a, which is clear from the picture below.
2. (3 points) Show that
a·b=
1
ka + bk2 − ka − bk2
4
for all a, b ∈ R3
Math 153 (26) - Midterm Test 2
Spring Quarter 2016
Page 6 of 6
Solution. Similar to homework / examples in class. From the properties of the dot
product we know that
ka + bk2 = kak2 + 2a · b + kbk2
ka − bk2 = kak2 − 2a · b + kbk2
so taking the difference of these identities gives
ka + bk2 − ka − bk2 = 4a · b,
as required.