Download Solutions to Math 412 Week #7 Practice Quiz: Limits of Complex Sequences - Prof. Scott Ann and more Quizzes Mathematics in PDF only on Docsity!
March 9-13, 2009 Week #7 Practice Quiz Name:
Math 412
SOLUTIONS
Problem 1. (4 points each) Evaluate each limit below, if it exists. If it does not exist,
explain why not.
(a):
lim n→∞
8 − 3 i
)n
SOLUTION: DOES NOT CONVERGE. Since
∣ ∣ ∣ ∣
8 − 3 i
we have that (^) ∣
∣ ∣ ∣
8 − 3 i
n
is unbounded as n → ∞. Hence, the sequence
∣ 8 −^3 i 7
n}^ cannot converge, and thus,
8 − 3 i 7
)n}
cannot converge.
(b):
lim n→∞
(3ni − 4) 3
(4i + 2)n 3 − 3 n 2
SOLUTION: Dividing all terms of both the numerator and denominator by n
3 , the resulting
fraction contains terms that all tend to zero except for
(3i)
3
4 i + 2
− 27 i
4 i + 2
27 i(2 − 4 i)
so this is the answer.
Problem 2. (5 points) Suppose that
∞ ∑
n=
zn and
∞ ∑
n=
wn converge. Prove that
∞ ∑
n=
(zn + wn)
converges, and ∞ ∑
n=
(zn + wn) =
∞ ∑
n=
zn +
∞ ∑
n=
wn.
SOLUTION: Assume that
∞ ∑
n=
zn = ζ and
∞ ∑
n=
wn = ψ.
We claim that ∞ ∑
n=
(zn + wn) = ζ + ψ.
Let > 0 be given. By definition, there exists N 1 ∈ N such that n > N 1 implies that |(z 1 + z 2 +... +
zn) − ζ| <
, and there exists N 2 ∈ N such that n > N 2 implies that |(w 1 + w 2 +... + wn) − ψ| <
Letting N := max{N 1 , N 2 }, we see that whenever n > N ,
|(z 1 + w 1 ) + (z 2 + w 2 ) +... + (zn + wn) − (ζ + ψ)| = |(z 1 + z 2 +... + zn) − ζ + (w 1 + w 2 +... + wn) − ψ|
≤ |z 1 + z 2 +... + zn − ζ| + |w 1 + w 2 +... + wn − ψ| <
Problem 3. (7 points) Let {αn} and {βn} be sequences of complex numbers such that
∞ ∑
n=
|αn|
2 and
∞ ∑
n=
|βn|
2 converge. Show that
∞ ∑
n=
αnβn converges.
SOLUTION: Observe that
0 ≤ (|αn| − |βn|)
2 = |αn|
2 − 2 |αn||βn| + |βn|
2 ,
so that
0 ≤ 2 |αn||βn| ≤ |αn|
2
2 .
Hence,
∞ ∑
n=
|αn||βn| ≤
∞ ∑
n=
|αn|
2
∞ ∑
n=
|βn|
2 .
Using the fact that the series of non-negative real numbers on the right (consisting of two convergent
series of real numbers) converges, the Comparison Test implies that the series in the middle con-
verges. Dividing by 2, we conclude that
∞ ∑
n=
|αn||βn| converges. This shows that
∞ ∑
n=
αnβn converges
absolutely, hence converges.
