Math F401: Homework 4
Due: September 29, 2016
1. Abbott 2.2.6
Show that limits, if they exist, must be unique. In other words, assume lim a n = l1 and
lim a n = l2 , and prove l1 = l2 .
Solution:
Let є > 0. Since lim a n = l1 , there exists an N1 such that if n ≥ N, then ∣a n − l1 ∣ < є.
Similarly, there is an N2 such that if n ≥ N2 , then ∣a n − l1 ∣ < є. Let N = max(N1 , N2 ). Then
∣l1 − l2 ∣ = ∣l1 − x N + x N − l2 ∣ ≤ ∣l1 − x N ∣ + ∣x N − l2 ∣ < є + є.
Hence,
∣l1 − l2 ∣ < 2є
for every є > 0. By Theorem 1.2.6, we conclude that l1 = l2 .
2. Abbott 2.3.1
Let x n ≥ 0 for all n ∈ N.
√
a) If (x n ) → 0, show that ( x n ) → 0.
√
√
b) If (x n ) → x, show that ( x n → x.
Solution, part a:
Let є > 0. Pick N so that if N ≥ n, √
then ∣x n ∣ < є2 . We have shown in class that if 0 ≤ x < y,
√
√
then 0 ≤ x < y. In particular, ∣x n ∣ < є if n ≥ N. So
√
√
√
√
∣ x n − 0∣ = x n = ∣x n ∣ ≤ є2 = є
√
if n ≥ N. Hence lim x n = 0.
Solution, part b:
√
Suppose (x n ) → x ≠ 0. Let є > 0. Pick N so that if n ≥ N then ∣x n − x∣ < є x. Then
√
√ √
√
∣x n − x∣
xn − x
∣ x n − x∣ = ∣ √
√ ≤ √
< є x/ x = є∣ .
xn + x
x
3. Abbott 2.3.3 Show that if x n ≤ y n ≤ z n for all n ∈ N, and if lim x n = lim z n = l, then
lim y n = l as well.
Solution:
We need to show that lim y n exists and that it equals l. To do this, let є > 0. There exists
N1 so that if n ≥ N1 , then l − є < x n < l + є, and there exists N2 so that if n ≥ N2 , then
l − є < z n < l + є. Let N = max(N1 , N2 ). Then if n ≥ N we have
l − є < x n ≤ y n ≤ z n < l + є.
Hence, if n ≥ N, then
∣y n − l∣ < є.
Math F401: Homework 4
Due: September 29, 2016
In conclusion, given є > 0, there exists N so that if n ≥ N, then ∣y n − l∣ < є. In other
words, lim y n = l.
4. Abbott 2.3.5 (W) (Hand this one in to David.)
Let (x n ) and (y n ), and define (z n ) to be the shuffled sequence (x1 , y1 , x2 , y2 , . . .). Prove
that (z n ) is convergent if an only if (x n ) and (y n ) are both convergent with lim x n =
lim y n .
Solution:
Suppose that (x n ) and (y n ) are both convergent with lim x n = lim y n = L. Let є > 0. There
exists an N ∈ N such that if n ≥ N, then ∣x n − L∣ < є and ∣y n − L∣ < є. (We have proved this
carefully before by finding an N1 for the sequence (x n ) and an N2 for the sequence (y n )
and letting N = max(N1 , N2 ). Having seen this argument a couple of times, we can now
safely skip over it). But then if n ≥ 2N we have z n is equal to some x k or y k with k ≥ N
(indeed if n is even then z n = y n/2 and if n is odd then z n = x(n+1)/2 ). Hence, ∣z n − L∣ < є
and we conclude that lim z n = L.
Conversely, suppose lim z n = L for some number L. We wish to show that limn→∞ x n = L
and limn→∞ y n = L.
Let є > 0. Pick N so that if n ≥ N, then ∣z n − L∣ < є.
Suppose n ≥ N. Then x n = z2n . Since 2n ≥ n ≥ N, we know that
∣z2n − L∣ < є.
Hence if n ≥ N, then
∣x n − L∣ = ∣z2n − L∣ < є
and limn→∞ x n = L.
Again, suppose n ≥ N. Then y n = z2n+1 . Since 2n + 1 > n ≥ N, we know that
∣z2n+1 − L∣ < є.
Hence if n ≥ N, then
∣y n − L∣ = ∣z2n+1 − L∣ < є
and limn→∞ y n = L.
5. Abbott 2.3.6
√
Consider the sequence b n = n − n2 + 2n. Compute its limit.
Solution:
Observe that
bn =
2n
2
√
√
=
.
n + n2 + 2n 1 + 1 + (2/n)
Then
lim b n = 2
as a consequence of Problem 2.3.1, the Algebraic Limit Theorem, and the fact lim 1/n = 0.
2
Math F401: Homework 4
Due: September 29, 2016
6. Abbott 2.3.9 (a)
Suppose (a n ) is bounded and b n → 0. Show a n b n → 0 and explain why the Algebraic
Limit Theorem does not apply.
Solution:
Since (a n ) is bounded, there exists a number M > 0 such that ∣a n ∣ ≤ M for every n. Now
let є > 0. There exists an N such that if n ≥ N, then ∣b n ∣ < є/M. But then if n ≥ N we have
∣a n b n ∣ = ∣a n ∣∣b n ∣ ≤ M∣b n ∣ < M
є
= є.
M
Hence lim a n b n = 0. We could not use the Algebraic Limit Theorem, since one of its
hypotheses is that lim a n exists, and we do not know this.
7. Abbott 2.3.10 Prove or provide a counterexample.
(a) If lim(a n − b n ) = 0 then lim a n = lim b n
(b) If b n → b then ∣b n ∣ → ∣b∣
(c) If a n → a and (b n − a n ) → 0 then b n → a.
(d) If a n → 0 and ∣b n − b∣ ≤ a n for all n then b n → b.
Solution, part a:
False. This is true only if one the limits of (a n ) or (b n ) exist (in which case they both do).
As a counterexample, consider a n = (−1)n and b n = −a n .
Solution, part b:
This is true. Let є > 0. Pick N so that if n ≥ Nthen ∣b − b n ∣ < є. Exercise 1.2.6(d) then
implies that if n ≥ N, then
∣∣b∣ − ∣b n ∣∣ ≤ ∣b − b n ∣ < є.
Solution, part c:
This is true and simply a consequence of the Algebraic Limit Theorem: b n = a n +(b n −a n ).
Solution, part d:
This is true. Indeed,
−a n ≤ b − b n ≤ a n
for every n. Now apply the squeeze theorem to conclude b n − b → 0.
8. Abbott 2.4.1
a) Prove that the sequence defined by x1 = 3 and
x n+1 =
converges.
3
1
4 − xn
Math F401: Homework 4
Due: September 29, 2016
b) Now that we know that limn x n exists, explain why limn x n+1 exists.
c) Take the limit of each side of the recursive equation and explicitly compute lim x n .
Solution, part a:
√
√
We first show that the sequence is bounded
between
√ 2 + 3 and 2 − 3. Certainly this is
√
true for x1 . Suppose for some k, 2 − 3 < x k < 2 + 3. Then
√
√
2 + 3 > 4 − xk > 2 − 3
and
1
1
1
√ <
√ .
<
2 + 3 4 − xk 2 − 3
Notice,
√
√
1
1
√ = 2 − 3 and
√ = 2 + 3.
2+ 3
2− 3
√
√
√
√
Hence 2 − 3 < x k+1 < 2 + 3. So 2 − 3 < x n < 2 + 3 for every n ∈ N.
√
√
Now notice that if 2 − 3 < x < 2 + 3, then
√
√
x 2 − 4x + 1 = (x − (2 − 3))(x − (2 + 3) < 0.
So
x>
1
.
4−x
In particular, for every n,
1
= x n+1
4 − xn
and the sequence is monotone decreasing.
xn >
Solution, part b:
Since (x n+1 ) is a subsequence of the convergent sequence (x n ), it has the same limit.
Solution, part c:
Let x = limn x n . Taking the limit of both sides of the equation results in
x=
1
.
4−x
Hence x is a root of the quadratic x 2 − 4x + 1, so x = 2 ±
√
√
3. But x ≤ 3, so x = 2 − 3.
9. Abbott 2.4.3
a) Show that
√ √ √
2, 2 + 2,
converges and find the limit.
4
√ √
√
2 + 2 + 2, . . .
Math F401: Homework 4
Due: September 29, 2016
b) Show that
√ √ √
2, 2 2,
√ √
√
2 2 2, . . .
converges and find the limit.
Solution,√part a:
√
Let a1 = 2 and define a k+1 = 2 + a k for every k. We claim the sequence is monotone
increasing and bounded above by 2.
√
√
Certainly a1 ≤ 2. Suppose some a k ≤ 2. Then a k+1 = 2 + a k < 2 + 2 = 2. Hence we
have shown by induction that a k ≤ 2 for every k.
We now show that the sequence is monotone increasing. Indeed for any k,
√
√
√
√
a k+1 = 2 + a k ≥ a k + a k = 2a k ≥ a k ⋅ a k = a k .
The Monotone Convergence Theorem implies (a k ) converges to a limit L. Taking the
limit of the recursive equation we conclude
√
L = 2+L
√
and hence L = 2 or L = −1. But the sequence is increasing from 2, which rules out
L = −1 Hence L = 2.
Solution,√
part b:
√
Let a1 = 2 and define a k+1 = 2a k for k ≥ 1. We claim this sequence is monotone
increasing and bounded above by 2. Certainly a1 ≤ 2. And if a k ≤ 2, then a2k+1 = 2a k ≤
4 = 22 . So a k ≤ 2 for every k.
√
√
Now for any k, 2a k ≥ a k a k , and hence a k+1 = 2a k ≥ (a k )2 = a k . Hence the sequence
is monotone increasing and bounded above by 2. But then the sequence converges to
some number a. Now a 2n+1 = 2a n . Since lim a 2n+1 = a 2 and since lim 2a n = 2a we have
a 2 = 2a.
So a = 0 or a = 2. Since the sequence is monotone increasing and a1 > 0, we conclude
that a = 2.
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