Math 446
Advanced Calculus
Spring 2007
Problem Set 3 Solutions - Due Thursday, February 8, 2007
Complete all of the following problems. You must show all necessary steps in order to receive
full credit.
1. Prove that if lim(xn ) = x and if x > 0, then there exists a natural number M such that
xn > 0 for all n ≥ M .
Proof. (Greg Henselman) For all ε > 0, there exists m ∈ IN such that for all n ≥ m,
0 ≤ |x − xn | < ε. Let ε = x. Thus we have
|x − xn | < x ⇔ −x < x − xn < x ⇔ 0 < xn < 2x.
Combining the last inequality with the first statement of this proof, we have it that there
exists m ∈ IN such that for all n ≥ m, xn > 0.
2. Show that lim
n2
n!
= 0.
Proof. (Jessica Fujii) n2 /n! = n/(n − 1)!. Ratio Test:
n+1
n!
n
(n−1)!
=
n + 1 (n − 1)!
n+1
·
=
.
n!
n
n2
lim(1/n) + lim(1/n2 ) = 0.
Therefore, (xn ) is convergent and lim(n2 /n!) = 0.
3. Prove that if lim(xn ) = 0 and (yn ) is a bounded sequence (but not necessarily convergent),
then lim(xn yn ) = 0.
Proof. (Daniel Gossard) Let ε > 0. By the definition of a convergent sequence, we know that
lim(xn ) = 0 implies that there exists K ∈ IN such that for all n ≥ K, |xn | < ε/M . Here, M
is the bound on (yn ) according to the definition. We also know that |yn | ≤ M for all n ∈ IN .
Now, we can see that for all n ≥ K,
|xn yn | = |xn ||yn | ≤ |xn |M <
ε
M = ε.
M
Therefore, since ε is always greater than 0, we see that lim(xn yn ) = 0.
4. Let X = (xn ) be a sequence of positive real numbers such that lim
xn+1
xn
= L > 1. Show
that X is not a bounded sequence and hence is not convergent.
Proof. (Steve Lester) Let (xn ) be a sequence of positive real numbers such that lim(xn+1 /xn ) >
1. Define L := lim(xn+1 /xn ). Let r ∈ IR such that L > r > 1. Define ε := L − r > 0. By
definition, there exists K ∈ IN such that for all n ≥ K
L−ε<
xn+1
so (L − ε)xn < xn+1 .
xn
Continue applying this relation to get
(L − ε)xn > (L − ε)2 xn−1 > · · · > (L − ε)n−K+1 xK ⇒ rxn > rn−K+1 xK ⇒ xn > rn−K xK .
Let c := xK /rK . Note that c is constant. So xn > rn c for all n ≥ K. But r > 1 so (crn )
tends to +∞. Thus (xn ) tends to +∞ and is unbounded.
5. Suppose that X = (xn ) is a convergent sequence and Y = (yn ) is another sequence such that
for any ε > 0 there exists M such that |xn − yn | < ε for all n ≥ M . Prove that Y converges
and lim X = lim Y .
Proof. (Paige Cudworth) Since (xn ) is convergent, there exists x ∈ IR such that for ε > 0
there exists N ∈ IN such that |x − xn | < ε/2 for all n ≥ N . So if we let p := max{N, M }, we
have |x − xn | < ε/2 and |xn − yn | < ε/2 for all n ≥ p. Using the Triangle Inequality, we get
|x − yn | = |x − xn + xn − yn | ≤ |x − xn | + |xn + yn | <
ε ε
+ = ε.
2 2
So we have |x − yn | < ε for all n ≥ p. So x is the limit of (yn ).
6. Let x1 ≥ 2 and xn+1 := 1 +
and find the liimit.
√
xn − 1 for n ∈ IN . Prove that the sequence (xn ) is convergent
Proof. (Elizabeth Alford) Using induction: For n = 1, x1 ≥ 2 by assumption. Then inductively, we assume that xk ≥ 2. We want to show that xk+1 ≥ 2.
√
xk+1 ≥ 2 ⇔ 1 + xk − 1 ≥ 2 ⇔ xk − 1 ≥ 1 ⇔ xk ≥ 2.
Therefore, xn ≥ 2 for all n ∈ IN . Then we need to show that (xn ) is decreasing.
√
xk+1 ≤ xk ⇔ 1 + xk − 1 ≤ xk ⇔ 0 ≤ (xk − 1)(xk − 2).
Since xk ≥ 2 for all k ∈ IN , the last inequality holds and thus (xn ) is decreasing.
Therefore, by MCT, x := lim(xn ) exists. Then to find what this limit is:
x := lim(xn ) = lim(xn+1 )
by tail-sequences. We then can solve for x to find our limit.
√
x = 1 + x − 1 ⇔ x2 − 3x + 2 = 0 ⇔ (x − 1)(x − 2) = 0.
Thus x = 1 or x = 2. But since xn ≥ 2 for all n ∈ IN , x = 2.
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7. Let (xn ) be a bounded sequence and for each n ∈ IN let sn := sup{xk : k ≥ n} and
tn := inf{xk : k ≥ n}. Prove that (sn ) and (tn ) are convergent sequences and prove that if
lim(sn ) = lim(tn ), then (xn ) is convergent. (One calls lim(sn ) the limit superior of (xn )
and lim(tn ) the limit inferior of (xn ).)
Proof. (Andrea Walker) Let Am := {xk : k ≥ M }. Then Am+1 ⊂ Am and thus
sup Am+1 ≤ sup Am and inf Am+1 ≥ inf Am .
Therefore, (sn ) is a decreasing sequence and (tn ) is an increasing sequence. We know that
t1 ≤ tn ≤ sn ≤ s1
for all n ∈ IN and thus t1 is a lower bound for (sn ) and s1 is an upper bound for (tn ).
Therefore, by the MCT, (sn ) and (tn ) are convergent sequences.
Suppose lim(sn ) = lim(tn ). By definition of sn and tn , we have
tn ≤ xn ≤ sn
for all n ∈ IN . By the Squeeze Theorem, lim(tn ) = lim(xn ) = lim(sn ).
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