Math 431 - Real Analysis I
Solutions to Homework due December 5
Question 1.
(a) Let ak , bk ≥ 0 for all k. Show that if
∞
X
ak converges and bk is a bounded sequence, then
k=0
∞
X
ak bk
k=0
converges as well.
(b) Find a counterexample to above statement if the hypothesis “ak , bk ≥ 0” is removed.
Solution 1.
(a) Since bk is a bounded sequence, we have that |bk | ≤ M for some M . Since bk ≥ 0¡ we have that bk ≤ M
for all k. Since ak ≥ 0, we have that
0 ≤ ak bk ≤ M · ak .
P∞
P∞
Notice that since k=0 ak converges, then k=0 M · ak converges as well. Thus, by the Comparison
Test, we have that
∞
X
ak bk
k=0
converges as well.
(b) Consider the convergent sequence
∞
X
ak =
k=1
∞
X
(−1)k
k=1
1
k
and the bounded sequence bk = (−1)k . Notice that the sequence ak bk = k1 , but
∞
X
1
k
k=1
diverges.
Question 2. Show that
∞
X
k=2
1
k(ln k)p
converges if and only if p > 1.
Solution 2. We will show that
Consider the function
P∞
1
k=2 k(ln k)p
converges if and only if p > 1 by using the integral test.
f (x) =
1
x(ln x)p
defined for x ≥ 2. Notice that f (x) ≥ 0 for all x ≥ 2 and that
0≤
1
1
≤ .
x(ln x)p
x
Thus, by the Squeeze Theorem, we have that limx→∞ f (x) = 0. Furthermore, notice that if x ≤ y, then
(ln(x))p ≤ (ln(y))p . Thus, we get that x(ln x)p ≤ y(ln y)p . Thus,
f (y) =
1
1
≤
= f (x).
y(ln y)p
x(ln x)p
1
Thus, f is a decreasing function. So, we can apply the integral test. Notice that letting u = ln x and du =
we have that
Z ∞
Z b
Z ln b
dx
dx
du
=
lim
=
lim
.
p
p
b→∞ 2 x(ln x)
b→∞ ln 2 up
x(ln x)
2
dx
x ,
If p = 1, then the above integral evaluates to
lim ln | ln b| − ln 2 = ∞.
b→∞
If p 6= 1, then the above integral evaluates to
b1−p
1
+
.
b→∞ 1 − p
p−1
lim
This will converge if and only if p < 1.
Thus, by the integral test, our series converges if and only if p > 1.
Question 3.
P∞
P∞
(a) Give an example of a divergent series k=1 ak for which k=1 a2k converges.
P∞
P∞
(b) Prove that if ak ≥ 0 and k=1 ak converges, then k=1 a2k converges as well.
(c) Find a counterexample to the above statement if the hypothesis “ak ≥ 0” is removed.
Solution 3.
(a) Consider the divergent series
∞
X
1
.
k
k=1
Notice that
∞
X
1
k2
k=1
converges.
P∞
(b) Since k=1 ak converges, then ak → 0 by the Divergence Test. Thus, there exists some N such that for
all n > N , |an − 0| < 1, and thus anP< 1. So, for all n > N ,P
an < 1 and thus a2n < an . Thus, by the
∞
∞
Generalized Comparison Test, since k=1 ak converges, then k=1 a2k converges as well.
(c) Consider the alternating series
∞
X
(−1)k
√ ,
k
k=1
which converges by the Alternating Series Test. Squaring the terms, we get that
X1
k
k=1
diverges.
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Question 4. In this question, we will show that if both
∞
X
a2k
and
k=0
converges, then
∞
X
b2k
k=0
∞
X
ak bk
k=0
converges absolutely.
(a) Show that 2|ak bk | ≤ a2k + b2k for all k.
P∞
P∞
P∞
(b) Use (a) to show that if k=0 a2k and k=0 b2k converges, then k=0 ak bk converges absolutely.
Solution 4.
(a) Notice that (ak − bk )2 ≥ 0. Thus, we have that a2k − 2ak bk + b2k ≥ 0. Rewriting, we get that
2ak bk ≤ a2k + b2k .
Similarly, consider (ak + bk )2 ≥ 0. Thus, we have that a2k + 2ak bk + b2k ≥ 0. Re-writing, we have that
−(a2k + bk )2 ≤ 2ak bk .
Putting these two inequalities together, we have that
−(a2k + bk )2 ≤ 2ak bk ≤ a2k + b2k ,
which is equivalent to our desired result.
P∞
P∞
P∞
(b) Since k=0 a2k and k=0 b2k converges, their sum also converges and is, in fact, equal to k=0 a2k + b2k .
Since 0 < 2|ak bk | ≤ a2k + b2k , by the Comparison Theorem, we have that
∞
X
2|ak bk |
k=0
converges, and thus
∞
X
|ak bk |
k=0
converges. Thus,
∞
X
ak bk
k=0
converges absolutely.
Question 5. Let an be a sequence of non-zero real numbers such that the sequence
P∞
constant sequence. Show that k=1 ak is a geometric series.
Solution 5. Let’s assume that
an+1
an
an+1
an
of ratios is a
is equal to the constant sequence r. Then, we have that for all n,
an+1
=r
an
3
and thus an+1 = ran . This gives a recursive definition of an . So, assume that a0 = c. Then, a1 = ra0 = c · r.
Similarly
a2P
= r · a1 = c · r2 . Continuing in this way (using induction), one can see that an = crn . Thus,
P∞
∞
k
k=0 ak =
k=0 cr , a geometric series.
In class, we learned of the following definition for the limit superior and limit inferior of a sequence xn .
Suppose a number U has the following two properties:
· For all ε > 0, there exists an N such that for all n > N , xn < U + ε; and
· For all ε > 0 and m > 0, there exists an n > m such that xn > U − ε.
Then, lim sup xn = U .
Similarly, suppose that a number L has the following two properties:
· For all ε > 0, there exists an N such that for all n > N , xn > L − ε; and
· For all ε > 0 and m > 0, there exists an n > m such that xn < L + ε.
Then lim inf xn = L.
Question 6. Let xn be a sequence. If lim inf xn = A = lim sup xn , show that xn converges and that, in
fact, xn → A.
Solution 6. Let ε > 0. Since lim inf xn = A, then there exists an N− such that for all n > N− , xn >
A − ε. Since lim sup xn = A, then there exists an N+ such that for all n > N+ , xn < A + ε. Choose
N = max{N− , N+ }. Then, for all n > N , we have that A − ε < xn < A + ε, which is equivalent to
|xn − A| < ε. So, xn → A.
Question 7. Consider the series
∞
X
√
k+1−
√
k.
k=1
(a) Show that
√
k+1−
√
k=√
1
√ .
k+1+ k
(b) Use (a) to decide if the series converges or diverges.
Solution 7.
(a) If we take our term and multiply by a special form of 1, we get
√
√
√
√
k+1+ k
k+1−k
1
√ =√
√ =√
√ .
k+1− k· √
k+1+ k
k+1+ k
k+1+ k
4
(b) We can now limit compare our terms to
√
Thus, since
1 √
k+1+ k
√1
k
√1
k
to get
√
=√
1
1
k
√ =q
= .
2
k+1
k+1+ k
k +1
∞
X
1
√ diverges by the p-series test, we have that our original series diverges as well.
k
k=1
Question 8. Consider the series
∞
X
ak where
k=1
ak =
2
(−1)k − 3
k
.
(a) Compute aak+1
for general k. Use this to compute
k
ak+1 ak+1 .
and lim inf lim sup ak ak Can you use the Ratio Test to reach a conclusion about the convergence or divergence of the series?
(b) Compute |ak |
1/k
for general k. Use this to compute
1/k
lim sup |ak |
.
Can you use the Root Test to reach a conclusion about the convergence or divergence of the series?
(c) In spite of the powerful Ratio Test and Root Tests, at times the more rudimentary tests are more helpful.
Show that our series diverges using the Divergence Test.
Solution 8.
(a) Consider
ak+1
. Performing this division, we get
ak
ak+1
=
ak
2
k+1
(−1)k+1 −3
2
(−1)k −3
k
=2
(−1)k − 3
k
((−1)k+1 − 3)
k+1
.
In absolute values, when k is even, this gives
k+1
(−2)k 2k+1
1
2
=
=
.
(−4)k+1 4k+1
2
In absolute values, when k is odd, this gives
k
(−4)k 4
2
= 2k .
(−2)k+1 = 2
1
Thus, the sequence of ratios alternates between k+1 and 2k . Thus, the limit superior is ∞ and the
2
limit inferior is 0. So, the Ratio Test does not yield an answer.
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1/k
(b) Consider |ak |
to get
1/k
|ak |
= 2
.
k
(−1) − 3 This sequence of ratios is equal to 1 when k is even and 21 when k is odd. Thus, the limit superior of
this sequence is 1, in which case the Root Test also does not yield an answer.
(c) Notice that when k is even, we get that |ak | = 1. Thus, there is no way that ak → 0. So, by the
Divergence Test, the series diverges.
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