Math 318 HW #3 Solutions
1. Exercise 6.5.2. Find suitable coefficients (an ) so that the resulting power series
P
an xn
(a) converges absolutely for all x ∈ [−1, 1] and diverges off of this set;
Answer. Consider the series
∞
X
xn
.
n2
n=1
xn P 1
1
converges, the Weierstrass M-test implies
Then for all x ∈ [−1, 1], n2 ≤ n2 ; since
n2
that the above series converges absolutely on [−1, 1]. On the other hand, for any x ∈ R
such that |x| > 1, the sequence of terms (xn /n) diverges, so we see that the above series
diverges away from [−1, 1].
(b) converges conditionally at x = −1 and diverges at x = 1;
Answer. Consider the series
∞
X
xn
.
n
n=1
We’ve already seen in class that this series converges conditionally at x = −1 and
diverges at x = 1.
(c) converges conditionally at both x = −1 and x = 1.
Answer. Consider the series
∞
X
(−1)n x2n
.
n
n=1
(It may look like I’m cheating
Pby fiddling with the exponent on x, but the above is really
just shorthand for the series an xn where an = 0 when n is odd and an = (−1)n/2 /(n/2)
when n is even.)
Notice that this series satisfies the hypotheses of the Alternating Series
Test for any
P (−1)n x2n x ∈ [−1, 1], so it converges on this interval. Notice that the series
reduces
n
to the harmonic series when x = ±1, so the series only converges conditionally at x = ±1.
(d) Is it possible to find an example of a power series that converges conditionally at x = −1
and converges absolutely at x = 1?
P
Answer. No. If there were such a series
an xn , then, when x = −1, we would have
∞
X
n
|an x | =
n=0
∞
X
n
|an ||(−1) | =
n=0
∞
X
|an | =
n=0
∞
X
|an (1)n |,
n=0
which converges by hypothesis. But this contradicts the fact that
conditionally at x = −1.
P
an xn only converges
2. (a) Exercise 2.7.12 (Summation by Parts). Let (xn ) and (yn ) be sequences, and let sn =
x1 + x2 + · · · + xn . Use the observation that xj = sj − sj−1 to verify the formula
n
X
xj yj = sn yn+1 − sm ym+1 +
j=m+1
n
X
j=m+1
1
sj (yj − yj+1 ).
Proof. Using the fact that xj = sj − sj−1 , we have
n
X
j=m+1
n
X
xj yj =
(sj − sj−1 )yj = sm+1 ym+1 − sm ym+1 + . . . + sn yn − sn−1 yn
j=m+1
= sn yn − sm ym+1 +
n−1
X
sj (yj − yj+1 ).
j=m+1
The above isn’t quite what we want, but adding and subtracting sn yn+1 yields
n−1
X
sn yn − sm ym+1 + sn yn+1 − sn yn+1 +
sj (yj − yj+1 )
j=m+1
= sn yn+1 − sm ym+1 + sn (yn − yn+1 ) +
n−1
X
sj (yj − yj+1 )
j=m+1
n
X
= sn yn+1 − sm ym+1 +
sj (yj − yj+1 ),
j=m+1
as desired.
(Notice that, when m = 0, we have that x1 = s1 , so the sm ym+1 term fails to appear.)
(b) Use part (a) to prove Abel’s Lemma (this is very similar to Exercise 2.7.14(b)).
P
Proof. Suppose |sn | = nj=1 aj ≤ A for all n and that b1 ≥ b2 ≥ . . . ≥ 0. Then, by
part (a),
X
n
n
X
aj bj = sn bn+1 +
sj (bj − bj+1 )
j=1
j=1
≤ |sn bn+1 | +
= |sn |bn+1 +
n
X
j=1
n
X
|sj (bj − bj+1 )|
|sj |(bj − bj+1 )
j=1
since bj ≥ bj+1 ≥ 0 for all j. Since each |sj | ≤ A, the above is less than or equal to
Abn+1 +
n
X
A(bj − bj+1 ) = Abn+1 + A(b1 − bn+1 ) ≤ 2Ab1
j=1
since b1 ≥ bn+1 ≥ 0. Putting this together with (1), then, yields
X
n
a
b
j j ≤ 2Ab1
j=1
for any choice of n.
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(1)
3. Exercise 6.5.8. Let
P
an xn be a power series with an 6= 0, and assume
an+1 L = lim n→∞
an exists.
(a) Show that if L 6= 0, then the series converges for all x in (−1/L, 1/L). (The advice in
Exercise 2.7.9 may be helpful.)
P
Proof. Suppose L 6= 0. Applying the Ratio Test to the power series
an xn , we have
an+1 an+1 an+1 xn+1 = |x|L.
= lim x
= |x| lim lim
n→∞
n→∞ an xn n→∞ an an The number |x|L is strictly less than 1 precisely when |x| < 1/L or, equivalently, when
x ∈ (−1/L, 1/L). Therefore, the Ratio Test implies that the series converges for such
x.
(b) Show that if L = 0, then the series converges for all x ∈ R.
Proof. Suppose L = 0. Then, the same calculation done in (a) shows that
an+1 xn+1 = |x|L = 0 < 1
lim
n→∞ an xn for any x ∈ R. Therefore, the Ratio Test implies that the series converges for all
x ∈ R.
(c) Show that (a) and (b) continue to hold if L is replaced by the limit
ak+1 :k≥n .
L0 = lim sn where sn = sup n→∞
ak The value L0 is called the “limit superior” or “lim sup” of the sequence |an+1 /an |. It
exists if and only if the sequence is bounded (Exercise 2.4.6).
Proof. Suppose the sequence (|an+1 /an |) is bounded (meaning that the limit superior
exists) and that
an+1 = L0 .
lim sup an Pick x ∈ (−1/L0 , 1/L0 ) and choose c such that |x| < c < 1/L0 . Then 1/c > L0 and so
there exists N ∈ N such that
an+1 1
an < c
for all n ≥ N . In particular, |aN +k | <
|aN +k | <
|aN +k−1 |
c
for all k ≥ 1, and so
|aN +k−1 |
|aN +k−2 |
|aN |
<
< ··· < k
c
c2
c
3
for all k ≥ 1. Hence,
∞
X
|aN +k xN +k | =
k=1
∞
X
|aN +k ||x|N +k ≤
k=1
∞
X
|aN |
k=1
ck
|x|N +k = |aN ||x|N
∞
X
|x|k
k=1
ck
,
which converges since |x| < c. Therefore,
∞
X
n
an x =
n=1
N
X
k
ak x +
k=1
∞
X
aN +k xN +k
k=1
converges absolutely.
Since the choice of x ∈ (−1/L0 , 1/L0 ) was arbitrary, we see that the series converges for
all x ∈ (−1/L0 , 1/L0 ), as desired.
As for (b), note that an+1
an ≥ 0 for all n, so
an+1 ≥ 0.
lim inf an Therefore, if lim sup |an+1 /an | = 0, we have that lim sup |an+1 /an | = lim inf |an+1 /an |.
But this implies that the limit exists and equals zero, so the hypotheses of (b) are
satisfied.
P
(d) Show that if |an+1 /an | is unbounded, then the original series
an xn converges only
when x = 0.
P
Answer. This is actually not true. Consider the series ∞
n=1 an with terms given by
a3k =
1
,
2k
a3k−1 =
1
,
22k
a3k−2 =
1
22k−1
;
i.e., the series
1
1
1
1
1
1 1 1 1
+ + + +
+ +
+
+ + ....
2 4 2 8 16 4 32 64 8
Now, notice that the first three terms are all ≤ 1/2, the second three terms are all ≤ 14 ,
the next three terms are all ≤ 18 , etc. In other words,
∞
X
n=1
∞
an ≤ 3 ·
X 1
1
1
1
+ 3 · + 3 · + ··· = 3
.
2
4
8
2n
n=1
The
the comparison test, the series
P∞ series on the right hand side converges to 3, so,
Pby
∞
n
n=1 an converges, meaning that the power series
n=1 an x converges at x = 1.
However,
a3k 1/2k 22k k
=
=
a3k−1 1/22k 2k = 2 ,
which can be arbitrarily large, so the sequence |an+1 /an | is unbounded, yet the power
series converges at x = 1.
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4. Exercise 6.5.9. Use Theorem 6.5.7 to argue that power series are unique. If
∞
X
an xn =
∞
X
n=0
bn xn
0
for all x in an interval (−R, R), prove that an = bn for all n = 0, 1, 2, . . .. (Start by showing
that a0 = b0 .)
P
P
Proof. Suppose
an xn =
bn xn for all x ∈ (−R, R). Then, in particular, the two series
are equal at x = 0:
∞
∞
X
X
n
a0 =
an (0) =
bn (0)n = b0 .
n=0
n=0
Moreover, by Theorem 6.5.7, both series are infinitely differentiable on (−R, R) and their
derivatives are given by differentiating term-by-term. Since the derivative of any function is
completely determined by the values that function takes (just look at the definition of the
derivative), it follows that the derivatives of both series must be equal:
∞
X
n−1
nan x
=
n=1
∞
X
nbn xn−1 .
n=1
In particular, evaluating at x = 0 yields
a1 =
∞
X
nan (0)n−1 =
n=1
∞
X
nbn (0)n−1 = b1 .
n=1
In general, the kth derivatives of both series must be equal
∞
X
n=k
∞
X
n!
n!
n−k
an x
=
bn xn−k
(n − k)!
(n − k)!
n=k
and, in particular, their values at x = 0 must be equal:
k!ak =
∞
X
n=k
∞
X
n!
n!
an (0)n−k =
bn (0)n−k = k!bk ,
(n − k)!
(n − k)!
n=k
so ak = bk for all k = 0, 1, 2, . . ..
5. Exercise 6.5.11. A series
P∞
n=0 an
is said to be Abel-summable to L if the power series
f (x) =
∞
X
an xn
n=0
converges for all x ∈ [0, 1) and L = limx→1− f (x).
(a) Show that any series that converges to a limit L is also Abel-summable to L.
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Proof. Suppose the series
P
an converges to L. Then
L=
∞
X
an =
n=0
∞
X
an · 1n ,
n=0
so f (1) = L. But then, by Theorem 6.5.7, since the power series converges at x = 1 it
is continuous at least on (−1, 1], so
lim f (x) = f (1) = L,
x→1−
so indeed the series is Abel-summable to L.
P
n
(b) Show that ∞
n=0 (−1) is Abel-summable and find the sum.
Proof. We need to consider the power series
f (x) =
∞
X
(−1)n xn =
n=0
∞
X
(−x)n .
n=0
So long as |x| < 1, the above is a geometric series that converges to
1
1
=
.
1 − (−x)
1+x
In other words, f (x) =
1
1+x
at least on (−1, 1). Therefore,
lim f (x) = lim
x→1−
x→1−
1
1
1
=
= ,
1+x
1+1
2
1
since the function g(x) = 1+x
is continuous at x = 1.
Hence, we see that the Abel sum of the given series is 1/2. (Note: this makes some
sense, since the sequence of partial sums of the series is 1, 0, 1, 0, 1, 0, . . ., so 1/2 is the
average of the partial sums.)
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