Homework 5 Solution
Yikun Zhang 1
Chapter 2. Ex.13 The purpose of this exercise is to prove that Abel summability is stronger
than the standard or Cesáro methods of summation.
∞
P
(a) Show that if the series
cn of complex numbers converges to a finite limit s, then the
n=1
series is Abel summable to s.
(b) However, show that there exist series which are Abel summable, but that do not converge.
∞
P
(c) Argue similarly to prove that if a series
cn is Cesáro summable to σ, then it is Abel
n=1
summable to σ.
(d) Give an example of a series that is Abel summable but not Cesáro summable.
The results above can be summarized by the following implications about series:
convergent =⇒ Ceráro summable =⇒ Abel summable,
and the fact that none of the arrows can be reversed.
Proof. (a) Without loss of generality, we assume that s = 0. In fact, if
simply define dn = cn −
s
2n
∞
P
and
∞
P
cn = s 6= 0, we can
n=1
dn = 0.
n=1
Letting s1 = c1 , sN = c1 + c2 + · · · + cN , we have
N
X
n
cn r =
n=1
=
N
−1
X
(sn+1 − sn )rn+1 + c1 r
n=1
N
X
n
sn r −
n=1
= (1 − r)
N
−1
X
sn rn+1
(1)
n=1
N
X
sn rn + sN rN +1 .
n=1
With the assumption that s = 0, we will obtain that
∞
P
n=1
cn rn = (1 − r)
∞
P
sn rn as N → ∞ in
n=1
the preceding equation.
For any > 0, we may choose N large enough such that |sn | < when n > N . Also, there
exists an M > 0 such that sup |sn | ≤ M .
n∈N
1
School of Mathematics, Sun Yat-sen University
1
Then we have
|
∞
X
n
cn r | = |(1 − r)
n=1
≤ |(1 − r)
∞
X
n=1
N
X
sn r n |
sn r | + |(1 − r)
n=1
< (1 − r)
N
X
∞
X
n
sn r n |
(2)
n=N +1
M rn + (1 − r)
n=1
N
rN
1−r
≤ M r(1 − r ) + .
Thus, lim− sup |
r→1
∞
P
n=1
cn rn | ≤ . Since is arbitrary, we have lim−
r→1
(b) Consider cn = (−1)n . The partial sum sN =
N
P
∞
P
cn rn = 0.
n=1
cn is −1 when N is odd and 0 when N is
n=1
even. Thus it does not converge. However, its Abel limit is lim−
r→1
∞
P
(−r)n = lim−
r→1
n=1
−r
1+r
= − 21 .
N
and obtain that s1 = σ1 , sN = N σN − (N − 1)σN −1 .
(c) First we let σN = s1 +s2 +···+s
N
Then assuming σ = 0 and by (a), we obtain that
N
X
n
cn r = (1 − r)
n=1
N
X
sn rn + sN rN +1
n=1
= (1 − r)
= (1 − r)
(1 − r)2
N
X
n=1
N
X
[nσn − (n − 1)σn−1 ]rn + [N σN − (N − 1)σN −1 ]rN +1
nσn rn − (1 − r)
n=1
N
X
N
−1
X
(3)
nσn rn + [N σN − (N − 1)σN −1 ]rN +1
n=1
nσn rn + (2 − r)N σN rN +1 − (N − 1)σN −1 rN +1 .
n=1
Since {σn } is bounded and the series
∞
P
rn converges for any 0 < r < 1, the derivative of
n=1
∞
P
rn
n=1
with respect to r also converges, yielding that the single items N rN −1 and (N − 1)rN −2 tend
to 0 as N → ∞.
∞
∞
P
P
Letting N → ∞ in the previous equation, we have
cn rn = (1 − r)2
nσn rn .
n=1
n=1
With the assumption σ = 0, we can choose N large enough such that |σn | < when n > N for
2
any > 0. Meanwhile, there exists a B > 0 such that |σn | ≤ B for all n ∈ N. Therefore,
|
∞
X
n
∞
X
2
cn r | = |(1 − r)
n=1
2
≤ (1 − r)
nσn rn |
n=1
N
X
n
Bnr + (1 − r)
n=1
N +1
2
∞
X
nrn
n=N +1
N +2
N +1
r − (N + 1)r
+ Nr
− (1 + N )rN +2
2 (N + 2)r
+
(1
−
r)
(1 − r)2
(1 − r)2
= B[r − (N + 1)rN +1 + N rN +2 ] + [(N + 2)rN +1 − (1 + N )rN +2 ].
(4)
= (1 − r)2 B
∞
P
As r → 1− , we have lim− sup |
r→1
n=1
cn rn | ≤ , i.e., lim−
r→1
∞
P
cn rn = 0 by the arbitrariness of .
n=1
In the case when σ 6= 0, we let d1 = c1 − σ, dn = cn for n > 1.
∞
∞
P
P
Then lim−
dn rn = 0 = lim−
cn rn − σ.
r→1
r→1
n=1
n=1
∞
P
(d) Consider cn = (−1)n−1 n. Then
Thus its Abel limit is lim−
r→1
∞
P
(−1)n−1 nrn = −[
n=1
(n + 1)(−r)n −
n=1
n−1 n
(−1)
n=1
∞
P
r = lim−
r→1
r
(1+r)2
=
Therefore,
(−r)n ] =
n=1
r
.
(1+r)2
1
.
4
Note that the Cesáro sum has the property σn − ( n−1
)σn−1 =
n
∞
P
series
an , lim ann must be 0.
n=1
∞
P
an
.
n
Hence for a Cesáro summable
n→∞
∞
P
(−1)n−1 n is not Cesáro summable.
n=1
3