MA342A (Harmonic Analysis 1) Tutorial sheet 7
[December 3, 2015]
Name: Solutions
P∞
z n for z ∈ C, |z| < 1.
P
Solution: The partial sums sn = nj=0 z j = (1−z n+1 )/(1−z) since we are not considering
z = 1.
1. Find the Cesàro sum of
n=0
But for |z| < 1, the series is summable, that is s = limn→∞ sn = 1/(1 − z) exists. The
Cesàro sum must then be the same.
P
n
2. Find the Cesàro sum of ∞
n=0 z for z ∈ C, |z| = 1, z 6= 1.
P
Solution: Again sn = nj=0 z j = (1 − z n+1 )/(1 − z) but for |z| = 1, this will not have a
limit.
If we compute the average of the first n partial sums
s0 + s1 + · · · + sn−1
n
n−1
=
1 X 1 − z j+1
n j=0 1 − z
=
1 X z j+1
1X 1
−
n j=0 1 − z n j=0 1 − z
n−1
n−1
1
1
1
z − z n+1
(n)
−
n
1 − z n(1 − z) 1 − z
1
z(1 − z n )
−
=
1 − z n(1 − z)2
1
as n → ∞
→
1−z
=
So the Cesàro sum is 1/(1 − z) (the same formula as for |z| < 1).
3. What would happen if you try to find the Cesàro sum of a divergent series
positive terms (an ≥ 0∀n)?
P∞
n=1
an with
Solution: If we write sn = a1 + a2 + · · · + an for the partial sums of the series, we must
have limn→∞ sn = ∞ (because series of positive terms either converge or diverge to ∞;
recall that the partial sums are increasing with n and are either bounded above or not —
if bounded above they have a least upper bound which will be limn→∞ sn ∈ R and if not
bounded above limn→∞ sn = ∞).
We can argue that
s1 + s2 + · · · + s2n
sn+1 + sn+2 + · · · + s2n
nsn+1
sn+1
≥
≥
=
2n
2n
2n
2
must tend to ∞ as n → ∞.
So there cannot be any (finite) Cesàro sum.
Richard M. Timoney
2