Brandon Behring
Complex Variables # 4
1 (i) Assume we have a finite z, |z| < R. Then we have |n + |z||| ≤ n + |z| ≥ 2n
for n ≥ R we have
|
1
1
|| ≥
n + |z|
2n
As are series is less than a series that diverges, it to diverges for all
n
2
(ii) For z = −n
P∞ for 1some n ∈ N the series clearly diverges. Assume
2
z 6= −n
n=1 n2 +z . Assume z is in the annulus r ≤ |Z| ≤ R where
j 2 < r < R < (j + 1)2 for some integer j. Then |n2 + z| ≥ n2 − |z| ≥
n2 − R. For sufficiently large n are series is bounded absolutely by
P
1
2n2 so it is converges uniformly in the annulus by the Weierstrass
M-Test.
P∞
1
1
2 Take n=1 (z2 +1)
n then for z such that |z 2 +1| < 1 we have
∞
X
1
2 + 1)n
(z
n=1
1
=
=
=
1−
1
z 2 +1
z2 + 1
+1−1
1
1 + 2.
z
z2
P∞
P∞
3 (i) We have n=0 an z 2n = n=0 bn z n where bn = 0 for n odd and bn/2 = an
for n even. Then using the root test
1
R̄
=
lim sup
p
n
|bn |
n→∞
=
=
2 1
lim sup(bn/2 ) n 2
n→∞
q
1
lim sup (an ) n
n→∞
=
1
1
√ .
R
Using the ratio test
an+1 z 2n+2
|<1
an z 2n
√
Implies |z|2 < lim sup |an+1 /an | = R or R̄ = R.
P∞
P∞
(ii) We have n=0 a2n z n = n=0 cn z n where cn = a2n
lim sup |
1
R̄
=
p
n
|an |2
lim sup
n→∞
=
lim sup
p
n
n→∞
2
1
|an | = 2
R
where in the last step we observed we are just taking the supremum
over n of our original sequence with each term squared.
Or, using the ratio test
lim sup |
a2n+1 z n+1
|<1
a2n z n
Implies |z| < lim sup |an+1 /an |2 = R2 . Thus again R̄ = R2 .
4 (i) Our sequence is xn = 1, 0, −1, 0, 1, 0, . . .. Then
lim inf xn
=
sup inf xm
n≥0 m≥n
= −1
and
lim sup xn
=
=
inf sup xm
n≥0 m≥n
1.
(ii) Our sequence can be rewritten as an alternating sequence xn =
(−1)n (−1 + 1/n).
Then
lim inf xn
=
=
lim inf xm
n→∞ m≥n
lim {−1 + 1/n}
n→∞
= −1
and
2
lim sup xn
=
=
=
lim sup xm
n→∞ m≥n
lim {1 − 1/n}
n→∞
1
(iii) Assume |yn | < M for all n. Then for |xn | < M/n and for every > 0
and n > M/ we have |xn | < so lim xn = 0. Thus both the lim sup
and lim inf also are zero because if the limit exists both limits exist
and are equal.
(iv) This sequence is convergent so lim sup xn = lim inf xn = lim xn = 1.
Pn
Pn
P
5 Let us define fn (z) = j=0 aj z j , gn (z) = j=0 bj z j and hn (z) = k=0 ck z k
Pn
where cn = j=0 aj bn−j . By arrangement we can rewrite
hn =
n X
k
X
aj bk−j z k =
k=0 j=0
n
X
gk an−k z k .
k=0
We can see this explicitly if we just use our imagination.
Or, expanding and collecting terms in an .
hn (z)
=
n X
k
X
aj bk−j z k
k=0 j=0
= a0 b0
+(a0 b1 + a1 b0 )z
+(a0 b2 + a1 b1 + a2 b0 )z 2
+(a0 b3 + a1 b2 + a2 b1 + a2 b0 )z n + · · ·
+(a0 bn + a1 bn−1 + . . . + an b0 )z n
= a0 (b0 + b1 z + . . . + bn z n ) + a1 z(b0 + b1 z + . . . + bn−1 z n−1 ) + . . . + an z n
=
=
a0 gn + a1 gk−1 z + . . . + an g0 z n
n
X
ak gn−k z k .
k=0
Now
hn (z)
=
=
n
X
ak gn−k z k .
k=0
n
X
(gn−k (z) − g(z) + g(z))ak z k
k=0
3
=
=
n
X
k=0
n
X
(gn−k (z) − g(z))ak z k + g(z)
n
X
ak z k
k=0
(gn−k (z) − g(z))ak z k + g(z)fn (z).
k=0
as desired.
Now, we know from gk → g that for any > 0 for sufficiently
P∞ large N and
|z| < R we have g(z) − gN −k (z) < and that the sum k=0 ak z k = f (z)
converges so it’s tails are bounded by say M
∞
X
(gn−k (z) − g(z))ak z k < k=N
we have limn→∞
∞
X
ak z k < M
k=N
Pn
k=0 (gn−k (z)
h(z)
− g(z))ak z k → 0 and
=
=
lim hn (z)
n→∞
lim fn g(z)
n→∞
= f (z)g(z)
for |z| < R.
4