Esercizi svolti di analisi reale e complessa - Dipartimento di ...

∞
n=2
1 − 1
n2
= 1
2 .
an = n
k=2
|z| < 1
1 1 − k2
(1 + z)(1 + z 2 )(1 + z 4 )(1 + z 8 ) · . . . · (1 + z 2n
) . . . = 1
1 − z .
m
n=1
θ(z) =
∞
n=1
(1 + z 2n
2
) =
m −1
z 2n .
n=0
(1 + h 2n−1 e z )(1 + h 2n−1 e −z )
|h| < 1
θ(z + 2 log h) = h −1 e −z θ(z) .
f(z) = sin πz g(z)
{an}n ⊂ C ∗ an → ∞ m ∈ N
{an}n m m > 0
f
f(z) = z m
g(z)
e 1 −
n
z
e
an
z
an + 1
z 2+...+
1 z
mn
2 an
mn an
g mn
mn
k
n
1
< ∞ .
|an| k+1
h h
g(z)
f g