ABSTRACT
Proof of Theorem 2.3 Assume |z| > 1. Choose R , 1 < R < ρ. We have by (2.4)
sr·2(n+1)−1 = 1
2πi
f(t) t− z
( 1−
( z
t
)r·2(n+1)) dt (4.1)
and
hr.·2(n+1)−1;α;β(f ; z) = 12πi ∮ |t|=R
( 1− (ω2(n+1);α;β(z))
) dt
) dt
) dt
(4.2)
We have for |z| > 1 and |t| > 1
= 1− ( zt )r·2(n+1) ( 1− re2πiα/3
)) ×
× ( 1− re2πiβ/3
)) ×
× ( 1 + re
)) ×
× ( 1 + re
))
= (
+O (
) +O
) +O
)) (4.3)
Combining (4.1), (4.2) and (4.3) we get
sr·2(n+1)−1 −hr.·2(n+1)−1;α;β(f ; z) = 12πi ∮ |t|=R
+
) +O
) +O
))
Considering for 1 < R < ρ the cases 1 < |z| < R and |z| > R we find
lim supn→∞ |sr·2(n+1)−1(f ; z)− hr.·2(n+1)−1;α;β(f ; z)| 1
= ( |z|
)2r ·max ( 1|z| , 1R) =
when 1 < |z| < R and 1 < R < ρ |z|2r
R2r+1 when |z| > R and 1 < R < ρ.
