RePoSS #11: The Mathematics of Niels Henrik Abel: Continuation ...

12.5. ABEL’s “exception” 239
A modern reconstruction of DIRICHLET’s proof. Let δ > 0 be given and choose (by
convergence) n such that
Then chose ε1 > 0 and ε2 > 0 such that
|sm − A| < δ
for all m ≥ n.
6
nεk < δ
3 for ε < ε1, and
|ρ n − 1| < δ
for all ε = 1 − ρ < ε2.
3k
Then, if ε < min {ε1, ε2}, the equation (12.12) becomes
n−1
|S − A| ≤ (1 − ρ) ∑ |sm| ρ m �
�
�
+ �(1
− ρ)
�
m=0
∞
∑
m=n
In the first sum, the interesting reconstructed inequality is
(1 − ρ)
n−1
∑
m=0
When the second sum is rewritten as
(1 − ρ)
|sm| ρ m ≤ ε
n−1
∑
m=0
smρ m − A
|sm| ≤ εnk < δ
3 .
∞
∑ smρ
m=n
m ∞
− A = P (1 − ρ) ∑ ρ
m=n
m − A = Pρ n − A
�
�
where P ∈ infm≥n sm, supm≥n sm , the inequalities of interest obtained from the requirements
are
|Pρ n − A| ≤ |Pρ n − P| + |P − A|
≤ k × δ δ
+ 2 ×
3k 6
= 2δ
3 .
Combining the inequalities show that for ε < min {ε1, ε2},
|S − A| ≤ δ 2δ
+
3 3
Thus, as this modern reconstruction illustrates, it can be proved along the lines of
DIRICHLET’S proof that for ε → 0 (i.e. ρ = 1 − ε → 1), the power series S (ρ) converges
to A. The interrelation among the limit processes is contained in the specification of ε1
and ε2. ✷
= δ.
Box 3: A modern reconstruction of DIRICHLET’s proof.
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