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

17.1. Infinite representations 323
17.1.2 Infinite sums
In order to express φ (α) by an infinite series, ABEL set β = α
2n+1 . Thus,
φ (α) = φ ((2n + 1) β) = 1
2n + 1
n
∑
n
∑ (−1)
m=−n µ=−n
m+µ φ
�
β +
�
mω + µ ¯ωi
.
2n + 1
Following a string of manipulations designed to group the terms of the right hand
side, ABEL reduced the problem to the search for the limit of the double sum
with
n−1 n−1
∑ ∑
m=0 µ=0
ψ (m, µ) = 1
2n + 1
(−1) m+µ ψ (m, µ) (17.4)
2φ � α
φ 2 � α
2n+1
2n+1
in which ζ (x) = f (x) F (x) and εµ =
In order to find the limit of (17.4), ABEL remarked,
and
� ζ
� − φ 2
�
m + 1
2
� �
εµ
2n+1
�
εµ
2n+1
�
�
ω +
�
µ + 1
�
2
“one attempts to put the preceding quantity [here (17.4)] on the form
P + v,
in which P is independent of n and v is a quantity which has the limit zero, because
then the quantity P is exactly the limit which is sought.” 2
ABEL had a candidate in mind for the expression P when he defined
θ (m, µ) = 2α
α 2 − ε 2 µ
ψ (m, µ) − θ (m, µ) =
2α
(2n + 1)
2 Rµ.
His candidate was the double sum ∑ ∞ m=0 ∑ ∞ µ=0 (−1) m+µ θ (m, µ) and he proceeded in
the following way in obtaining this limit.
For each value of m, ABEL argued, the difference was
n−1
∑
µ=0
(−1) µ n−1
(ψ (m, µ) − θ (m, µ)) = 2α ∑
µ=0
2 “il faut essayer de mettre la quantité précédente sous la forme
P + v,
(−1) µ Rµ
,
2
(2n + 1)
où P est indépendant de n, et v une quantité qui a zéro pour limité, car alors la quantité P sera
précisément la limite dont il s’agit.” (N. H. Abel, 1827b, 156).
¯ωi.