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

12.9. ABEL’s proof of the binomial theorem 255
where the notation i has been adopted for ABEL’S
binomial coefficients in polar form,
which meant
δµ
m − µ + 1
µ
= δµ
� cos γµ + i sin γµ
√ −1. ABEL wrote the factors of the
� �
cos γµ + i sin γµ ,
� k + ik
= ′ − µ + 1
,
µ
and for each given µ, the values of δµ and γµ could be found. When these factors were
multiplied to produce the binomial coefficients, ABEL found
� �
�
mµ =
With the conventions
�
µ
∏ δn
n=1
×
cos
x = α (cos φ + i sin φ) , λµ =
�
µ
∑ γn
n=1
+ i sin
�
µ
∑ γn
n=1
µ
∏ δn, and θµ = µφ +
n=1
��
.
µ
∑ γn,
n=1
ABEL had thus decomposed the general term of the binomial series into the form
mµx µ = λµ
� � µ
cos θµ + i sin θµ α ,
thereby reducing the binomial series to its real and imaginary parts,
φ (x) =1 +
∞
∑
µ=1
λµα µ cos θµ +i
� �� �
=p
∞
∑
µ=1
λµα µ sin θµ . (12.24)
� �� �
=q
Convergence of the binomial series. Having obtained the decomposition of the bi-
nomial series into real and imaginary parts (12.24), ABEL claimed that it converged if
α < 1 and diverged if α > 1. In order to prove this claim, he applied his version of the
ratio test, observing that because
�
� �2 � �
k − µ k ′ 2
δµ+1 =
+ → 1 as µ → ∞,
µ + 1 µ + 1
the ratio of consecutive terms converged to α,
λµ+1α µ+1
λµα µ
= δµ+1α → α for µ → ∞.
ABEL took care of the trigonometric factors of the general terms, cos θµ and sin θµ
by applying his own version of the ratio test as expressed in his Lehrsätze I&II. How-
ever, he did not provide any details of the argument. In the simplest case, α < 1, the
absolute convergence of both the series p and q can be obtained directly from Lehrsatz