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

230 Chapter 12. ABEL’s reading of CAUCHY’s new rigor and the binomial theorem
Proof of Lehrsatz I In order to present a proof of Lehrsatz I, we write ε = α−1
2
and chose n ∈ N such that
Then, by iteration,
ρm+1 ≥ (α − ε) ρm for all m ≥ n.
ρm+1 ≥ (α − ε) m−n ρn.
The choice of ε ensures that ρm+1 increases beyond all bounds. Now assume that the
series in (12.6), ∑ εmρm, is to be convergent. Then, in particular, a k ∈ N exists such
that
If m ≥ max (k, n),
meaning
|εm+1ρm+1| < ε for all m ≥ k.
ε > |εm+1| · ρm+1 ≥ |εm+1| (α − ε) m−n ρn
|εm+1| ≤ ε
ρn
1
m−n → 0 for m → ∞
(α − ε)
because α − ε > 1. In conclusion, if ∑ εmρm is be convergent, the sequence {εm} has
to converge toward zero. And since this is not the case, the sum (12.6) cannot be
convergent. ✷
Box 2: Proof of Lehrsatz I
for which the ratio of consecutive terms converges toward α > 1,
any linear combination
ρm+1
ρm
→ α > 1,
∞
∑ εmρm
m=0
will be divergent, provided the sequence {εm} does not converge to zero.
> 0
(12.6)
The contents of this Lehrsatz I is thus a generalization of one part of CAUCHY’S ra-
tio test of convergence. It is remarkable from a conceptual viewpoint that ABEL’S first
theorem would be one of divergence when his entire theory was so focused on conver-
gent series. The Lehrsatz I is thus — as it stands — a negative demarcation criterion.
This apparent imbalance was leveled by the second theorem. ABEL’S Lehrsatz II
is a counterpart to the Lehrsatz I describing analogous — but this time sufficient —
conditions for convergence. ABEL found that if
ρm+1
ρm
→ α < 1