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

13.4. More characterizations and tests of convergence 273
would be convergent if α > 0. Thus, from a divergent series, ABEL had prescribed
means of obtaining two derived series, one of which was divergent, the other conver-
gent.
In another section of the note, ABEL devised another way of obtaining a divergent
series, which would lead him to a new test of convergence. ABEL found that for any
continuous function φ (n) which increased without bounds for n → ∞, the series of
derived terms,
would be divergent.
∞
∑ φ
n=1
′ (n) , (13.4)
ABEL’S proof proceeded from the Taylor series expansion of φ (to the second term
and with remainder),
φ (n + 1) = φ (n) + φ ′ (n) + φ′′ (n + θ)
, for some 0 < θ < 1.
2
At this point, ABEL’S draft style made the precise assumptions of the ensuing de-
ductions difficult to interpret. However, if ABEL’S requirements interpreted to mean
φ ′′ (n) < 0, we obtain what was his next line,
φ (n + 1) − φ (n) < φ ′ (n) .
Then, the divergence of (13.4) followed by summation,
φ ′ (n) > φ (n + 1) − φ (0) → ∞ as n → ∞.
Subsequently, ABEL applied this procedure to prove the divergence of the series
∞
∑
n=2
1
n ∏ m k=1 logk for m integral,
n
where log k n = log log k−1 n. ABEL did so by defining
and differentiating it to obtain
φ ′ m (n) =
φm (n) = log m (n + a)
1
(n + a) ∏ m−1
k=1 logk (n + a) .
As a consequence of the theorem stated above, the series (corresponding to a = 0)
was divergent.
∞
∑ φ
n=2
′ m (n) =
∞
∑
n=2
1
n ∏ m−1
k=1 logk n