Problem 11014
(American Mathematical Monthly, Vol.110, May 2003) Proposed by A. Sinefakopoulos (USA).
Letf : R → R be differentiable, with f (0) = 1, and with f and f′ both nowhere zero on R. Let a1 be a positive real number, and forn ≥ 1 let an+1= anf (an). Prove thatP∞
n=1an is divergent.
Solution proposed by Paolo Roselli and Roberto Tauraso,
Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
The function f is injective otherwise there are two points x1< x2 such that f (x1) = f (x2) and, by Rolle’s theorem, there would be a point inside the interval (x1, x2) where f′ vanishes. Since f is continuous this means that f is positive and strictly monotone.
However the statement holds in a more general setting: it suffices that f is positive and lim inf
x→0+
f (x) − 1
x 6= −∞.
Note that f (x) = 1/(1 + x1/t)t generates the sequence an = 1/nt and the above limit is equal to
−∞ iff t > 1 that is when the series is convergent.
Assume that the positive sequence {an} goes to zero (otherwise the series is trivially divergent ).
Then there are α, δ > 0 such that f (x) ≥ 1 − αx > 0 for 0 < x < δ and 1
x f (x) < 1 x· 1
1 − αx ≤ 1
x· (1 + α x + o(x)) ≤ 1
x+ α + o(1) < 1 x+ 2 α.
Now, let n0≥ 1 such that 0 < an< δ for all n > n0 then 1
an
= 1
an−1f (an−1) < 1 an−1
+ 2 α < 1 an−2
+ 4 α < · · · < 1 an0
+ 2 (n − n0) α.
Thus the series is divergent because for n > n0
an > 1 1 an0
+ 2 (n − n0) α
∼ 1 n.
