n) is convergent for all ǫ >0

Problem 12064

(American Mathematical Monthly, Vol.125, October 2018) Proposed by Cesar Adolfo Hernandez Melo (Brazil).

Let f be a convex, continuously differentiable function from[1, ∞) to R such that f^′(x) > 0 for all x ≥1. Prove that the improper integral

Z ∞ 1

dx f^′(x) is convergent if and only if the series

X∞ n=1

(f⁻¹(f (n) + ǫ) − n)

is convergent for all ǫ >0.

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

Solution. Since f is convex with positive derivative, it follows that f is strictly increasing, f^′ is increasing and

f(x) ≥ f^′(1)(x − 1) + f (1)

which implies that limx→+∞f(x) = +∞. Hence f ([1, +∞)) = [f (1), +∞), and the inverse d f⁻¹ is a continuous, differentiable, and strictly increasing function in [f (1), +∞).

Given ǫ > 0, for any positive integer n, we define xⁿ = f⁻¹(f (n) + ǫ) > n. By the Mean Value Theorem applied to f⁻¹, there is sⁿ∈(f (n), f (xⁿ)) such that

f⁻¹(f (n) + ǫ) − n = f⁻¹(f (xn)) − f⁻¹(f (n)) = D(f⁻¹)(sn) · (f (xn) − f (n)) = ǫ f^′(tⁿ) where tn = f⁻¹(sn) ∈ (n, xn).

Hence f^′(tn) ≥ f^′(n) ≥ f^′(x) for x ∈ [1, n], and if the improper integral is convergent then X∞

n=2

(f⁻¹(f (n) + ǫ) − n) = ǫ X∞ n=2

1 f^′(tn)≤ǫ

X∞ n=2

Z ⁿ

n−1

dx f^′(x) = ǫ

Z ^∞

1

dx f^′(x), and we may conclude that also the series is convergent.

On the other hand, if the series is convergent, the term f⁻¹(f (n) + ǫ) − n = xn−n → 0 and therefore xn+1−xn = (xn+1−(n + 1)) − (xn−n) + 1 → 1. It follows that for any positive integer n, 0 < xn+1−xn≤M for some constant M ≥ 1.

Moreover f^′(tn) ≤ f^′(xn) ≤ f^′(x) for x ∈ [xn,+∞) implies that M

ǫ X∞ n=1

(f⁻¹(f (n) + ǫ) − n) = M X∞ n=1

1 f^′(tn) ≥

X∞ n=1

Z ^xn+1

xn

dx f^′(x) =

Z ^∞

x₁

dx f^′(x),

and we have that the improper integral is convergent. 

Remark. If the series

X∞ n=1

(f⁻¹(f (n) + ǫ) − n) is convergent for one ǫ > 0 then it converges for all ǫ > 0.