(1)Problem 11989 (American Mathematical Monthly, Vol.124, June-July 2017) Proposed by S

Problem 11989

(American Mathematical Monthly, Vol.124, June-July 2017) Proposed by S. P. Andriopoulos (Greece).

Let x be a number between0 and 1. Prove

∞

Y

n=1

(1 − xⁿ) ≥ exp 1

2− 1

2(1 − x)²

 .

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

Solution. Let x ∈ (0, 1). Then

f(x) := − 1

2 − 1

2(1 − x)²



=1 2

 1

(1 − x)² −1



=

∞

X

k=1

(k + 1) 2 x^k=

∞

X

k=1

x^k k

k

X

j=1

j.

Moreover

g(x) := − ln

∞

Y

n=1

(1 − xⁿ)

!

=

∞

X

n=1

ln

 1

1 − xⁿ



=

∞

X

n=1

∞

X

j=1

(xⁿ)^j

j =

∞

X

k=1

x^kX

j|k

1 j =

∞

X

k=1

x^k k

X

j|k

j.

SincePk

j=1j ≥P

j|kj, it follows that f (x) ≥ g(x) and

∞

Y

n=1

(1 − xⁿ) = exp(−g(x)) ≥ exp(−f (x)) = exp 1

2 − 1

2(1 − x)²

 .

 Addendum. Note that for x ∈ (0, 1),

g(x) =

∞

X

j=1

1 j

∞

X

n=1

(x^j)ⁿ=

∞

X

j=1

1 j

x^j 1 − x^j ≥ 1

2 x²

1 − x² = − 1

2 − 1

2(1 − x²)

 .

Hence we have also a similar reversed inequality

∞

Y

n=1

(1 − xⁿ) = exp(−g(x)) ≤ exp 1

2− 1

2(1 − x²)

 .