Problem 12225
(American Mathematical Monthly, Vol.128, January 2021) Proposed by P. Jiradilok (USA).
Let Γ denote the gamma function.
(a) Prove that ⌈Γ(1/n)⌉ = n for every positive integer n.
(b) Find the smallest constant c such that Γ(1/n) ≥ n − c for every positive integer n.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. For x > 0, from Γ(x) =R∞
0 e−ttx−1dt, we find that Γ′′(x) =
Z ∞ 0
e−ttx−1(ln(t))2dt > 0,
which implies that the gamma function is strictly convex in (0, +∞).
Since Γ(1) = Γ(2) = 1 and Γ′(1) = −γ, by the strict convexity of Γ, it follows that
∀x ∈ (1, 2], −γ(x − 1) + 1 < Γ(x) ≤ 1 (1)
where y = Γ′(1)(x − 1) + Γ(1) = −γ(x − 1) + 1 is the tangent line to Γ at x = 1.
(a) Given any positive integer n, we have that ⌈Γ(1/n)⌉ = n if and only if
n − 1 < Γ(1/n) ≤ n ⇔ 1 − 1 n < 1
nΓ 1 n
= Γ
1 + 1
n
≤ 1
which holds because by letting x = 1 +n1 ∈ (1, 2] in (1), we get
1 − 1 n < −γ
n+ 1 < Γ
1 + 1
n
≤ 1.
(b) By the Mean Value Theorem, for any positive integer n,
Γ(1/n) ≥ n − c ⇔ c ≥ −Γ 1 +n1 − Γ(1)
1 + n1 − 1 = −Γ′(xn) for some xn∈
1, 1 + 1
n
.
Therefore c is the smallest constant such that Γ(1/n) ≥ n − c for all n ∈ N+, if and only if c = sup
n∈N+
(−Γ′(xn)) = − inf
n∈N+Γ′(xn) = − lim
x→1+Γ′(x) = −Γ′(1) = γ
where we applied the fact that, by the strict convexity of Γ, Γ′ is a strictly increasing continuous
function.
