A Remarkable Monotonic Property of the Gamma Function
by 'Literka'.
We can say that the Euler's gamma function is a generalization of a factorial. It has a property that
for x>0
and for a positive integer k>0
We'll restrict our attention to values of argument x>0. Our question is: if we increase an argument of function
by 1, then function is multiplied by x, then what would happen if the argument is increased, for example by 1/2?
The natural answer for this question, without thinking very much, would be that function is multiplied by
.
More or less this is the right answer, although we do not have the same nice formula as for the increment equal 1.
What we mean is that the function
is a bounded function. It has a limit 0 as x tends to 0 and a limit 1 when x tends to infinity. This is a very valuable function, having applications for example in random walks, but we are not interested in this now. 'Literka' learned from a paper of U. Haagerup
(see [1] or [2])
that this is an increasing function (this was only a step in the proof of a certain theorem). Let us define a function
Notice that, for a positive integer n,
because
We see that
This shows that the sequence F(2), F(4), F(6),…. is increasing.
We would like to present and prove a stronger theorem than remark of U. Haagerup. Our proof is different than a proof shown by U. Haagerup.
Theorem (Literka). Let k be a positive integer and
Each term of the last summation is positive. To see this set y=x+p, a=1/k. Then 1>a>0 and p-th term is
and is positive because of convexity of the function 1/y.
We showed that the derivative of the function P(x) is positive, which means that P(x) is increasing, which means that the function S(x) is increasing.
Theorem (Literka). Let 1>a>0 be a positive real number and
Then S(x) is an increasing function for x>0.
After completing this page 'Literka' received an email from Prof. John Roumeliotis with a conjecture that, for every 1>a>0, the function S(x) has a limit 1, when x tends to infinity. We already know that this is true for the case a=1 and a=1/2. 'Literka' thanks Prof. J. Roumeliotis for his email, since 'Literka' could answer positively his question and could enrich the current considerations.
Theorem. Let 1>a>0 be a positive real number and
Then
Sketch of proof. We assume that a=1/k. The general form will be easily deduced.
Since S(x) is increasing, it must have a limit at the infinity - denote it by h. Hence S(x)
h (it means S(x) is close to h) for large x. Since
We have
for large numbers x. This can hold only if h=1.
Use this to prove theorem for the case a is rational. Just split corresponding ratio into a product of finite number of ratios of the form written previously.
References: 1. U. Haagerup, The best constants in the Khintchine inequality, Studia
Math. 70, 231-283 (1982). 2. U. Haagerup, Les Meilleurs constantes de l'inequalite de Khintchine,
Comptes Rendus de l'academie de Science de Paris, Serie A, 286, 259-262
(1978).