1 Introduction

The arithmetic-geometric mean which was first investigated by Lagrange and Gauss [1] has found many applications due to its rapid convergence properties. It has played an important role in the calculation of the number π, as well as various elliptic integrals.

Let a and b denote real numbers such that \(a >b >0\). A sequence of arithmetic means and a sequence of geometric means can be constructed by letting \(a_{0}=a\) and \(b_{0} =b\) and defining the recursions [2]

$$ a_{n+1} = \frac{1}{2} ( a_{n} + b_{n}), \qquad b_{n+1} = \sqrt{a_{n} b_{n} } , \quad n=0,1,2, \ldots. $$
(1)

The sequence \((b_{n})\) is increasing and bounded above by a while the sequence \((a_{n})\) is decreasing and bounded below by b since inductively

$$ b < \cdots < b_{n} < a_{n} < \cdots < a. $$
(2)

Therefore, each sequence converges by the monotone convergence theorem. Since it is the case that [3]

$$ 0 \leq a_{n+1} - b_{n+1} \leq \frac{1}{2} \cdot \frac{(a_{n} - b_{n} )^{2}}{( \sqrt{a_{n}} + \sqrt{b_{n}})^{2}}, $$

we observe that \(a_{n}\) and \(b_{n}\) converge to a common limit determined uniquely by \(a_{0}\) and \(b_{0}\). The arithmetic-geometric mean \(AG (a,b)\) is defined as the common limit of these two sequences

$$ AG (a,b) = \lim_{n \rightarrow \infty } a_{n} = \lim _{n \rightarrow \infty } b_{n}. $$

Gauss was able to prove that \(AG ( a,b)\) can be calculated by means of an integral

$$ \frac{1}{AG (a,b)} = \frac{2}{\pi } \int _{0}^{\infty } \frac{dx}{\sqrt{(x ^{2} + a^{2})(x^{2} + b^{2})}}. $$
(3)

Another type of mean which also has many applications is the logarithmic mean \(L (a,b)\) defined to be

$$ \frac{1}{L (a,b)} = \frac{\log (a) - \log (b)}{a - b} = \int _{0}^{ \infty } \frac{dx}{(x+a)(x+b)}. $$
(4)

The similarity of expressions (3) and (4) motivates the introduction of a parametrized set of means which depends on a positive parameter and incorporates these two means. Let \(a,b>0\) be given real numbers and \(p \in (0, \infty )\) then define \(M_{p} (a,b)\) by the integral [4]

$$ \frac{1}{M_{p} (a,b)} = c_{p} \int _{0}^{\infty } \frac{dx}{[ ( x ^{p} + a^{p}) (x^{p} + b^{p}))]^{1/p}}, \quad 0 < p < \infty . $$
(5)

The arithmetic-geometric mean is obtained from (5) by taking \(p=2\) and the logarithmic mean is the case \(p=1\). The constant \(c_{p}\) depends only on the single parameter p and is defined to satisfy the condition \(M_{p} (a,a)=a\). Expressed as an integral it takes the form

$$ \frac{1}{c_{p}} = a \int _{0}^{\infty } \frac{dx}{(x^{p} + a^{p})^{2/p}} = \int _{0}^{\infty } \frac{\,dt}{(t ^{p} + 1)^{2/p}}. $$
(6)

It is the intention here to study some of the properties of \(1/ c_{p}\) and \(M_{p} (a,b)\) [5, 6]. It will be shown that \(1/ c_{p}\) increases monotonically for all \(p \in (0, \infty )\). Some bounds for the means \(M_{p} (a,b)\) are obtained. Two additional means \(M_{0} (a,b)\) and \(M_{\infty } (a,b)\) are defined by

$$ M_{0} (a,b) = \lim_{p \rightarrow 0^{+}} M_{p} (a,b), \qquad M_{\infty } (a,b) = \lim_{p \rightarrow \infty } M_{p} (a,b). $$
(7)

These two limits will be calculated in closed form. It is shown that \(M_{p} (a,b) \leq M_{\infty } (a,b)\) for all \(p \in (0, \infty )\).

In general, a binary symmetric mean \(M (a,b)\) of positive numbers a and b is a function that satisfies the following properties: (i) \(\min (a,b) \leq M (a,b) \leq \max (a,b)\); (ii) \(M (a,b) = M(b,a)\); (iii) \(M ( \lambda a, \lambda b) = \lambda M (a,b)\) for all \(\lambda >0 \); and (iv) \(M (a,b)\) is nondecreasing in a and b. It is the case that \(M_{p} (a,b)\) satisfies (ii)–(iv), and by the end it will be seen all (i)–(iv) hold [7].

To start let us establish some general bounds for \(M_{p} (a,b)\) in an elementary way.

Theorem 1

Let \(a,b >0\), then

$$ \min (a,b) \leq \sqrt{a b} \leq M_{p} (a,b) \leq \biggl( \frac{a^{p} + b ^{p}}{2} \biggr)^{1/p} \leq \max (a,b). $$
(8)

Proof

The first inequality on the left follows from the fact that \(\min (a,b) \leq \sqrt{ \min (a,b)^{2}} \leq \sqrt{a b}\). The last inequality on the right follows from the fact that \(((a^{p} + b ^{p} )/2)^{1/p}\) is strictly increasing with p and the fact that

$$ \lim_{p \rightarrow \infty } \biggl( \frac{a^{p} + b^{p}}{2} \biggr)^{1/p} = \max (a,b). $$

Since \((a^{p/2} - b^{p/2})^{2} \geq 0\), it follows that \(2 (a b)^{p/2} \leq a ^{p} + b^{p}\). Squaring this, we have further \(4 ( a b )^{p} < (a^{p} + b^{p} )^{2}\). The following inequalities follow by first adding \(x^{2p}\) and then \(a^{p} b^{p}\) to both sides of this result:

$$\begin{aligned} x^{2p} + 2 (\sqrt{ a b})^{p} x^{p} + a^{p} b^{p} &\leq x^{2p} + \bigl( a ^{p} + b^{p} \bigr) x^{p} + a^{p} b^{p} \\ & = \bigl( x^{p} + a^{p}\bigr) \bigl( x^{p} + b^{p}\bigr) \leq x^{2p} + \bigl( a^{p} + b^{p}\bigr) x ^{p} + \frac{1}{4} \bigl( a^{p} + b^{p}\bigr)^{2}. \end{aligned}$$

This is equivalent to the inequalities

$$ \bigl( x^{p} + ( \sqrt{a b})^{p} \bigr)^{2/p} \leq \bigl[\bigl(x^{p} + a^{p} \bigr) \bigl(x^{p} + b ^{p}\bigr)\bigr]^{1/p} \leq \biggl( x^{p} + \frac{a^{p} + b^{p}}{2} \biggr)^{2/p}. $$
(9)

Inverting these inequalities and then integrating with respect to x, we get

$$ \int _{0}^{\infty } \frac{dx}{( x^{p} + \frac{a^{p} + b^{p}}{2} )^{2/p}} \leq \int _{0}^{\infty } \frac{dx}{[(x ^{p} + a^{p})(x^{p} + b^{p})]^{1/p}} \leq \int _{0}^{\infty } \frac{dx}{( x^{p} + ( \sqrt{a b})^{p})^{2/p}}. $$
(10)

Multiply through (10) by \(c_{p}\), invert what results, and then use definition (5) of \(M_{p} (a,b)\) to obtain the result

$$ M_{p} ( \sqrt{ab}, \sqrt{a b}) \leq M_{p} (a,b) \leq M_{p} \biggl( \biggl( \frac{a ^{p} + b^{p}}{2}\biggr)^{1/p}, \biggl( \frac{a^{p} + b^{p}}{2}\biggr)^{1/p}\biggr). $$
(11)

Using \(M_{p} (z,z) =z\) for any \(z>0\), the two inner inequalities in (8) are obtained. □

2 Series representations for the reciprocal of \(M_{p} (a,b)\)

Some useful integral and series representations related to these means will be developed next. First, make the substitution \(t = ( y^{p} +1)^{-1}\) with \(dy =- p^{-1} t^{-1-1/p} (1- t)^{1/p -1} \,dt\) so \(c_{p}\) can be put in the form of the beta function integral

$$ \frac{1}{c_{p}} = \frac{1}{p} \int _{0}^{1} t^{1/p -1} ( 1 -t)^{1/p -1} \,dt = \frac{\varGamma (\frac{1}{p})^{2}}{p \varGamma (\frac{2}{p})}. $$
(12)

Returning to the integral for \(M_{p} (a,b)\), let us make the following change of variable:

$$ x^{p} = a^{p} \biggl( \frac{1}{t} -1\biggr) = a^{p} \biggl( \frac{1-t}{t}\biggr) , \qquad dx =- \frac{a}{p} t^{-1-1/p} (1- t)^{1/p -1} \,dt. $$

In this case, the integral for \(M_{p} (a,b)\) takes the following form:

$$ \frac{1}{M_{p} (a,b)} = c_{p} \frac{a}{p} \int _{0}^{1} \frac{t^{1- 1/p} (1-t)^{1/p-1}}{ \frac{a}{t^{1/p}} (a^{p} \frac{1-t}{t} + b ^{p})^{1/p}} \,dt = \frac{c_{p}}{p} \int _{0}^{1} \frac{t^{1/p-1} ( 1 - t)^{1/p-1}}{(a^{p} (1-t) + b^{p} t)^{1/p}} \,dt. $$
(13)

Theorem 2

The function \(1/ c_{p}\) increases monotonically for all \(p \in (0, \infty )\). Moreover, as \(p \rightarrow \infty \), this function admits the asymptotic expansion

$$ \frac{1}{c_{p}} = 2 - \frac{\pi ^{2}}{2 p^{2}} + \frac{4 \zeta (3)}{p ^{3}} + O \biggl( \frac{1}{p^{4}} \biggr). $$
(14)

Proof

Beginning with (12) and setting \(h (p) =1/ c _{p}\), we have

$$ \log \bigl( h (p)\bigr) = \log \biggl( \frac{\varGamma (\frac{1}{p})^{2}}{p \varGamma ( \frac{2}{p})} \biggr). $$

Differentiating both sides of this with respect to p, it follows that

$$ \frac{h' (p)}{h (p)} = \frac{1}{p^{2}} \biggl( 2 \psi \biggl(\frac{2}{p} \biggr) - 2 \psi \biggl( \frac{1}{p}\biggr) -p\biggr). $$

It suffices to show the quantity in brackets is always positive. To this end, substitute the series form of \(\psi (z)\) to obtain

$$\begin{aligned} &2 \psi \biggl( \frac{2}{p}\biggr) - 2 \psi \biggl( \frac{1}{p} \biggr) -p \\ &\quad = -p + \frac{2}{p} \sum_{n=1}^{\infty } \frac{1}{n (n + \frac{2}{p})} + 2 p - \frac{1}{p} \sum_{n=1}^{\infty } \frac{1}{n (n + \frac{1}{p})} - p \\ &\quad = \frac{1}{p} \sum_{n=1}^{\infty } \frac{1}{ (n + \frac{1}{p})(n + \frac{2}{p})} >0. \end{aligned}$$

Since \(h (p) \rightarrow 0\) as \(p \rightarrow 0^{+}\) and \(h' (p) >0\) on \((0, \infty )\), it follows that the function \(h (p)\) is positive and strictly increasing on \((0, \infty )\). □

The reciprocal of \(M_{p} (a,b)\) can be expanded into an infinite series. This expansion will be seen to have several uses. Let us introduce the following expressions:

$$ (a, k) = a (a+1) \cdots (a+k-1), \qquad (a,0)=1, \qquad (a,-1)=0, \quad a \neq 0. $$

Theorem 3

Given \(a,b >0\) and \(p \in (0, \infty )\), the following expansion holds:

$$ \frac{1}{M_{p} (a,b)} = \frac{1}{ \max (a,b)} \sum_{k=0}^{\infty } \frac{ ( \frac{1}{p} , k)}{(\frac{2}{p} , k) k!} \biggl[ 1 - \biggl( \frac{ \min (a,b)}{\max (a,b)} \biggr)^{p} \biggr]^{k}, $$
(15)

with \(0! =1\).

Proof

Both sides of (15) are equal to \(1/a\) when \(a=b\). Assume without loss of generality that \(a > b >0\). Set \(\beta = 1 - (b/a)^{p}\) so we have \(0 < \beta <1\) and

$$ a^{p} (1-t) + b^{p} t = a^{p} (1 - \beta t). $$

Expanding the denominator of (13) into power series gives

$$ \frac{1}{[ a^{p} (1-t) + b^{p} t]^{1/p}} = \frac{1}{a} ( 1 - \beta t)^{-1/p} = \frac{1}{a} \sum_{k=0}^{\infty } \biggl( \frac{1}{p} ,k \biggr) \frac{\beta ^{k}}{k!} t^{k}. $$
(16)

The series on the right-hand side of (16) converges when \(\beta \in (0,1)\), and so integration of the series term by term is justified. Substitute (16) into (13) to obtain

$$\begin{aligned} \frac{1}{M_{p} (a,b)} =& \frac{1}{a B (\frac{1}{p}, \frac{1}{p})} \int _{0}^{1} \sum_{k=0}^{\infty } \prod_{m=0}^{k-1} \biggl( \frac{1}{p} + m\biggr) \frac{\beta ^{k}}{k!} t^{k+1/p-1} (1- t)^{1/p-1} \,dt \\ =& \frac{1}{a B( \frac{1}{p}, \frac{1}{p})} \sum_{k=0}^{\infty } \biggl( \frac{1}{p}, k\biggr) \frac{\beta ^{k}}{k!} \int _{0}^{1} t^{k+1/p-1} (1-t)^{1/p-1} \,dt \\ =& \frac{1}{a} \sum_{k=0}^{\infty } \biggl( \frac{1}{p} , k\biggr) \frac{\alpha ^{k}}{k!} \frac{B ( k + \frac{1}{p}, \frac{1}{p})}{B ( \frac{1}{p}, \frac{1}{p})}. \end{aligned}$$
(17)

Expressing the beta functions in (17) in terms of the gamma function, we find that

$$ \frac{B ( k + \frac{1}{p}, \frac{1}{p})}{B ( \frac{1}{p}, \frac{1}{p})} = \frac{\varGamma (k + \frac{1}{p}) \varGamma (\frac{1}{p})}{\varGamma ( \frac{1}{p} ) \varGamma ( k + \frac{2}{p})} = \prod_{m=0}^{k-1} \frac{( \frac{1}{p} +m)}{( \frac{2}{p} + m)} = \frac{( \frac{1}{p}, k)}{( \frac{2}{p}, k)}. $$
(18)

Substituting (18) into (17), the required result (15) is obtained. □

Another expansion which is relevant to \(M_{p} (a,b)\) is given in the following theorem.

Theorem 4

Given \(a, b>0\) and \(p \in (0, \infty )\),

$$ \frac{1}{M_{p} (a,b)} = \biggl( \frac{2}{a^{p} + b^{p}} \biggr)^{1/p} \sum _{k=0} ^{\infty } \frac{( \frac{1}{p}, k)}{k!} \frac{(\frac{1}{p} , 2k)}{( \frac{2}{p} ,2k) } \cdot \biggl( \frac{a^{p} - b^{p}}{a^{p} + b^{p}} \biggr)^{2k}. $$
(19)

Proof

Completing the square, we can write

$$\begin{aligned} \bigl( x^{p} + a^{p} \bigr) \bigl( x^{p} + b^{p}\bigr) =& \biggl( x^{p} - \frac{a^{p} + b^{p}}{2} \biggr)^{2} - \biggl( \frac{a^{p} - b^{p}}{2} \biggr)^{2} \\ =& \biggl(x^{p} + \frac{a^{p} + b ^{p}}{2} \biggr)^{2} \bigl( 1 + \tau (x)\bigr), \end{aligned}$$
(20)

where \(\tau (x)\) is defined to be

$$ \tau (x) = \frac{ \frac{a^{p} - b^{p}}{2}}{x^{p} + \frac{a ^{p} + b^{p}}{2}}. $$

Clearly, \(| \tau (x)| < 1\) and using (20), the following expansion holds:

$$\begin{aligned} \bigl[ \bigl( x^{p} + a^{p} \bigr) \bigl( x^{p} + b^{p} \bigr) \bigr]^{-1/p} =& \biggl(1 + \frac{a^{p} + b ^{p}}{2} \biggr)^{-2/p} \bigl(1- \tau (x)^{2}\bigr)^{-1/p} \\ =& \biggl( x^{p} + \frac{a^{p} + b^{p}}{2} \biggr)^{-2/p} \sum _{k=0}^{\infty } \frac{( \frac{1}{p} ,k )}{k!} \tau (x)^{k}. \end{aligned}$$
(21)

Substituting expansion (21) into the integral (13) for \(M_{p} (a,b)^{-1}\), we obtain

$$\begin{aligned} \frac{1}{M_{p} (a,b)} =& c_{p} \int _{0}^{\infty } \frac{1}{(x^{p} + \frac{a^{p} + b^{p}}{2} )^{2/p}} \sum _{k=0}^{\infty } \frac{( \frac{1}{p} ,k ) }{k!} {\tau (x)^{2k}} dx \\ =& c_{p} \int _{0}^{\infty } \frac{dx}{(x^{p} + \frac{a^{p} + b ^{p}}{2})^{2/p}} + c_{p} \sum_{k=1}^{\infty } \frac{( \frac{1}{p},k)}{k!} \int _{0}^{\infty } \frac{\tau (x)^{2k}}{(x^{p} + \frac{a^{p} + b^{p}}{2} )^{2/p}} dx \\ =& \biggl( \frac{a^{p} + b^{p}}{2}\biggr)^{-1/p} + c_{p} \sum _{k=1}^{\infty } \frac{( \frac{1}{p} ,k ) }{k!} \int _{0}^{\infty } \frac{ \tau (x)^{2k}}{(x^{p} + \frac{a^{p} + b^{p}}{2} )^{2/p}} dx \\ =& \biggl( \frac{2}{a^{p} + b^{p}}\biggr)^{1/p} + c_{p} \sum _{k=1}^{\infty } \frac{( \frac{1}{p},k)}{k!} \biggl( \frac{a^{p} - b^{p}}{2} \biggr)^{2k} \int _{0} ^{\infty } \frac{dx}{(x^{p} + \frac{a^{p} + b^{p}}{2} )^{2k +2/p}}. \end{aligned}$$
(22)

Consider the integral apart from (22) and make use of the substitution

$$ x = \biggl( \frac{a^{p} + b^{p}}{2} \biggr)^{1/p} y. $$

The integral in (22) takes the form

$$ \int _{0}^{\infty } \frac{dx}{(x^{p} + \frac{a^{p} + b^{p}}{2})^{2k+2/p}} = \biggl( \frac{a ^{p} + b^{p}}{2} \biggr)^{-2k-1/p} \int _{0}^{\infty } \frac{dy}{(y ^{p} +1)^{2k+2/p}}. $$

Finally, introduce the change of variable \(y^{p} +1 = t^{-1}\) into the integral so it takes the form

$$ \int _{0}^{\infty } \frac{dx}{(x^{p} + \frac{a^{p} + b^{p}}{2})^{2k+2/p}} = \biggl( \frac{a ^{p} + b^{p}}{2}\biggr)^{-2k-1/p} \frac{1}{p} B \biggl( 2k + \frac{1}{p}, \frac{1}{p}\biggr). $$
(23)

Multiply (23) by \(c_{p}\) from (12) to obtain

$$ c_{p} \int _{0}^{\infty } \frac{dx}{(x^{p} + \frac{a^{p} + b ^{p}}{2})^{2k+2/p}} = \biggl( \frac{a^{p} + b^{p}}{2}\biggr)^{-2k-1/p} \frac{B (2k + \frac{1}{p}, \frac{1}{p})}{B ( \frac{1}{p}, \frac{1}{p})} = \biggl( \frac{2}{a ^{p} + b^{p}}\biggr)^{2k +1/p} \frac{(\frac{1}{p} ,2k)}{( \frac{2}{p} , 2k)}. $$

Substituting this integral into (22), we arrive at the desired expansion

$$ \frac{1}{M_{p} (a,b)} = \biggl( \frac{2}{a^{p} + b^{p}}\biggr)^{1/p} + \biggl( \frac{2}{a ^{p} + b^{p}}\biggr)^{1/p} \sum_{k=1}^{\infty } \frac{( \frac{1}{p} ,k )}{k!} \frac{( \frac{1}{p} , 2k)}{(\frac{2}{p} ,2k)} \biggl( \frac{a ^{p} - b^{p}}{a^{p} + b^{p}} \biggr)^{2k}. $$

 □

3 Bounds for \(M_{p} (a,b)\)

Theorem 5

Given \(a,b >0\), the following set of inequalities holds:

$$ M_{0} (a,b) = \sqrt{ab} \leq M_{p} (a,b) , \qquad M_{\infty } (a,b) = \frac{2 \max (a,b)}{2 + \log ( \frac{ \max (a,b)}{\min (a,b)} )} \leq \frac{a+b}{2}. $$
(24)

Proof

It can be verified that

$$ \lim_{p \rightarrow 0^{+}} \biggl( \frac{a^{p} + b^{p}}{2} \biggr)^{1/p} = \sqrt{a b}, $$

so the equality \(M_{0} (a,b) = \sqrt{a b}\) follows directly from Theorem 1. The inequalities on either side of \(M_{p} (a,b)\) follow from monotonicity of \(M_{p} (a,b)\) as a function of p. The last two inequalities follow easily if \(a=b\). Suppose then that \(a>b>0\). Since it has been shown that \(c_{p} \rightarrow 1/2\) as \(p \rightarrow \infty \), we conclude that

$$ \lim_{p \rightarrow \infty } \int _{0}^{\infty } \frac{dy}{(y ^{p} +1)^{2/p}} =2. $$

Moreover,

$$\begin{aligned} \lim_{p \rightarrow \infty } \int _{0}^{\infty } \frac{dx}{[(x ^{p} + a^{p})(x^{p} + b^{p})]^{1/p}} =& \int _{0}^{b} \frac{dx}{ab} + \int _{b}^{a} \frac{dx}{xa} + \int _{a}^{\infty } \frac{dx}{x^{2}}\\ =& \frac{2 + ( \log (a) - \log (b))}{a} = \frac{2 + \log ( \frac{a}{b})}{a}. \end{aligned}$$

Therefore,

$$ \lim_{p \rightarrow \infty } M_{p} (a,b) = \frac{2a}{2 + \log ( \frac{a}{b})}. $$

Finally, it is the case that the mean \(M_{1} (a,b)\) satisfies the inequality

$$ \frac{a-b}{\log (a) - \log (b)} \leq \frac{a+b}{2}. $$

Consequently, this implies the following inequality:

$$ 2 (a - b) \leq (a+b) \bigl( \log (a) - \log (b)\bigr). $$

Collecting terms on the right and adding 2a to both sides, we get

$$ 4a \leq 2 (a + b)+ (a+b) \bigl( \log (a) - \log (b)\bigr). $$

This result implies the upper bound for \(M_{\infty } (a,b)\),

$$ \frac{2a}{2 + \log ( \frac{a}{b}) } \leq \frac{a+b}{2}. $$

This is the final inequality on the right in (24), so we are done. □

Lemma 1

For fixed \(p >0\) and \(k \in \mathbb{N}\),

$$ \frac{(\frac{1}{p} +k)^{2}}{k ( \frac{2}{p} +k)} \geq 1. $$
(25)

Proof

Since \((1/p)^{2} \geq 0\), it follows that

$$ \biggl( \frac{1}{p}\biggr)^{2} + 2 \frac{k}{p} + k^{2} \geq 2 \frac{k}{p} + k^{2}. $$

Consequently,

$$ \biggl( \frac{1}{p}+ k\biggr)^{2} \geq k \biggl( \frac{2}{p} +k\biggr). $$

Dividing both sides of this by the right-hand side, (25) is obtained. □

Theorem 6

For \(a,b >0\) and all \(p \in (0, \infty )\), the following bound holds:

$$ M_{p} (a,b) \leq M_{\infty } (a,b). $$
(26)

Proof

The proof relies on expressing series (15) given in Theorem 3 and (24) in a certain way. In fact, the reciprocal of (26) will be shown. Let \(p>0\) and define

$$ r= r_{p} = 1 - \biggl( \frac{\min (a,b)}{\max (a,b)} \biggr)^{p}. $$

Clearly, \(r \in (0,1)\) and this can be solved for the ratio on the right as a function of r,

$$ \frac{\max (a,b)}{\min (a,b)} = (1- r)^{-1/p}. $$
(27)

From Theorem 3, the following expansion holds for any \(p \in (0, \infty )\):

$$ \frac{\max (a,b)}{M_{p} (a,b)} = 1 + \sum_{k=1}^{\infty } \frac{( \frac{1}{p}, k)^{2}}{(\frac{2}{p},k) k!} r^{k}. $$
(28)

Moreover, substituting (27) into (24), with \(r \in (0,1)\), expand the logarithm function in series to obtain

$$ \frac{\max (a,b)}{M_{\infty } (a,b)} = 1 + \frac{1}{2p} \sum_{k=1} ^{\infty } \frac{r^{k}}{k}. $$
(29)

Therefore, it suffices to show the reciprocal of (26) holds,

$$ \sum_{k=1}^{\infty } \frac{(\frac{1}{p}, k)^{2}}{(\frac{2}{p}, k) k!} r^{k} \geq \frac{1}{2 p} \sum_{k=1}^{\infty } \frac{r^{k}}{k}. $$

This is equivalent to the inequality

$$ \sum_{k=1}^{\infty } \biggl[ \frac{( \frac{1}{p}, k)^{2}}{( \frac{2}{q}, k) k!} - \frac{1}{2 p k} \biggr] r^{k} \geq 0. $$
(30)

If it can be shown that the coefficients in (30) are positive for each k, inequality in (30) will follow. This amounts to showing that

$$ \frac{(\frac{1}{p}, k)^{2}}{(\frac{2}{p},k) k!} \geq \frac{1}{2pk}. $$

This implies it has to be shown that

$$ 2 p \biggl( \frac{1}{p}, k\biggr)^{2} \geq \biggl( \frac{2}{p}, k\biggr) (k-1)!. $$
(31)

It is clear that when \(k=1\) and 2 are put in (31) the inequality holds. Suppose (31) holds up to some value of k, so the following statement holds

$$ 2p \biggl(\frac{1}{p}\biggr)^{2} \biggl( \frac{1}{p} +1 \biggr)^{2} \cdots \biggl( \frac{1}{p} +k-1\biggr)^{2} \geq \frac{2}{p} \biggl( \frac{2}{p} +1\biggr) \cdots \biggl( \frac{2}{p} +k -1\biggr) (k-1)!. $$
(32)

Multiply both sides of (32) by \((\frac{1}{p}+k)^{2}\) and use (25) from Lemma 1 to obtain that

$$\begin{aligned}& 2p \biggl(\frac{1}{p}\biggr)^{2} \cdots \biggl( \frac{1}{p} +k-1\biggr)^{2} \biggl(\frac{1}{p} +k \biggr)^{2} \\& \quad \geq \frac{2}{p} \biggl( \frac{2}{p} +1\biggr) \cdots \biggl( \frac{2}{p} +k-1\biggr) \biggl( \frac{2}{p} +k \biggr) k! \cdot \biggl( \frac{( \frac{1}{p} +k)^{2}}{k ( \frac{2}{p} +k)} \biggr) \\& \quad \geq \frac{2}{p} \biggl( \frac{2}{p} + 1 \biggr) \cdots \biggl( \frac{2}{p} + k\biggr) k!. \end{aligned}$$

This is exactly (32) but with k replaced by \(k+1\). By the Principle of Mathematical Induction, (32) holds. □

This serves to generalize the result given in [4] where it was shown that \(M_{2} (a,b) \leq M_{\infty } (a,b)\).