## 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 $$M_{2} (a,b) \leq M_{\infty } (a,b)$$.