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Some New Series Expansions of a Special Type of Functions Involving the Logarithmic

Function

Christophe Chesneau
Department of Mathematics, LMNO, University of Caen-Normandie, 14032 Caen, France

christophe.chesneau@gmail.com

Abstract. In this article, we present new series expansions for a certain family of functions thatdepend on the logarithmic function. A general result is demonstrated by considering a tunable inter-mediate function. This result has the interest of unifying several important results in the literature,including a well-known series expansion established by Srinivasa Ramanujan. Several precise ex-amples are given and discussed in detail. In addition, we recover the so-called Seidel formula andderive new product expansions, with an emphasis on the so-called Einstein function. Some inequali-ties involving logarithmic functions are also applications of our series expansion approach. Selectedresults are supported by graphical work.

1. Introduction
The logarithmic function, denoted log(x), plays a crucial role in several mathematical contexts,including calculus, number theory, and computer science. See [13], and the references therein.Understanding its properties, especially its various series expansions, is fundamental to manymathematical analyses. The classical (Taylor) expansion of log(x) is given by

log(x) =

+∞∑
k=1

(−1)k−1

k
(x − 1)k .

It is valid for x ∈ (0, 2] only (see [8], among others). From this expansion, we immediately get
x − 1− log(x) =

+∞∑
k=2

1

k
(1− x)k . (1)

A consequence of this result is the following inequality: log(x) ≤ x − 1 for x ∈ (0, 2]. However, asdiscussed in [4], we know that it actually holds for x > 0. Thus, the use of the classical logarithmicexpansion is somehow inadequate for a full understanding of this inequality. There is a kind
Received: 26 Sep 2024.
Key words and phrases. series expansions; logarithmic function; product expansions; inequalities.1

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https://doi.org/10.28924/ada/ma.5.12
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Eur. J. Math. Anal. 10.28924/ada/ma.5.12 2of "gap in understanding". A solution is given in [4]. Using a telescoping technique and carefuldevelopment, the following series expansion is demonstrated:
x − 1− log(x) =

+∞∑
k=1

2k−1(x2
−k − 1)2,

and it is valid for x > 0. From this result, we immediately observe that log(x) ≤ x−1 for x > 0; theconstraint x ∈ (0, 2] is relaxed. This key inequality is now fully understandable using the seriesexpansion tool. In addition, the underlying telescoping technique provides an original alternativeproof, making it very interesting from a mathematical point of view. It can also be used for otherpurposes; the proof of natural logarithmic inequalities is just one example.On the other hand, from a completely different perspective, a famous result of Srinivasa Ramanu-jan ensures that, for x > 0 with x 6= 1, we have
1

log(x)
+
1

1− x =
+∞∑
k=1

1

2k(1 + x2
−k
)
.

See [12, page 364]. The proof is based on an iteration technique, noting that 1/(1 − x) =
(1/2)[1/(1 +

√
x) + 1/(1 −

√
x)]. It received special attention in [2, Chapter 31, Entry 29, page399] and was the object of an in-depth study in [5]. As a new visual note, this extension can bereformulated as
1

x − 1 −
1

log(x)
= −

+∞∑
k=1

1

2k(1 + x2
−k
)
.

Based on this form, doing a parallel with the formula in Equation (1), a functional pattern seemsto be present. In fact, both expansions can be expressed as
φ(x − 1)− φ[log(x)] =

+∞∑
k=1

ak(x), (2)
where φ(t) = t and φ(t) = 1/t , respectively, and ak(x) = 2k−1(x2−k−1)2 and ak(x) = −1/2k(1+
x2
−k
), respectively. Given this, a unified approach seems possible.In this article, we formalize such an approach. It aims to generate a wide range of new seriesexpansions of certain functions that depend on the logarithmic function, i.e., functions of the form

φ(x − 1) − φ[log(x)]. The proof is based on telescoping techniques inspired by [4] and precisefactorization developments. The results established have interesting consequences, including thederivation of old and new product expansions. Among other things, a new product expansion ofthe Einstein function, i.e., E2(x) = x/(ex − 1) (see [1]), is established. In addition, inequalitiesinvolving logarithmic functions are obtained almost immediately, in the spirit of [3, 6, 7, 9–11, 16].Some graphics illustrate the results.The following sections structure the article: Section 2 is devoted to the general result on theseries expansion of φ(x − 1) − φ[log(x)] and emphasizes several examples. Section 3 deals with

https://doi.org/10.28924/ada/ma.5.12


Eur. J. Math. Anal. 10.28924/ada/ma.5.12 3some of its implications, including product expansions and inequalities. A conclusion is given inSection 4.
2. Results

Our general and specific findings are presented in this section.
2.1. A general result. The theorem below suggests a series expansion for the difference function
φ(x − 1)− φ[log(x)], defined with a certain function φ.
Theorem 2.1. Let x > 0 and φ be a continuous function such that |φ(x−1)| < +∞ and |φ[log(x)]| <
+∞. Then we have

φ(x − 1)− φ[log(x)] =
+∞∑
k=1

αk(φ)(x),

where

αk(φ)(x) = φ
[
2k−1(x2

−(k−1) − 1)
]
− φ

[
2k(x2

−k − 1)
]
.

Proof. Introducing an integer n ≥ 1 and using the telescoping technique, we get
φ(x − 1)− φ

[
2n(x2

−n − 1)
]
= φ

[
20(x2

−0 − 1)
]
− φ

[
2n(x2

−n − 1)
]

=

n∑
k=1

{
φ
[
2k−1(x2

−(k−1) − 1)
]
− φ

[
2k(x2

−k − 1)
]}

=

n∑
k=1

αk(φ)(x).

Since limn→+∞ 2n(x2−n−1) = limn→+∞ 2n(e2−n log(x)−1) = limn→+∞ 2n {[1 + 2−n log(x)]− 1} =
log(x), thanks to the continuity of φ, we obtain

φ(x − 1)− φ[log(x)] = φ(x − 1)− φ
[
lim

n→+∞
2n(x2

−n − 1)
]

= lim
n→+∞

{
φ(x − 1)− φ

[
2n(x2

−n − 1)
]}

= lim
n→+∞

n∑
k=1

αk(φ)(x) =

+∞∑
k=1

αk(φ)(x).

This ends the proof of the theorem. �

From this theorem, if φ is bijective, then the following series expansion of the logarithmic functionholds:
log(x) = φ−1

{
φ(x − 1)−

+∞∑
k=1

αk(φ)(x)

}
.

It can be useful in several mathematical contexts.

https://doi.org/10.28924/ada/ma.5.12


Eur. J. Math. Anal. 10.28924/ada/ma.5.12 4In fact, this theorem has the advantage of being general and tunable thanks to the function φ.It also unifies several results in the literature, as will be developed in the next part.
2.2. Specific results. The proposition below shows some consequences of Theorem 2.1, includingsome new results.
Proposition 2.2. The series expansions below are valid.(1) For x > 0, we have

x − 1− log(x) =
+∞∑
k=1

2k−1(x2
−k − 1)2.

As mentioned in the introduction, this result is not new; it was established in [4].(2) For x > 0, we have

1

x − 1 −
1

log(x)
= −

+∞∑
k=1

1

2k(1 + x2
−k
)
.

As mentioned in the introduction, this result is not new; it is a famous result proved by
Srinivasa Ramanujan, as highlighted in [2, Chapter 31, page 399] and [5].

To the best of our knowledge, the eight results below are new.(3) For x > 0, we have

(x − 1)2 − [log(x)]2 =
+∞∑
k=1

22(k−1)(x2
−k − 1)3(3 + x2−k ).

(4) For x ≥ 1, we have

√
x − 1−

√
log(x) =

+∞∑
k=1

2(k−1)/2(x2
−k − 1)3/2√

1 + x2
−k
+
√
2

.

(5) For x > 1, we have

1√
x − 1

−
1√
log(x)

= −
+∞∑
k=1

1

2k/2
[√
1 + x2

−k
+
√
2
]√x2

−k − 1
1 + x2

−k .

(6) For x > 1, we have

log(x − 1)− log [log(x)] =
+∞∑
k=1

log

(
1 + x2

−k

2

)
.

(7) For x > 0, we have

sin(x − 1)− sin[log(x)] = 2
+∞∑
k=1

sin
[
2k−2(x2

−k − 1)2
]
cos
[
2k−2(x2

−k − 1)(3 + x2−k )
]
.

(8) For x > 0, we have

cos(x − 1)− cos[log(x)] = −2
+∞∑
k=1

sin
[
2k−2(x2

−k − 1)2
]
sin
[
2k−2(x2

−k − 1)(3 + x2−k )
]
.

https://doi.org/10.28924/ada/ma.5.12


Eur. J. Math. Anal. 10.28924/ada/ma.5.12 5(9) For x > 0, we have

sinh(x − 1)− sinh[log(x)] = 2
+∞∑
k=1

sinh
[
2k−2(x2

−k − 1)2
]
cosh

[
2k−2(x2

−k − 1)(3 + x2−k )
]
.

(10) For x > 0, we have

cosh(x − 1)− cosh[log(x)] = 2
+∞∑
k=1

sinh
[
2k−2(x2

−k − 1)2
]
sinh

[
2k−2(x2

−k − 1)(3 + x2−k )
]
.

Proof. Let us prove each result, one by one.(1) For x > 0, by applying Theorem 2.1 with φ(t) = t , we obtain
x − 1− log(x) = φ(x − 1)− φ[log(x)] =

+∞∑
k=1

αk(φ)(x),

where
αk(φ)(x) = φ

[
2k−1(x2

−(k−1) − 1)
]
− φ

[
2k(x2

−k − 1)
]

= 2k−1(x2
−(k−1) − 1)− 2k(x2−k − 1)

= 2k−1(x2
−k − 1)(1 + x2−k )− 2k(x2−k − 1)

= 2k−1(x2
−k − 1)

[
(1 + x2

−k
)− 2

]
= 2k−1(x2

−k − 1)2.

The used factorization arguments are the same as those in [4]. Hence, we have
x − 1− log(x) =

+∞∑
k=1

2k−1(x2
−k − 1)2.

(2) For x > 0, applying Theorem 2.1 to φ(t) = 1/t , we establish that
1

x − 1 −
1

log(x)
= φ(x − 1)− φ[log(x)] =

+∞∑
k=1

αk(φ)(x),

where
αk(φ)(x) = φ

[
2k−1(x2

−(k−1) − 1)
]
− φ

[
2k(x2

−k − 1)
]

=
1

2k−1(x2−(k−1) − 1)
−

1

2k(x2
−k − 1)

= −
2k−1(x2

−(k−1) − 1)− 2k(x2−k − 1)
22k−1(x2−(k−1) − 1)(x2−k − 1)

= −
2k−1(x2

−k − 1)2

22k−1(x2−(k−1) − 1)(x2−k − 1)
= −

(x2
−k − 1)2

2k(1 + x2
−k
)(x2

−k − 1)2

= −
1

2k(1 + x2
−k
)
.

https://doi.org/10.28924/ada/ma.5.12


Eur. J. Math. Anal. 10.28924/ada/ma.5.12 6Hence, we have
1

x − 1 −
1

log(x)
= −

+∞∑
k=1

1

2k(1 + x2
−k
)
.

(3) For x > 0, applying Theorem 2.1 to φ(t) = t2, we get
(x − 1)2 − [log(x)]2 = φ(x − 1)− φ[log(x)] =

+∞∑
k=1

αk(φ)(x),

where
αk(φ)(x) = φ

[
2k−1(x2

−(k−1) − 1)
]
− φ

[
2k(x2

−k − 1)
]

=
[
2k−1(x2

−(k−1) − 1)
]2
−
[
2k(x2

−k − 1)
]2

=
[
2k−1(x2

−(k−1) − 1)− 2k(x2−k − 1)
] [
2k−1(x2

−(k−1) − 1) + 2k(x2−k − 1)
]

= 2k−1(x2
−k − 1)22k−1(x2−k − 1)(3 + x2−k )

= 22(k−1)(x2
−k − 1)3(3 + x2−k ).

Hence, we have
(x − 1)2 − [log(x)]2 =

+∞∑
k=1

22(k−1)(x2
−k − 1)3(3 + x2−k ).

(4) For x ≥ 1, using Theorem 2.1 with φ(t) = √t , we obtain
√
x − 1−

√
log(x) = φ(x − 1)− φ[log(x)] =

+∞∑
k=1

αk(φ)(x),

where
αk(φ)(x) = φ

[
2k−1(x2

−(k−1) − 1)
]
− φ

[
2k(x2

−k − 1)
]

=

√
2k−1(x2−(k−1) − 1)−

√
2k(x2

−k − 1)

=
2k−1(x2

−(k−1) − 1)− 2k(x2−k − 1)√
2k−1(x2−(k−1) − 1) +

√
2k(x2

−k − 1)

=
2k−1(x2

−k − 1)2√
2k−1(x2−k − 1)(1 + x2−k ) +

√
2[2k−1(x2−k − 1)]

=
2(k−1)/2(x2

−k − 1)3/2√
1 + x2

−k
+
√
2

.

Hence, we have
√
x − 1−

√
log(x) =

+∞∑
k=1

2(k−1)/2(x2
−k − 1)3/2√

1 + x2
−k
+
√
2

.

https://doi.org/10.28924/ada/ma.5.12


Eur. J. Math. Anal. 10.28924/ada/ma.5.12 7(5) For x > 1, applying Theorem 2.1 to φ(t) = 1/√t , we have
1√
x − 1

−
1√
log(x)

= φ(x − 1)− φ[log(x)] =
+∞∑
k=1

αk(φ)(x),

where
αk(φ)(x) = φ

[
2k−1(x2

−(k−1) − 1)
]
− φ

[
2k(x2

−k − 1)
]

=
1√

2k−1(x2−(k−1) − 1)
−

1√
2k(x2

−k − 1)

= −

√
2k−1(x2−(k−1) − 1)−

√
2k(x2

−k − 1)√
22k−1(x2−(k−1) − 1)(x2−k − 1)

= −
2k−1(x2

−(k−1) − 1)− 2k(x2−k − 1)[√
2k−1(x2−(k−1) − 1) +

√
2k(x2

−k − 1)
]√
22k−1(x2−(k−1) − 1)(x2−k − 1)

= −
2k−1(x2

−k − 1)2[√
2k−1(x2−k − 1)(1 + x2−k ) +

√
2[2k−1(x2−k − 1)]

]√
22k−1(x2−k − 1)2(1 + x2−k )

= −
1

2k/2
[√
1 + x2

−k
+
√
2
]√x2

−k − 1
1 + x2

−k .

Hence, we have
1√
x − 1

−
1√
log(x)

= −
+∞∑
k=1

1

2k/2
[√
1 + x2

−k
+
√
2
]√x2

−k − 1
1 + x2

−k .

(6) For x > 1, it follows from Theorem 2.1 with φ(t) = log(t) that
log(x − 1)− log[log(x)] = φ(x − 1)− φ[log(x)] =

+∞∑
k=1

αk(φ)(x),

where
αk(φ)(x) = φ

[
2k−1(x2

−(k−1) − 1)
]
− φ

[
2k(x2

−k − 1)
]

= log
[
2k−1(x2

−(k−1) − 1)
]
− log

[
2k(x2

−k − 1)
]

= log

[
2k−1(x2

−(k−1) − 1)
2k(x2

−k − 1)

]
= log

[
(x2

−k − 1)(1 + x2−k )
2(x2

−k − 1)

]

= log

(
1 + x2

−k

2

)
.

https://doi.org/10.28924/ada/ma.5.12


Eur. J. Math. Anal. 10.28924/ada/ma.5.12 8Hence, we have
log(x − 1)− log [log(x)] =

+∞∑
k=1

log

(
1 + x2

−k

2

)
.

(7) For x > 0, applying Theorem 2.1 to φ(t) = sin(t), we obtain
sin(x − 1)− sin[log(x)] = φ(x − 1)− φ[log(x)] =

+∞∑
k=1

αk(φ)(x),

where
αk(φ)(x) = φ

[
2k−1(x2

−(k−1) − 1)
]
− φ

[
2k(x2

−k − 1)
]

= sin
[
2k−1(x2

−(k−1) − 1)
]
− sin

[
2k(x2

−k − 1)
]
.

Using the standard trigonometric formula sin(t)− sin(u) = 2 sin[(t − u)/2] cos[(t + u)/2],we get
αk(φ)(x) = 2 sin

[
2k−2(x2

−(k−1) − 1)− 2k−1(x2−k − 1)
]
cos
[
2k−2(x2

−(k−1) − 1) + 2k−1(x2−k − 1)
]

= 2 sin
[
2k−2(x2

−k − 1)2
]
cos
[
2k−2(x2

−k − 1)(3 + x2−k )
]
.

Hence, we have
sin(x − 1)− sin[log(x)] = 2

+∞∑
k=1

sin
[
2k−2(x2

−k − 1)2
]
cos
[
2k−2(x2

−k − 1)(3 + x2−k )
]
.

(8) For x > 0, it follows from Theorem 2.1 with φ(t) = cos(t) that
cos(x − 1)− cos[log(x)] = φ(x − 1)− φ[log(x)] =

+∞∑
k=1

αk(φ)(x),

where
αk(φ)(x) = φ

[
2k−1(x2

−(k−1) − 1)
]
− φ

[
2k(x2

−k − 1)
]

= cos
[
2k−1(x2

−(k−1) − 1)
]
− cos

[
2k(x2

−k − 1)
]
.

Using the standard trigonometric formula cos(t)−cos(u) = −2 sin[(t−u)/2] sin[(t+u)/2],we obtain
αk(φ)(x) = −2 sin

[
2k−2(x2

−(k−1) − 1)− 2k−1(x2−k − 1)
]
sin
[
2k−2(x2

−(k−1) − 1) + 2k−1(x2−k − 1)
]

= −2 sin
[
2k−2(x2

−k − 1)2
]
sin
[
2k−2(x2

−k − 1)(3 + x2−k )
]
.

Hence, we have
cos(x − 1)− cos[log(x)] = −2

+∞∑
k=1

sin
[
2k−2(x2

−k − 1)2
]
sin
[
2k−2(x2

−k − 1)(3 + x2−k )
]
.

https://doi.org/10.28924/ada/ma.5.12


Eur. J. Math. Anal. 10.28924/ada/ma.5.12 9(9) For x > 0, applying Theorem 2.1 to φ(t) = sinh(t), we establish that
sinh(x − 1)− sinh[log(x)] = φ(x − 1)− φ[log(x)] =

+∞∑
k=1

αk(φ)(x),

where
αk(φ)(x) = φ

[
2k−1(x2

−(k−1) − 1)
]
− φ

[
2k(x2

−k − 1)
]

= sinh
[
2k−1(x2

−(k−1) − 1)
]
− sinh

[
2k(x2

−k − 1)
]
.

Using the standard hyperbolic formula sinh(t)− sinh(u) = 2 sinh[(t−u)/2] cosh[(t+u)/2],we get
αk(φ)(x) = 2 sinh

[
2k−2(x2

−(k−1) − 1)− 2k−1(x2−k − 1)
]
cosh

[
2k−2(x2

−(k−1) − 1) + 2k−1(x2−k − 1)
]

= 2 sinh
[
2k−2(x2

−k − 1)2
]
cosh

[
2k−2(x2

−k − 1)(3 + x2−k )
]
.

Hence, we have
sinh(x − 1)− sinh[log(x)] = 2

+∞∑
k=1

sinh
[
2k−2(x2

−k − 1)2
]
cosh

[
2k−2(x2

−k − 1)(3 + x2−k )
]
.

(10) For x > 0, using Theorem 2.1 with φ(t) = cosh(t), we find that
cosh(x − 1)− cosh[log(x)] = φ(x − 1)− φ[log(x)] =

+∞∑
k=1

αk(φ)(x),

where
αk(φ)(x) = φ

[
2k−1(x2

−(k−1) − 1)
]
− φ

[
2k(x2

−k − 1)
]

= cosh
[
2k−1(x2

−(k−1) − 1)
]
− cosh

[
2k(x2

−k − 1)
]
.

Using the standard trigonometric formula cosh(t) − cosh(u) = 2 sinh[(t − u)/2] sinh[(t +
u)/2], we obtain

αk(φ)(x) = 2 sinh
[
2k−2(x2

−(k−1) − 1)− 2k−1(x2−k − 1)
]
sinh

[
2k−2(x2

−(k−1) − 1) + 2k−1(x2−k − 1)
]

= 2 sinh
[
2k−2(x2

−k − 1)2
]
sinh

[
2k−2(x2

−k − 1)(3 + x2−k )
]
.

Hence, we have
cosh(x − 1)− cosh[log(x)] = 2

+∞∑
k=1

sinh
[
2k−2(x2

−k − 1)2
]
sinh

[
2k−2(x2

−k − 1)(3 + x2−k )
]
.

All the claimed expansions are established, ending the proof. �

https://doi.org/10.28924/ada/ma.5.12


Eur. J. Math. Anal. 10.28924/ada/ma.5.12 10Thus, Proposition 2.2 is derived from Theorem 2.1 by using various functions φ, namely φ(t) = t ,
φ(t) = 1/t , φ(t) = t2, φ(t) = √t , φ(t) = 1/√t , φ(t) = log(t), φ(t) = sin(t), φ(t) = cos(t),
φ(t) = sinh(t) and φ(t) = cosh(t), one for each sub-result, in order. In addition, some extendedresults can be proved. For example, for x > 0, based on the proof Theorem 2.1 with the function
φ(t) =

√
|t|, we can extend the item numbered 4 as√

|x − 1| −
√
| log(x)| = sign(x − 1)

+∞∑
k=1

2(k−1)/2|x2−k − 1|3/2√
1 + x2

−k
+
√
2

,

where sign(x − 1) =

1 if x > 1,
0 if x = 1,
−1 if x < 1.

Analogous extension of the item numbered 5 is possible.

With a little mathematical effort, we can get similar series expansions by considering the trans-lated version of φ, i.e., φ(t; a) = φ(t + a) for some a ∈ R. Of course, other interesting functionscan also be examined for φ, such as φ(t) = arctanh(t), which benefits from an interesting additionformula, among other things.It is important to note that the convergence of the series expansions in Proposition 2.2 hasbeen checked on the basis of theoretical and practical work. Let us illustrate graphically theconvergence of the series expansion in the item numbered 3. To do this, we consider the followingtruncated-series function:
ϕ(x ;m) = (x − 1)2 − [log(x)]2 −

m∑
k=1

22(k−1)(x2
−k − 1)3(3 + x2−k ),

where m denotes an integer such that m ≥ 1. Figure 1 displays the plots of ϕ(x ;m) for m =
1, 2, . . . , 15 and four arbitrary values of x .

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Eur. J. Math. Anal. 10.28924/ada/ma.5.12 11

0 5 10 15

−
0
.1

4
−

0
.1

0
−

0
.0

6
−

0
.0

2

m

0 5 10 15

0
.0

0
0
.0

1
0
.0

2
0
.0

3

m

(a) (b)

0 5 10 15

0
.0

0
.5

1
.0

1
.5

2
.0

m

0 5 10 15

0
5

1
0

1
5

2
0

2
5

3
0

3
5

m

(c) (d)
Figure 1. Plots of ϕ(x ;m) for m = 1, 2, . . . , 15 and (a) x = 0.5, (b) x = 1.5, (c)
x = 4, and (d) x = 18.

From this figure we see that, for all the values of x considered, ϕ(x ;m) converges very quicklyto 0; it begins to be very close to the y = 0 axis from m = 7.Let us mention that the proof of the item numbered 2 has a different construction than the onein [2, Chapter 31, page 399], even though it is based on the same functional basis.Also, from Proposition 2.2, several expansions of the logarithmic function can be deduced. Forinstance, based on the item numbered 4, for x ≥ 1, we have
log(x) =

[
√
x − 1−

+∞∑
k=1

2(k−1)/2(x2
−k − 1)3/2√

1 + x2
−k
+
√
2

]2
.

Such expansions are innovative, to the best of our knowledge.Concerning the item numbered 9 in Proposition 2.2, one can remark that
sinh(x − 1)− sinh[log(x)] =

1

2x
−
x

2
− sinh(1− x)

https://doi.org/10.28924/ada/ma.5.12


Eur. J. Math. Anal. 10.28924/ada/ma.5.12 12and, by using the series expansion of the hyperbolic sine function, the following alternative expres-sion is obtained:
sinh(x − 1)− sinh[log(x)] =

1

2x
−
x

2
−
+∞∑
k=0

(1− x)2k+1

(2k + 1)!
.

Clearly, it can be more manageable to use in comparison to the one in the item numbered 9,depending on the context. The same remark holds for the item numbered 10; we have
cosh(x − 1)− cosh[log(x)] = −

1

2x
−
x

2
+

+∞∑
k=0

(1− x)2k

(2k)!
.

Thus, the items numbered 9 and 10 are mainly interesting because of their originality, i.e., thetrigonometric and hyperbolic functions involved, respectively.
3. Applications

Some consequences of our results are described in this part.
3.1. Product expansions. Proposition 2.2 can be used for many purposes, including product ex-pansions. The result below illustrates this claim with an example.
Proposition 3.1. The infinite product expansions below are valid.(1) For x > 1, we have

x − 1
log(x)

=

+∞∏
k=1

1 + x2
−k

2
.

This expansion is, in fact, valid for x ∈ (0,+∞) \ {1}, as discussed later. It "almost"
corresponds to the so-called Seidel formula (see [15]).(2) For x > 1, we have

log(x)

x − 1 =
+∞∏
k=1

2

1 + x2
−k .

This expansion is, in fact, valid for x ∈ (0,+∞) \ {1}, as discussed later.(3) For x > 0, we have

ex − 1
x

=

+∞∏
k=1

1 + e2
−kx

2
.

This expansion is, in fact, valid for x ∈ R \ {0}, as discussed later.(4) For x > 0, we have the following product expansion of the Einstein function:

E2(x) =
x

ex − 1 =
+∞∏
k=1

2

1 + e2
−kx

.

This expansion is, in fact, valid for x ∈ R \ {0}, as discussed later.

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Eur. J. Math. Anal. 10.28924/ada/ma.5.12 13

Proof. Let us prove each result, one by one.(1) We propose two different proofs.
Proof 1: use of Proposition 2.2.: For x > 1, it follows from the item numbered 6 inProposition 2.2 that

log(x − 1)− log [log(x)] =
+∞∑
k=1

log

(
1 + x2

−k

2

)
.

By the continuity of the logarithmic function over its domain, it can be rewritten as
log

[
x − 1
log(x)

]
= log

[
+∞∏
k=1

1 + x2
−k

2

]
.

Composing with the exponential function, we get the desired result, i.e.,
x − 1
log(x)

=

+∞∏
k=1

1 + x2
−k

2
.

Proof 2: iterative scheme.: To provide an alternative proof, we now revisit the originalproof of the Seidel formula in [15]. We can write x − 1 = (1 +
√
x)(
√
x − 1) =

(1 + x2
−1
)(x2

−1 − 1) and, with the same principle, we can write the last term as
x2
−1 − 1 = (1 + x2−2)(x2−2 − 1), and the same for x2−2 − 1, etc. So, for any integer

n ≥ 1, we have
x − 1 = (1 + x2−1)(x2−1 − 1) = (1 + x2−1)(1 + x2−2)(x2−2 − 1)

= . . . =

[
n∏
k=1

(1 + x2
−k
)

]
(x2

−n − 1) =

[
n∏
k=1

1 + x2
−k

2

]
2n(x2

−n − 1).

Therefore, by considering the limit when n → +∞, we obtain
x − 1 =

[
lim

n→+∞

n∏
k=1

1 + x2
−k

2

] [
lim

n→+∞
2n(x2

−n − 1)
]
=

[
+∞∏
k=1

1 + x2
−k

2

]
log(x),

which implies that
x − 1
log(x)

=

+∞∏
k=1

1 + x2
−k

2
.

Thus, Proof 1 offers an alternative to these known developments by using Proposition2.2. Proof 2, however, has the advantage of being valid for x ∈ (0,+∞) \ {1}, not just
x > 1. We will show later why this is actually not a problem.(2) For x > 1, by using the previous result, we get

log(x)

x − 1 =
1

(x − 1)/ log(x) =
1∏+∞

k=1

[
(1 + x2

−k
)/2
]

=

+∞∏
k=1

1

(1 + x2
−k
)/2
=

+∞∏
k=1

2

1 + x2
−k .

https://doi.org/10.28924/ada/ma.5.12


Eur. J. Math. Anal. 10.28924/ada/ma.5.12 14(3) For y > 0, by applying the result in the item numbered 1 with x = ey > 1, we establishthat
ey − 1
y

=
x − 1
log(x)

=

+∞∏
k=1

1 + x2
−k

2
=

+∞∏
k=1

1 + e2
−ky

2
.

(4) For x > 0, by using the result in the previous item, we have
E2(x) =

x

ex − 1 =
1

[(ex − 1)/x ] =
1∏+∞

k=1[(1 + e
2−kx)/2]

=

+∞∏
k=1

1

(1 + e2
−kx)/2

=

+∞∏
k=1

2

1 + e2
−kx

.

This ends the proof. �

From the item numbered 2, for x ≥ 1, we get the following product expansion of the logarithmicfunction:
log(x) = (x − 1)

+∞∏
k=1

2

1 + x2
−k , (3)

which corresponds to the Seidel formula restricted to (1,+∞). We can complete it for x ∈ (0, 1)based on the case x ≥ 1. Indeed, for x ∈ (0, 1), since 1/x > 1 and ∑+∞k=1 2−k = 1, we get
log(x) = − log

(
1

x

)
= −

(
1

x
− 1
) +∞∏
k=1

2

1 + x−2−k

= (x − 1)
1

x

[
+∞∏
k=1

2

1 + x2
−k

][
+∞∏
k=1

x2
−k

]
= (x − 1)

[
+∞∏
k=1

2x2
−k

1 + x2
−k

] [
x−1+

∑+∞
k=1 2

−k
]

= (x − 1)
+∞∏
k=1

2

1 + x2
−k .

We thus find the Seidel formula in its entirety, as mentioned in the second proof of the itemnumbered 1 in Proposition 3.1. The advantages of this decomposition are that it has no constraintson the natural domain of definition, i.e., x > 0, to satisfy log(x) = − log(1/x), which is not the casefor most series expansions of log(x), and also that log(x) and x − 1 have the same sign accordingto x ∈ (0, 1) and x > 1. It is also underexploited in the literature to determine sharp logarithmicinequalities. We will emphasize this aspect in the next section. Thus, in a sense, the results inTheorem 2.1 unify three known results, one by Srinivasa Ramanujan in [12], one by Ludwig Seidelin [15], and a more recent one by David M. Bradley in [4].As an additional numerical contribution, let us illustrate this expansion by considering thefollowing function:
ζ(x ;m) = log(x)− (x − 1)

m∏
k=1

2

1 + x2
−k ,

https://doi.org/10.28924/ada/ma.5.12


Eur. J. Math. Anal. 10.28924/ada/ma.5.12 15where m denotes an integer such that m ≥ 1. Figure 2 displays the plots of ζ(x ;m) for m =
1, 2, . . . , 15 and four arbitrary values of x , including x = 0.5 ∈ (0, 1) to check the previousstatement.

0 5 10 15

−
0
.1

0
−

0
.0

6
−

0
.0

2

m

0 5 10 15

−
0
.0

4
−

0
.0

3
−

0
.0

2
−

0
.0

1
0
.0

0

m

(a) (b)

0 5 10 15

−
0
.6

−
0
.4

−
0
.2

0
.0

m

0 5 10 15

−
3
.5

−
2
.5

−
1
.5

−
0
.5

m

(c) (d)
Figure 2. Plots of ζ(x ;m) for m = 1, 2, . . . , 15 and (a) x = 0.5, (b) x = 1.5, (c)
x = 4, and (d) x = 18.

This figure shows that ζ(x ;m) converges very quickly to 0 for all the considered values of x ; itstarts very close to the axis y = 0 from m = 7.Sophisticated infinite product formulas can be derived from the natural properties of the loga-rithmic function and Equation (3). In particular, the formulas below are true.
• For x > 0, we have x = e log(x), which is equivalent to

x = e
(x−1)

∏+∞
k=1

2

1+x2
−k .

• For x > 0 and y ∈ R, we have log(xy ) = y log(x), which yields
(xy − 1)

+∞∏
k=1

2

1 + x2
−ky
= y(x − 1)

+∞∏
k=1

2

1 + x2
−k .

https://doi.org/10.28924/ada/ma.5.12


Eur. J. Math. Anal. 10.28924/ada/ma.5.12 16Similarly, for x > 0 and y ∈ R \ {0}, we have log(x) = (1/y) log(xy ), giving the followingexpansions:
log(x) =

1

y
(xy − 1)

+∞∏
k=1

2

1 + x2
−ky

.

In particular, for y = 2m with an arbitrary m > 0, we have
log(x) =

1

2m
(x2

m − 1)
+∞∏
k=1

2

1 + x2
m−k .

• For x > 0 and y > 0, we have log(xy) = log(x) + log(y), which is equivalent to
(xy − 1)

+∞∏
k=1

2

1 + (xy)2
−k = (x − 1)

+∞∏
k=1

2

1 + x2
−k + (y − 1)

+∞∏
k=1

2

1 + y2
−k .

In addition, it is known that the Einstein function can be expressed as a series expansion involvingBernoulli numbers as
E2(x) =

+∞∑
k=0

B−k
k!
xk ,

where
B−k =

k∑
`=0

∑̀
v=0

(−1)v
(
`

v

)
v k

`+ 1

and (`v) = `!/[v !(`− v)!].In some sense, item numbered 4 completes this result by investigating a simple product expansionfor x > 0, given as
E2(x) =

+∞∏
k=1

2

1 + e2
−kx

. (4)
It is interesting to note that this formula is also valid for x < 0. Indeed, in this case, we can remarkthat

E2(x) = e
−xE2(−x) = e−x

+∞∏
k=1

2

1 + e−2−kx
= e−x

[
+∞∏
k=1

2

1 + e2
−kx

][
+∞∏
k=1

e2
−kx

]

=

[
+∞∏
k=1

2

1 + e2
−kx

] [
e−x+x

∑+∞
k=1 2

−k
]
=

+∞∏
k=1

2

1 + e2
−kx

.

Thus, for x ∈ R\{0}, the formula in Equation (4) is true. To the best of our knowledge, this specialrepresentation of the Einstein function is a new result.The next part is about some inequalities derived from our findings.

https://doi.org/10.28924/ada/ma.5.12


Eur. J. Math. Anal. 10.28924/ada/ma.5.12 173.2. Inequalities. Inequalities of various kinds can be derived from our previous results. The propo-sition below proposes an original one.
Proposition 3.2. For x ≥ 1, we have

√
x − 1−

√
log(x) ≤

1

4

+∞∑
k=1

2k/2(x2
−k − 1)3/2.

This last series expansion converges.

Proof. Thanks to the item numbered 4 in Proposition 2.2, for x ≥ 1, we have
√
x − 1−

√
log(x) =

+∞∑
k=1

2(k−1)/2(x2
−k − 1)3/2√

1 + x2
−k
+
√
2

.

Since x ≥ 1, for any integer k ≥ 1, we have x2−k ≥ 1, so √1 + x2−k ≥ √2, which implies that√
1 + x2

−k
+
√
2 ≥ 2

√
2. Furthermore, it is clear that 2(k−1)/2(x2−k − 1)3/2 ≥ 0. Therefore, weobtain

√
x − 1−

√
log(x) ≤

1

2
√
2

+∞∑
k=1

2(k−1)/2(x2
−k − 1)3/2 =

1

4

+∞∑
k=1

2k/2(x2
−k − 1)3/2.

The convergence of this series expansion can be shown by the equivalence technique. Moreprecisely, when k → +∞, we have
2k/2(x2

−k − 1)3/2 = 2−k [2k(x2−k − 1)]3/2 ∼ 2−k [log(x)]3/2,

and 2−k is the term of a convergent geometric series. The desired result is demonstrated. �

This inequality may be more interesting for the lower bound of the series term than for the upperbound of √x − 1−√log(x). In fact, it is difficult to capture the analytic function associated withthe series term.The proposition below is a general inequality setting based on Theorem 2.1.
Proposition 3.3. In the framework of Theorem 2.1, the inequalities below are true.(1) If φ is non-decreasing, then, for any sets of integers M ⊆ {1, 2, . . .}, we have

φ(x − 1)− φ[log(x)] ≥
∑
k∈M

αk(φ)(x).

(2) If φ is non-increasing, then, for any sets of integers M ⊆ {1, 2, . . .}, we have

φ(x − 1)− φ[log(x)] ≤
∑
k∈M

αk(φ)(x).

Proof. For x > 0, let us consider the following function:
ψ(y) = y(x1/y − 1), y ∈ (0,+∞).

https://doi.org/10.28924/ada/ma.5.12


Eur. J. Math. Anal. 10.28924/ada/ma.5.12 18Then we have
ψ′(y) = x1/y −

1

y
x1/y log(x)− 1 = x1/y − x1/y log(x1/y )− 1.

Using the well-known logarithmic identity log(t) ≥ t−1(t − 1) for t > 0, with t = x1/y , we obtain
ψ′(y) ≤ x1/y − x1/yx−1/y (x1/y − 1)− 1 = 0.

As a result, ψ is a non-increasing function. Then, for any integer k ≥ 1, since 2k−1 ≤ 2k , we have
ψ(2k) ≤ ψ(2k−1). Let us now distinguish two cases:

• If φ is non-decreasing, then we have φ [ψ(2k)] ≤ φ [ψ(2k−1)], implying that αk(φ)(x) =
φ
[
ψ(2k−1)

]
− φ

[
ψ(2k)

]
≥ 0. It follows from Theorem 2.1 that, for any sets of integers

M ⊆ {1, 2, . . .}, we have
φ(x − 1)− φ[log(x)] =

+∞∑
k=1

αk(φ)(x) =
∑
k∈M

αk(φ)(x) +
∑
k 6∈M

αk(φ)(x) ≥
∑
k∈M

αk(φ)(x).

• With similar arguments, if φ is non-increasing, then we have φ [ψ(2k)] ≥ φ
[
ψ(2k−1)

],implying that αk(φ)(x) = φ [ψ(2k−1)]− φ [ψ(2k)] ≤ 0. It follows from Theorem 2.1 that,for any sets of integers M ⊆ {1, 2, . . .},
φ(x − 1)− φ[log(x)] =

+∞∑
k=1

αk(φ)(x) =
∑
k∈M

αk(φ)(x) +
∑
k 6∈M

αk(φ)(x) ≤
∑
k∈M

αk(φ)(x).

The desired inequalities are demonstrated. �

Let us exemplify this general result with an immediate application. Taking M = {m, . . . , n},where m and n are integers such that n ≥ m ≥ 1, with regard to the item numbered 3 inProposition 2.2 using φ(t) = t2, the following inequalities hold:
• For x > 1, we have

(x − 1)2 − [log(x)]2 ≥
n∑

k=m

22(k−1)(x2
−k − 1)3(3 + x2−k ).

• For x ∈ (0, 1), we have
(x − 1)2 − [log(x)]2 ≤

n∑
k=m

22(k−1)(x2
−k − 1)3(3 + x2−k ).

We can also remark that, for any integer k ≥ 1, if x > 1, then we have (x2−k − 1)3 > 0, and if
x ∈ (0, 1), then we have (x2−k − 1)3 < 0, and the above inequalities follow.Another simple application is the inequality formulated in the lemma below.
Lemma 3.4. For x > 0, we have

log(x) ≤ 2(
√
x − 1).

Proof. We propose three different proofs.

https://doi.org/10.28924/ada/ma.5.12


Eur. J. Math. Anal. 10.28924/ada/ma.5.12 19

Proof 1: judicious approach.: The following inequality is well known: log(y) ≤ y − 1 for
y > 0. Applying it to the judicious choice y = √x , we get log(x) = 2 log(√x) ≤ 2(√x−1).

Proof 2: use of our series expansion.: We can apply the first items in Propositions 2.2 and3.3 with M = {1}. Indeed, since φ(t) is non-decreasing (or the coefficients of the relatedseries expansion are clearly non-negative), we have
x − 1− log(x) ≥

∑
k∈M
2k−1(x2

−k − 1)2 = (
√
x − 1)2,

implying that
log(x) ≤ x − 1− (

√
x − 1)2 = 2(

√
x − 1).

Proof 3: use of differentiation.: In the proof of Proposition 3.3, we showed that, for x > 0,the function ψ(y) = y(x1/y − 1) is non-increasing. This implies that, for any θ > 0, wehave limy→0+ ψ(y) ≤ ψ(θ), i.e.,
log(x) ≤ θ(x1/θ − 1).

The desired result is just a special case; it is enough to take θ = 2.This completes the proof. �

This lemma is not new; it has been demonstrated with other differentiation techniques in [17],and its sharpness has also been illustrated.Another logarithmic inequality is highlighted in the lemma below.
Lemma 3.5. For x > 0 and any integers m and n such that n ≥ m ≥ 1, we have

log(x) ≤ (x − 1)
n∏

k=m

2

1 + x2
−k .

Proof. The proof is a consequence of the infinite product expansion in Equation (3). Let us distin-guish the cases x ≥ 1 and x ∈ (0, 1).
• For x ≥ 1 and any integer k ≥ 1, we have x2−k ≥ 1, implying that 2/(1 + x2−k ) ≤ 1. Sowe have

log(x) = (x − 1)

[
m−1∏
k=1

2

1 + x2
−k

][
n∏

k=m

2

1 + x2
−k

][
+∞∏

k=n+1

2

1 + x2
−k

]

≤ (x − 1)
n∏

k=m

2

1 + x2
−k ,

with the convention ∏0k=1[2/(1 + x2−k )] = 1.
• For x ∈ (0, 1) and any integer k ≥ 1, we have x2−k ≤ 1, implying that 2/(1 + x2−k ) ≥ 1and x − 1 ≤ 0. The exact same inequality as above is obtained.The proof is therefore finished. �

https://doi.org/10.28924/ada/ma.5.12


Eur. J. Math. Anal. 10.28924/ada/ma.5.12 20This lemma generalizes the famous inequality log(x) ≤ x − 1 for x > 0. It also shows how theSeidel formula, i.e., Equation (3), can be applied to refine it.A similar inequality involving the Einstein function is examined below.
Lemma 3.6. For x > 0 and any integers m and n such that n ≥ m ≥ 1, we have

E2(x) ≤
n∏

k=m

2

1 + e2
−kx

.

For x < 0, the reversed inequality holds.

Proof. The proof follows from the infinite product expansion in Equation (4). Indeed, for x > 0 andany integer k ≥ 1, we have e2−kx > 1, implying that 2/(1 + e2−kx) < 1. As a result, we have
E2(x) =

[
m−1∏
k=1

2

1 + e2
−kx

][
n∏

k=m

2

1 + e2
−kx

][
+∞∏

k=n+1

2

1 + e2
−kx

]

≤
n∏

k=m

2

1 + e2
−kx

,

with the convention ∏0k=1[2/(1 + e2−kx)] = 1. For x < 0, for x > 0 and any integer k ≥ 1, wehave e2−kx < 1, implying that 2/(1 + e2−kx) > 1. Using this, the reversed inequality above isimmediately established. The desired results are obtained. �

The inequalities above are just a sample of what can be derived from our results; other explo-rations are left for future studies.
4. Conclusion

In conclusion, the logarithmic function log(x) is central to several mathematical disciplines. Inparticular, its series expansions play a crucial role in mathematical analysis. While the classicalseries expansion initially presented limitations for the natural domain, i.e., x ∈ (0,+∞), a re-fined telescoping technique elaborated in [4] allows to relax the constraint. This allows a broaderunderstanding of the logarithmic function. In our first investigations, based on this result and awell-known series expansion established by Srinivasa Ramanujan (see [2] and [5]), we discovereda unified functional pattern. This connection can be expressed in the form "φ(x − 1) − φ[log(x)]".In light of this, using telescoping techniques and thorough factorization developments, we generatenew series extensions for such functions. Several examples are given and discussed. As illustrated,these results lead to new product expansions, including one for the Einstein function, and to in-equalities involving the logarithmic function. This article thus contributes to a better understandingof the logarithmic function and series expansions in general, and lays some foundations for futurework.

https://doi.org/10.28924/ada/ma.5.12


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https://doi.org/10.28924/ada/ma.5.12
https://doi.org/10.1007/978-1-4612-0879-2
https://doi.org/10.1007/978-1-4612-0879-2
https://doi.org/10.4169/math.mag.90.5.353
https://doi.org/10.1007/s11139-017-9961-y
https://doi.org/10.1080/00029890.1987.12000678
https://doi.org/10.1080/00029890.1987.12000678
https://doi.org/10.21608/ejmaa.2020.312813
https://doi.org/10.1080/00029890.2019.1605799
https://doi.org/10.1080/00029890.2019.1605799
https://doi.org/10.21608/ejmaa.2020.312859
https://doi.org/10.2307/3615890
https://doi.org/10.1080/07468342.2022.2039553
https://doi.org/10.1080/07468342.2022.2039553
https://doi.org/10.37418/amsj.10.5.14
https://doi.org/10.48550/arXiv.2309.02479

	1. Introduction
	2. Results
	2.1. A general result
	2.2. Specific results

	3. Applications
	3.1. Product expansions
	3.2. Inequalities

	4. Conclusion
	References

