**R E S E A R C H**

**Open Access**

## New approximation inequalities for

## circular functions

### Ling Zhu

1*_{and Marija Nenezi´c}

2
*_{Correspondence:}

1_{Department of Mathematics,}

Zhejiang Gongshang University, Hangzhou City, China Full list of author information is available at the end of the article

**Abstract**

In this paper, we obtain some improved exponential approximation inequalities for
the functions (sin*x)/x*and sec(x), and we prove them by using the properties of
Bernoulli numbers and new criteria for the monotonicity of quotient of two power
series.

**MSC:** Primary 26D05; 26D15; secondary 33B10

**Keywords:** Circular functions; Bernoulli numbers; Mitrinovic–Adamovic inequality;
Exponential approximation inequalities

**1 Introduction**

The following result is known as the Mitrinovic–Adamovic inequality [1,2]:

sin*x*
*x*

3

>cos*x*, 0 <*x*<*π*

2. (1.1)

Nishizawa [3] gave the upper bound of the function ((sin*x*)/*x*)3_{in the form of the above}

inequality (1.1) and obtained the following power exponential inequality:

sin*x*
*x*

3

< (cos*x*)1–2x/*π*, 0 <*x*<*π*

2. (1.2)

Chen and Sándor [4] looked into the bounds for the functionsec*x*and obtain the following
result for 0 <*x*<*π*/2:

*π*2

*π*2_{– 4}* _{x}*2 <sec

*x*<

4*π*

*π*2_{– 4}* _{x}*2. (1.3)

Nishizawa [3] obtained the following inequality with power exponential functions derived from the right-hand inequality side of (1.3):

4*π*
*π*2_{– 4}* _{x}*2

4x2_{/}* _{π}*2

<sec*x*, 0 <*x*<*π*

2. (1.4)

The purpose of this article is to establish some exponential approximation inequalities which improve the ones of (1.1)–(1.4). We prove these results for circular functions by us-ing the properties of Bernoulli numbers and new criteria for the monotonicity of quotient of two power series.

**Theorem 1.1** *Let*0 <*x*<*π*/2,*a*= 2/15≈0.13333*and b*= 4/*π*2_{≈}_{0.40528.}_{Then we have}

(cos*x*)1–ax2<

sin*x*
*x*

3

< (cos*x*)1–bx2, (1.5)

*where a and b are the best constants in*(1.5).

**Theorem 1.2** *Let*0 <*x*<*π*/2,*c*= 19/945≈0.02011*and d*= 8(30 –*π*2)/(15*π*4)≈0.11022.

*Then we have*

(cos*x*)1–2x2/15–cx4<

sin*x*
*x*

3

< (cos*x*)1–2x2/15–dx4, (1.6)

*where c and d are the best constants in*(1.6).

**Theorem 1.3** *Let*0 <*x*<*π*/2,*b*= 4/*π*2_{≈}_{0.40528}_{and p}_{= 1/(2}_{ln}_{(4/}_{π}_{))}_{≈}_{2.0698.}_{Then}*we have*

4*π*
*π*2_{– 4}* _{x}*2

*bx*2

<sec*x*<

4*π*
*π*2_{– 4}* _{x}*2

*px*2

, (1.7)

*where b and p are the best constants in*(1.7).

**Theorem 1.4** *Let*0 <*x*<*π*/2,

*α*= 1

12ln* _{π}*4 –

2

*π*2_{ln}2 4

*π*

≈–3.1277, *β*= 16

*π*4

1 –1 8

*π*2

ln* _{π}*4

≈–0.67462.

*Then we have*

4*π*
*π*2_{– 4}* _{x}*2

*x*2/(2ln(4/*π*))+*αx*4

<sec*x*<

4*π*
*π*2_{– 4}* _{x}*2

*x*2/(2ln(4/*π*))+*βx*4

, (1.8)

*whereαandβare the best constants in*(1.8).

We note that the right-hand side of the inequality (1.5) is stronger than that one in (1.2) due to

1 – 4

*π*2*x*
2_{=}

1 +2*x*

*π*

1 –2*x*

*π*

> 1 –2*x*

*π*

while the double inequality (1.6) and (1.8) are sharper than the (1.5) and (1.7), respectively.

**2 Lemmas**

**Lemma 2.1**([5–8]) *Let B2nbe the even-indexed Bernoulli numbers*,*n*= 1, 2, . . . .*Then*

2(2*n*)!
(2*π*)2n

22n

22n_{– 1}<|*B2n*|<

2(2*n*)!
(2*π*)2n

22n

22n_{– 2}, (2.1)

22n–1_{– 1}

22n+1_{– 1}

(2*n*+ 2)(2*n*+ 1)

*π*2 <

|*B*2n+2|
|*B2n*| <

22n_{– 1}

22n+2_{– 1}

(2*n*+ 2)(2*n*+ 1)

**Lemma 2.2** *Let B2nbe the even-indexed Bernoulli numbers*.*Then the following power *
*se-ries expansion*:

lnsin*x*

*x* = –

∞

*n=1*

22n

2*n*(2*n*)!|*B2n*|*x*

2n_{,} _{0 <}_{|}_{x}_{|}_{<}_{π}_{,} _{(2.3)}

*and*

ln cos*x*= –

∞

*n=1*

22n– 1
2*n*(2*n*)!2

2n_{|}_{B2n}_{|}* _{x}*2n

_{,}

_{|}

_{x}_{|}

_{<}

*π*

2, (2.4)

*hold*.

*Proof* The following power series expansions can be found in [9, 1.3.1.4(2)(3)]:

cot*x*=1

*x*–

∞

*n=1*

22n
(2*n*)!|*B2n*|*x*

2n–1_{,} _{(2.5)}

tan*x*=

∞

*n=1*

22n_{– 1}

(2*n*)! 2

2n_{|}_{B2n}_{|}* _{x}*2n–1

_{.}

_{(2.6)}

By (2.5) and (2.6) we have

lnsin*x*

*x* =

*x*
0

lnsin*t*
*t*

*dt*=

*x*
0

cot*t*–1

*t*

*dt*

= –

∞

*n=1*

22n

2*n*(2*n*)!|*B2n*|*x*

2n

and

ln cos*x*=

*x*
0

(ln cos*t*)*dt*= –

*x*
0

tan*t dt*

= –

∞

*n=1*

22n_{– 1}

2*n*(2*n*)!2

2n_{|}_{B2n}_{|}* _{x}*2n

_{.}

**Lemma 2.3**([10]) *Let anand bn*(*n*= 0, 1, 2, . . .)*be real numbers*,*and let the power series*
*A*(*t*) =∞_{n=0}antn*and B*(*t*) =

∞

*n=0bntn* *be convergent for*|*t*|<*R*(*R*≤+∞).*If bn*> 0*for*
*n*= 0, 1, 2, . . . ,*and ifεn*=*an*/*bnis strictly increasing*(*or decreasing*)*for n*= 0, 1, 2, . . . ,*then*
*the function A*(*t*)/*B*(*t*)*is strictly increasing*(*or decreasing*)*on*(0,*R*) (*R*≤+∞).

In order to prove Theorem1.4, we need the following lemma. We introduce a useful
auxiliary function*Hf,g*. For –∞ ≤*a*<*b*≤ ∞, let*f* and*g*be diﬀerentiable on (*a*,*b*) and

*g*= 0 on (*a*,*b*). Then the function*Hf,g*is deﬁned by

*Hf*,g=
*f*
*gg*–*f*.

**Lemma 2.4**([11]) *Let A*(*t*) =∞_{k=0}aktk_{and B}_{(}_{t}_{) =}∞

*k=0bktkbe two real power series *
*con-verging on*(–*r*,*r*)*and bk*> 0*for all k*.*Suppose that*,*for certain m*∈*N*,*the non-constant*
*sequence*{*ak*/*bk*}*is increasing*(*resp*.*decreasing*)*for*0≤*k*≤*m and decreasing*(*resp*.
*in-creasing*)*for k*≥*m*.*Then the function A*/*B is strictly increasing*(*resp*.*decreasing*)*on*(0,*r*)

*if and only if HA,B*(*r*–)≥*(resp.*≤*)*0.*Moreover*,*if HA,B*(*r*–) <*(resp.*>*)*0,*then there exists*
*t*0∈(0,*r*)*such that the function A*/*B is strictly increasing*(*resp*.*decreasing*)*on*(0,*t*0)*and*
*strictly decreasing*(*resp*.*increasing*)*on*(*t*0,*r*).

**3 Proof of Theorem1.1**
Let

*F*1(*x*) =
3lnsin_{x}x

ln cos*x* – 1

*x*2 =

3lnsin* _{x}x*–ln cos

*x*

*x*2

_{ln cos}

*=*

_{x}ln cos*x*– 3lnsin_{x}x

–*x*2_{ln cos}* _{x}* =

*A*(

*x*)

*B*(*x*), 0 <*x*<

*π*

2,

where

*A*(*x*) =ln cos*x*– 3lnsin*x*

*x* = –

∞

*n=1*

22n_{– 1}

2*n*(2*n*)!2

2n_{|}_{B2n}_{|}* _{x}*2n

_{+}∞

*n=1*

3·22n

2*n*(2*n*)!|*B2n*|*x*

2n

= –

∞

*n=1*

22n_{– 4}

2*n*(2*n*)!2

2n_{|}_{B2n}_{|}* _{x}*2n

_{= –}∞

*n=2*

22n_{– 4}

2*n*(2*n*)!2

2n_{|}_{B2n}_{|}* _{x}*2n

= –

∞

*n=1*

22n+2_{– 4}

(2*n*+ 2)(2*n*+ 2)!2

2n+2_{|}_{B}

2n+2|*x*2n+2=
∞

*n=1*
*anx*2n

and

*B*(*x*) = –*x*2ln cos*x*=

∞

*n=1*

22n_{– 1}

2*n*(2*n*)!2

2n_{|}_{B2n}_{|}* _{x}*2n+2

_{=}∞

*n=1*
*bnx*2n

by Lemma2.2. Let

*an*

*bn*= –

16*n*

(2*n*+ 2)(2*n*+ 1)(*n*+ 1)

|*B2n+2*|
|*B2n*| = –*en*,

where

*en*= 16*n*

(2*n*+ 2)(2*n*+ 1)(*n*+ 1)

|*B2n+2*|
|*B2n*| .

We now show that{*en*}is increasing for*n*≥1. Since

*en–1*= 16(*n*– 1)

(2*n*)(2*n*– 1)(*n*)

|*B2n*|
|*B*2n–2|

< 16(*n*– 1)
(2*n*)(2*n*– 1)(*n*)

1
(2*π*)2

2*n*(2*n*– 1)22n–1
22n–1_{– 1} ,

*en*> 16*n*

(2*n*+ 2)(2*n*+ 1)(*n*+ 1)
1
(2*π*)2

(2*n*+ 2)(2*n*+ 1)(22n–1– 1)
22n–1

by Lemma2.1, the proof of*en–1*<*en*for*n*≥2 can be completed when proving

*n*
*n*+ 1

(22n–1_{– 1)}

22n–1 >
*n*– 1

*n*

22n–1

In fact,

*n*222n–1– 12–*n*2– 122(2n–1)= 24n–2– 4*nn*2+*n*2> 0

for*n*≥2. So{*an*/*bn*}*n≥1*is decreasing, and*F*1(*x*) is decreasing on (0,*π*/2) by Lemma2.3.

In view of*F1*(0+) = –2/15, and*F1*((*π*/2)–) = –4/*π*2, the proof of Theorem1.1is complete.

**4 Proof of Theorem1.2**

(i) We ﬁrst prove the left-hand side inequality of (1.6). Let

*F*2(*x*) = 3ln
sin*x*

*x* –

1 – 2
15*x*

2_{–} 19

945*x*

4

ln cos*x*, 0 <*x*<*π*
2.

Then by Lemma2.2we have

*F*2(*x*) =
∞

*n=3*

*in*22n–2|*B*2n|*x*2n+2,

where

*in*= 16(2

2n+2_{– 4)}

(2*n*+ 2)(2*n*+ 2)!

|*B*2n+2|
|*B2n*| –

8(22n_{– 1)}

30*n*(2*n*)! –

19(22n–2_{– 1)}

945(2*n*– 2)(2*n*– 2)!

|*B*2n–2|
|*B2n*| .

By Lemma2.1, we have

*in*> 16(2

2n+2_{– 4)}

(2*n*+ 2)(2*n*+ 2)!

(2*n*+ 2)(2*n*+ 1)(22n–1_{– 1)}

*π*2_{(2}2n+1_{– 1)} –

8(22n_{– 1)}

30*n*(2*n*)!

– 19(2

2n–2_{– 1)}

945(2*n*– 2)(2*n*– 2)!

*π*2_{(2}2n–1_{– 1)}

(2*n*)(2*n*– 1)(22n–3_{– 1)}

= 1
(2*n*)!*jn*

with

*jn*=

16(22n+2_{– 4)}

(2*n*+ 2)

(22n–1_{– 1)}
*π*2_{(2}2n+1_{– 1)}–

8 15

22n_{– 1}

2*n* –

19 945

22n–2_{– 1}

(2*n*– 2)

*π*2_{(2}2n–1_{– 1)}

(22n–3_{– 1)}

> 16(2

2n+2_{– 4)}

(2*n*+ 2)

(22n–1_{– 1)}
79

8(22n+1– 1)

– 8 15

(22n_{– 1)}

2*n* –

19 945·

79 8

(22n–2_{– 1)}

(2*n*– 2)

(22n–1_{– 1)}

(22n–3_{– 1)}

= 1

1,194,480

*h*(*n*)

*n*(2·22n_{– 1)(2}2n_{– 8)(}_{n}_{– 1)(}_{n}_{+ 1)}

due to*π*2< 79/8, where

It is not diﬃcult to verify

*h1*(*n*) =1,061,146*n*2– 2,172,518*n*+ 637,05622n
–13,695,401*n*2– 22,830,487*n*+ 6,052,032
> 0

and

*h2*(*n*) =39,747,422*n*2– 52,928,098*n*+ 7,963,20022n
–27,468,904*n*2– 31,914,392*n*+ 2,548,224
> 0

for*n*≥3. So*in*> 0 for*n*≥3, and*F2*(*x*) > 0 for*x*∈(0,*π*/2).
(ii) Then we prove the right-hand side inequality of (1.6). Let

*F3*(*x*) = 3lnsin*x*

*x* –

1 – 2
15*x*

2_{–} 8

15
30 –*π*2

*π*4 *x*

4

ln cos*x*, 0 <*x*<*π*
2.

Then by Lemma2.2we have

*F3*(*x*) =

∞

*n=2*

*ln*22n–2|*B2n*|*x*2n+2,

where

*ln*= 16(2

2n+2_{– 4)}

(2*n*+ 2)(2*n*+ 2)!

|*B2n+2*|
|*B2n*| –

8 15

22n_{– 1}

2*n*(2*n*)!–
8
15

30 –*π*2
*π*4

22n–2_{– 1}

(2*n*– 2)(2*n*– 2)!

|*B2n–2*|
|*B2n*| .

By Lemma2.1we have

*ln*< 16(2

2n+2_{– 4)}

(2*n*+ 2)(2*n*+ 2)!
22n_{– 1}

22n+2_{– 1}

(2*n*+ 2)(2*n*+ 1)

*π*2 –

8 15

22n_{– 1}

2*n*(2*n*)!

– 8 15

30 –*π*2
*π*4

22n–2_{– 1}

(2*n*– 2)(2*n*– 2)!

*π*2(22n_{– 1)}

(2*n*)(2*n*– 1)(22n–2_{– 1)},

that is,

(2*n*)!*ln*<

16(22n+2_{– 4)}

(2*n*+ 2)

(22n_{– 1)}
*π*2_{(2}2n+2_{– 1)}–

8 15

22n_{– 1}

2*n* –

8 15

30 –*π*2
*π*4

22n–2_{– 1}

(2*n*– 2)

*π*2_{(2}2n_{– 1)}

(22n–2_{– 1)}

= 4 15

22n– 1 *t*(*n*)

*π*2_{n}_{(}* _{n}*2

_{– 1)(4}·

_{2}2n

_{– 1)},

where

(iii) Let

*F4*(*x*) =

3lnsin* _{x}x*
ln cos

*x*– (1 –

2
15*x*

2_{)}

*x*4 , 0 <*x*<
*π*

2.

Then

*F4*0+= –19

945, *F4*

*π*

2

–

= –8 15

30 –*π*2

*π*4 .

This complete the proof of Theorem1.2.

**5 Proof of Theorem1.3**
(1) Let

*G*1(*x*) =ln sec*x*–

2*x*
*π*

2

ln 4*π*

*π*2_{– 4}* _{x}*2, 0 <

*x*<

*π*

2.

Then we get

*G1*(*x*) =

∞

*n=0*
*knx*2n+2,

where

*k0*=1
2–

4

*π*2ln

4

*π* > 0,

*kn*= –

2

*π*

2n+2_{1}

*n*–

22n+2_{– 1}

(2*n*+ 2)(2*n*+ 2)!2

2n+2_{|}_{B2n+2}_{|}

, *n*= 1, 2, . . . .

We now show

*kn*= –

2

*π*

2n+2

1

*n*–

22n+2_{– 1}

(2*n*+ 2)(2*n*+ 2)!2

2n+2_{|}_{B2n+2}_{|}

< 0 (5.1)

for*n*≥1, that is,

2

*π*

2n+2

1

*n*–

22n+2_{– 1}

(2*n*+ 2)(2*n*+ 2)!2

2n+2_{|}_{B2n+2}_{|}_{> 0}

or

|*B2n+2*|< 1

*π*2n+2

(2*n*+ 2)!
22n+2_{– 1}

2*n*+ 2

*n*

holds for*n*≥1. In fact, by Lemma2.1we have

|*B2n+2*|<2(2*n*+ 2)!
(2*π*)2n+2

22n

22n_{– 2},

so (5.1) holds as long as we can prove that

2(2*n*+ 2)!
(2*π*)2n+2

22n

22n_{– 2}<

1

*π*2n+2

(2*n*+ 2)!
22n+2_{– 1}

2*n*+ 2

that is,

*n*22n+2– 1< 4(*n*+ 1)22n– 2,
which is equivalent to

4(*n*+ 1)22n– 2–*n*22n+2– 1= 4·22n– 7*n*– 8 > 0

for*n*≥1. So*kn*< 0 for*n*≥1, which leads to*G*_{1}(*x*) =∞* _{n=2}*2

*n*(2

*n*– 1)(2

*n*– 2)

*knx*2n–3

_{< 0,}

and*G*_{1}(*x*) is decreasing on (0,*π*/2). We can compute

*G*_{1}(*x*) =tan*x*– 8

*π*2*x*ln

–4 *π*
4*x*2_{–}* _{π}*2

+32

*π*2
*x*3

4*x*2_{–}* _{π}*2,

*G*_{1}(*x*) =tan2*x*– 8

*π*2ln

–4 *π*
4*x*2_{–}* _{π}*2

+160

*π*2
*x*2

4*x*2_{–}* _{π}*2–

256

*π*2
*x*4

(4*x*2_{–}* _{π}*2

_{)}2 + 1,

which give

*G*_{1}0+= 1 – 8

*π*2ln

4

*π* ≈0.80420 > 0, *G*
1

*π*

2

–

= –∞.

Then there exists an unique real number*x1*∈(0,*π*/2) such that*G*_{1}(*x*) > 0 on (0,*x1*) and

*G*_{1}(*x*) < 0 on (*x*1,*π*/2). So*G*1(*x*) is increasing on (0,*x*1) and decreasing on (*x*1,*π*/2). Since

*G*_{1}0+= 0, *G*_{1}

*π*

2

–

= –∞,

there exists an unique real number*x*2∈(*x*1,*π*/2) such that*G*1(*x*) > 0 on (0,*x*2) and*G*1(*x*) <

0 on (*x2*,*π*/2). So*G1*(*x*) is increasing on (0,*x2*) and decreasing on (*x2*,*π*/2). In view of

*G1*(0+_{) = 0 =}_{G1}_{((}_{π}_{/2)}–_{), the proof of the left-hand side inequality of (}_{1.7}_{) is complete.}

(2) Let

*G*2(*x*) =
*x*2

2ln* _{π}*4 ln

4*π*

*π*2_{– 4}* _{x}*2–ln sec

*x*, 0 <

*x*<

*π*

2.

Then we get

*G2*(*x*) =

∞

*n=1*

*wnx*2n+2,

where

*wn*= 1

2ln* _{π}*4

2

*π*

2n

1

*n*–

22n+2_{– 1}

(2*n*+ 2)(2*n*+ 2)!2

2n+2_{|}_{B2n+2}_{|}_{,} _{n}_{= 1, 2, . . . .}

We now show*wn*> 0 for*n*≥1, that is,

|*B2n+2*|< (*n*+ 1)(2*n*+ 2)!

holds for*n*≥1. In fact, by Lemma2.1we have

|*B2n+2*|<2(2*n*+ 2)!
(2*π*)2n+2

22n

22n_{– 2},

so (5.2) holds as long as we can prove that

2*n*ln 4

*π*

22n+2– 1<*π*2(*n*+ 1)22n– 2,

which is true for*n*≥1. So*G*_{2}(*x*) > 0, and*G2*(*x*) is increasing on (0,*π*/2). We can compute

*G2*(0+) = 0 and*G2*((*π*/2)–) = +∞, the proof of the right-hand side inequality of (1.7) is
complete.

(3) Let

*G3*(*x*) = ln sec*x*

*x*2_{ln} 4*π*
*π*2–4x2

, 0 <*x*<*π*
2.

Then

*G3*0+= 1

2ln4

*π*

≈2.0698, *G3*

*π*

2

–

= 4

*π*2≈0.40528,

this completes the proof of Theorem1.3.

**6 Proof of Theorem1.4**
Let

*G4*(*x*) =

ln sec*x*
ln 4*π*

*π*2–4*x*2

–1_{2}_{ln}*x*24

*π*

*x*4 =

ln sec*x*–1_{2}_{ln}*x*24

*π*

ln* _{π}*24

_{–4x}

*π*2

*x*4_{ln} 4*π*
*π*2_{–4x}2

=*f*(*x*)

*g*(*x*), 0 <*x*<

*π*

2,

where

*f*(*x*) =*p*1*x*4+
∞

*n=2*
*pnx*2n+2

and

*g*(*x*) =*q1x*4+

∞

*n=2*
*qnx*2n+2

with

*p*1=

1 12–

1 2

1

ln* _{π}*4

2

*π*

4

;

*pn*= 2

2n+2_{– 1}

(2*n*+ 2)(2*n*+ 2)!2

2n+2_{|}_{B2n+2}_{|}_{–}1

2 1

ln* _{π}*4

2

*π*

2n_{1}

*n*, *n*≥2.

*q1*=ln 4

*π* > 0;

*qn*=

2

*π*

2n–2

1

Since
*p1*
*q1* =
1
12–
1
2
1
ln* _{π}*4(

2
*π*)
4
ln4
*π*
≈–1.0624
and
*pn*
*qn*

=2(*n*– 1)

*π*2

4*π*2n

(2*n*+ 2)!

22n+2– 1

*n*+ 1 |*B2n+2*|–
1

ln* _{π}*4

1

*n*

, *n*≥2,

we can obtain

*p1*
*q*1

≈–1.0624 <*p2*

*q*2
= 1
*π*2
1
180*π*

4_{–} 1
ln* _{π}*4

≈–0.36461,
but
*pn*
*qn*
>*pn+1*
*qn+1*
(6.1)

for*n*≥2. The inequality (6.1) is equivalent to

2(*n*– 1)

*π*2

4*π*2n

(2*n*+ 2)!

22n+2_{– 1}

*n*+ 1 |*B2n+2*|–
1

ln* _{π}*4

1

*n*

>2*n*

*π*2

4*π*2n+2

(2*n*+ 4)!

22n+4_{– 1}

*n*+ 2 |*B*2n+4|–
1

ln4

*π*
1

*n*+ 1

, *n*≥2.

By Lemma2.1, we have

2(*n*– 1)

*π*2

4*π*2n

(2*n*+ 2)!

22n+2_{– 1}

*n*+ 1 |*B*2n+2|–
1

ln* _{π}*4

1

*n*

>2(*n*– 1)

*π*2

8

*π*2_{(}_{n}_{+ 1)}–

1

ln* _{π}*4

1
*n*
and
2*n*
*π*2

4*π*2n+2

(2*n*+ 4)!

22n+4_{– 1}

*n*+ 2 |*B2n+4*|–
1

ln* _{π}*4

1

*n*+ 1

<2*n*

*π*2

1

*π*2_{(2}2n+3_{– 1)}

22n+6_{– 4}

*n*+ 2 –

1

ln* _{π}*4

1

*n*+ 1

.

So (6.1) holds when we prove

*n*

1

*π*2_{(2}2n+3_{– 1)}

22n+6_{– 4}

*n*+ 2 –

1

ln* _{π}*4

1

*n*+ 1

< (*n*– 1)

8

*π*2_{(}_{n}_{+ 1)}–

1

ln* _{π}*4

1

*n*

,

or

*π*222n+3– 1(*n*+ 2) >

ln 4

*π*

*n*22n+7+ 4*n*2+ 4*n*– 16,

So

*p1*
*q*1

<*p2*

*q*2

>*p3*

*q*3

>*p4*

*q*4

>· · ·.

Since

*Hf*,g

*π*

2

–

= lim
*x→(π*

2)–

*f*
*gg*–*f*

= 0,

we see that*G4*(*x*) is increasing on (0,*π*/2) by Lemma2.4. In view of

*G4*0+=*α*= 1

12ln* _{π}*4 –

2

*π*2_{ln}2 4

*π*

≈–3.1277,

*G4*

*π*

2

–

=*β*=16

*π*4

1 –1 8

*π*2

ln* _{π}*4

≈–0.67462,

the proof of Theorem1.4is complete.

**7 Remark**

*Remark*7.1 The results of inequalities in Theorems1.1–1.4can be validated by methods

and algorithms developed in [12,13] and [14].

**Funding**

The ﬁrst author was supported by the National Natural Science Foundation of China (no. 11471285 and no. 61772025).

**Competing interests**

The authors declare that they have no competing interests.

**Authors’ contributions**

The authors provided the questions and gave the proof for the main results. They read and approved the manuscript.

**Author details**

1_{Department of Mathematics, Zhejiang Gongshang University, Hangzhou City, China.}2_{School of Electrical Engineering,}

University of Belgrade, Belgrade, Serbia.

**Publisher’s Note**

Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional aﬃliations.

Received: 24 April 2018 Accepted: 9 November 2018
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