An Approximation Formula for Nielsen’s Beta Function Involving the Trigamma Function

Abstract

We prove that the function $\sigma \left(s\right)$ defined by $\beta \left(s\right)=\frac{6s^2+12s+5}{3s^2(2s+3)}-\frac{{\psi }^{\prime }\left(s\right)}{2}-\frac{\sigma \left(s\right)}{2s^5},s>0$, is strictly increasing with the sharp bounds $0<\sigma \left(s\right)<\frac{49}{120}$, where $\beta \left(s\right)$ is Nielsen’s beta function and ${\psi }^{\prime }\left(s\right)$ is the trigamma function. Furthermore, we prove that the two functions $s\mapsto {(-1)}^{1+\mu }\left[\beta \left(s\right)-\frac{6s^2+12s+5}{3s^2(2s+3)}+\frac{{\psi }^{\prime }\left(s\right)}{2}+\frac{49\mu }{240s^5}\right], \mu =0,1$ are completely monotonic for $s>0$. As an application, double inequality for $\beta \left(s\right)$ involving ${\psi }^{\prime }\left(s\right)$ is obtained, which improve some recent results.

1. Introduction and Preliminaries

The diagamma function is defined by $\psi ={\mathrm{\Gamma }}^{\prime }/\mathrm{\Gamma }$ and its derivative ${\psi }^{\prime }$ is called the trigamma function, where $\mathrm{\Gamma }$ is the Euler’s gamma function [1]. The Nielsen’s beta function [2,3] is defined by

$$\beta \left(s\right)=\frac{1}{2}\left[\psi \left(\frac{s}{2}+\frac{1}{2}\right)-\psi \left(\frac{s}{2}\right)\right],s\ne 0,-1,-2,\dots$$

(1)

and satisfies

$$\beta (s+1)+\beta \left(s\right)=\frac{1}{s},$$

(2)

$$\beta \left(s\right)={\int }_0^{\infty }\frac{e^{-st}}{1+e^{-t}}dt,s>0$$

(3)

and

$$\beta \left(s\right)=\frac{1}{2s}{}_2{F}_1\left(1,1;1+s;\frac{1}{2}\right),$$

(4)

where ${}_2{F}_1$ is Gauss hypergeometric function [4].

In some literature, one finds the function $G\left(s\right)=2\beta \left(s\right)$, which is called the Bateman’s $G-$function [5,6].

Qiu and Vuorinen [7] established the bounds

$$\frac{(3-2ln4)}{s^2}<\beta \left(s\right)-\frac{1}{2s}<\frac{1}{4s^2},s>1/2$$

(5)

and Mortici [8] established the relation

$$0<\psi (s+b)-\psi \left(s\right)\le \psi \left(b\right)+\gamma -b+b^{-1},s\ge 1;b\in (0,1),$$

(6)

where $\gamma$ is the Euler constant. Then, Mahmoud and Agarwal [6] established the asymptotic formula

$$2\beta \left(s\right)-s^{-1}\sim \sum _{k=1}^{\infty }\frac{(2^{2k}-1)B_{2k}}{k}{s}^{-2k},s\to \infty ,$$

(7)

where $B_{r}s$ are the Bernoulli numbers [9]. In addition, they obtained the relation

$$\frac{1}{2s^2+\frac{3}{2}}<2\beta \left(s\right)-s^{-1}<\frac{1}{2s^2},s>0$$

(8)

which improves the lower bound of the inequality (5) for $s>\sqrt{\frac{9-12ln2}{16ln2-11}}\approx 2.74$. In [10], Mahmoud and Almuashi presented the inequality

$$\sum _{k=1}^{2m}\frac{(2^{2k}-1)}{k}{B}_{2k}{s}^{-2k}<2\beta \left(s\right)-s^{-1}<\sum _{k=1}^{2m-1}\frac{(2^{2k}-1)}{k}{B}_{2k}{s}^{-2k},m\in \mathbb{N},$$

(9)

where the constants $\frac{(2^{2k}-1)}{k}{B}_{2k}$ are the best possible. Mahmoud, Talat, and Moustafa [11] presented the family

$$\varrho (\xi ,s)=ln\left(1+\frac{1}{s+\xi }\right)+\frac{2}{s(s+1)},\xi \in [1,2];s>0$$

which is asymptotically equivalent to $2\beta \left(s\right)$ for $s\to \infty$.

Recently, Nantomah [12] established the inequality

$$\frac{1}{4}{\psi }^{\prime }(s/2+1/4)<\beta \left(s\right)<\frac{1}{8}\left[{\psi }^{\prime }(s/2)+{\psi }^{\prime }((s+1)/2)\right],s>0.$$

(10)

For more inequalities and approximations of the Nielsen’s beta $\beta \left(s\right)$ or Bateman’s $G-$function $G\left(s\right)$, refer to [7,12,13,14,15] and the references therein.

Nielsen’s $\beta -$function is very useful in evaluating and estimating several integrals [1], as well as some mathematical constants such as

$$G=\frac{-{\beta }^{\prime }(1/2)}{4}\approx 0.9159655\dots ,\pi =2\beta (1/2),\zeta (r+1)=\frac{{(-2)}^r{\beta }^{\left(r\right)}\left(1\right)}{r!\left(2^r-1\right)};r\in \mathbb{N},$$

where G is the Catalan’s constant and $\zeta \left(s\right)$ is the Riemann zeta function. The function $\beta \left(s\right)$ is also related to the Euler’s beta function $B(s,v)$ by (see [16])

$$\frac{d}{ds}\left(lnB(s/2,1/2)\right)=-\beta \left(s\right)\mathrm{and}B(s,1-s)=\beta \left(s\right)+\beta (1-s).$$

Furthermore, the function $\beta \left(s\right)$ is related to the important alternating series [6]

$$\sum _{r=0}^{\infty }\frac{{(-1)}^r}{s+r}=\beta \left(s\right),s\ne 0,-1,-2,\dots .$$

Hence, our results enable us to estimate the errors of the numerical values of the Nielsen’s $\beta -$function and its related constants, integrals, series, and mathematical constants.

In this paper, we prove that the function

$$\sigma \left(s\right)=s^5\left[\frac{2\left(6s^2+12s+5\right)}{3s^2(2s+3)}-{\psi }^{\prime }\left(s\right)-2\beta \left(s\right)\right],s>0$$

is strictly increasing with the sharp bounds $0<\sigma \left(s\right)<\frac{49}{120}$. Furthermore, we prove the completely monotonicity property of two functions related to $\sigma \left(s\right)$. As an application, we obtain the double inequality

$$\frac{6s^2+12s+5}{3s^2(2s+3)}-\frac{1}{2}{\psi }^{\prime }\left(s\right)-\frac{49}{240s^5}<\beta \left(s\right)<\frac{6s^2+12s+5}{3s^2(2s+3)}-\frac{1}{2}{\psi }^{\prime }\left(s\right),s>0$$

whose upper and lower bounds are better than the ones in (10) for $s>0$ and $s\ge 2$, respectively. Furthermore, its upper (lower) bound is better than the counterpart in (8) for $s>1/2$ and $s\ge 2.4$, respectively.

2. Main Results

Recall that an infinite differentiable function $M\left(s\right)$ on $s>0$ is said to be completely monotonic [17] if, for $s>0$ and $r\ge 0$, we have ${(-1)}^r{M}^{\left(r\right)}\left(s\right)\ge 0$. The necessary and sufficient condition for $M\left(s\right)$ to be completely monotonic function on $s>0$ is the convergence of the integral [18]

$$M\left(s\right)={\int }_0^{\infty }{e}^{-st}d\theta \left(t\right),$$

(11)

where $\theta \left(t\right)$ is bounded and non-decreasing for $t\ge 0$.

Theorem 1.

The function

$$H\left(s\right)=2\beta \left(s\right)-\frac{2\left(6s^2+12s+5\right)}{3s^2(2s+3)}+\frac{49}{120s^5}+{\psi }^{\prime }\left(s\right)$$

(12)

is completely monotonic for $s>0$.

Proof. 

Using the integral representation (3), we have

$$H\left(s\right)={\int }_0^{\infty }\frac{e^{-st}h\left(t\right)}{8640\left(e^{2t}-1\right)}{e}^{-\frac{3t}{2}}dt,$$

where

$$\begin{array}{ccc}\hfill h\left(t\right)& =& -147e^{\frac{3t}{2}}{t}^4+147e^{\frac{7t}{2}}{t}^4+9600e^{\frac{3t}{2}}t+8640e^{\frac{5t}{2}}t-960e^{\frac{7t}{2}}t+16640e^{\frac{3t}{2}}\hfill \\ & & -17280e^{\frac{5t}{2}}+640e^{\frac{7t}{2}}-640e^{2t}+640\hfill \\ & =& \sum _{r=11}^{\infty }U\left(r\right)\frac{2^{4-r}{t}^r}{1323r!}\hfill \\ & & +\frac{290803t^{10}}{1120}+\frac{747299t^9}{1920}+\frac{25283t^8}{56}+\frac{10159t^7}{28}+153t^6\hfill \end{array}$$

with

$$\begin{array}{ccc}\hfill U\left(r\right)& =& 1512\left(189(r-5)5^r-\left(35\right)4^r-5(3r-7)7^r+70(5r+13)3^r\right)\hfill \\ & & -\left(\left(2401\right)3^r-\left(81\right)7^r\right)(r-3)(r-2)(r-1)r\hfill \\ & >& 3^r\left(81r^4-486r^3+891r^2-23166r+52920\right)\hfill \\ & & +3^r\left(-2401r^4+14406r^3-26411r^2+543606r+1375920\right)\hfill \\ & & +5^r(285768r-1428840)-\left(52920\right)5^r\hfill \\ & >& 3^r(1428840+520440r-25520r^2+13920r^3-2320r^4)\hfill \\ & & +5^r(285768r-1481760),r\ge 11.\hfill \end{array}$$

Using induction, we obtain

$$\frac{2320r^4-13920r^3+25520r^2-520440r-1428840}{285768r-1481760}<{\left(\frac{5}{3}\right)}^r,r\ge 11,$$

where

$$\begin{array}{ccc}& & \frac{5}{3}\left(\frac{2320r^4-13920r^3+25520r^2-520440r-1428840}{285768r-1481760}\right)\hfill \\ & & -\frac{2320{(r+1)}^4-13920{(r+1)}^3+25520{(r+1)}^2-520440(r+1)-1428840}{285768(r+1)-1481760}\hfill \\ & =& \frac{5Q\left(r\right)}{3969(27r-140)(27r-113)}>0,r\ge 11\hfill \end{array}$$

with

$$Q(r+11)=3132r^5+126266r^4+2004712r^3+15048292r^2+47492336r+25456347>0,r\ge 0.$$

Hence $U\left(r\right)>0$ for $r\ge 11$, which completes the proof. □

From Theorem 1, the function $H\left(s\right)$ is completely monotonic for $s>0$, and therefore it is positive, so we get the following result:

Corollary 1.

Nielsen’s beta function satisfies the inequality

$$\beta \left(s\right)>\frac{6s^2+12s+5}{3s^2(2s+3)}-\frac{49}{240s^5}-\frac{1}{2}{\psi }^{\prime }\left(s\right),s>0.$$

(13)

Theorem 2.

The function

$$F\left(s\right)=\frac{2\left(6s^2+12s+5\right)}{3s^2(2s+3)}-{\psi }^{\prime }\left(s\right)-2\beta \left(s\right)$$

(14)

is completely monotonic for $s>0$.

Proof. 

Using the integral representation (3), we have

$$F\left(s\right)={\int }_0^{\infty }\frac{e^{-st}f\left(t\right)}{27\left(e^{2t}-1\right)}{e}^{-\frac{3t}{2}}dt,$$

where

$$\begin{array}{ccc}\hfill f\left(t\right)& =& -27e^{\frac{5t}{2}}(t-2)+2e^{2t}+e^{\frac{7t}{2}}(3t-2)-2e^{\frac{3t}{2}}(15t+26)-2\hfill \\ & =& \frac{147t^5}{160}+\frac{291t^6}{160}+\frac{8469t^7}{4480}+\sum _{r=8}^{\infty }P\left(r\right)\frac{2^{1-r}{t}^r}{35r!}\hfill \end{array}$$

with

$$P\left(r\right)=-\left(350r\right)3^r-\left(189r\right)5^r+\left(15r\right)7^r-\left(910\right)3^r+\left(35\right)4^r+\left(189\right)5^{r+1}-\left(5\right)7^{r+1},r\ge 8.$$

Using

$$\begin{array}{ccc}& & 3\left(\left(910\right)3^r-\left(35\right)4^r-\left(189\right)5^{r+1}+\left(5\right)7^{r+1}\right)-7\left((-350)3^r-\left(189\right)5^r+\left(15\right)7^r\right)\hfill \\ & & =7\left(\left(740\right)3^r-\left(15\right)4^r-\left(216\right)5^r\right)<0,r\ge 8.\hfill \end{array}$$

Then,

$$\frac{\left(\left(910\right)3^r-\left(35\right)4^r-\left(189\right)5^{r+1}+\left(5\right)7^{r+1}\right)}{-\left(350\right)3^r-\left(189\right)5^r+\left(15\right)7^r}<\frac{7}{3}<r,r\ge 8$$

and hence $P\left(r\right)>0$ for $r\ge 8$, which completes the proof. □

From Theorem 2, the function $F\left(s\right)$ is completely monotonic for $s>0$, and therefore it is positive, so we obtain the following result:

Corollary 2.

Nielsen’s beta function satisfies the inequality

$$\beta \left(s\right)<\frac{6s^2+12s+5}{3s^2(2s+3)}-\frac{1}{2}{\psi }^{\prime }\left(s\right),s>0.$$

(15)

Theorem 3.

The function

$$\sigma \left(s\right)=s^5\left[\frac{2\left(6s^2+12s+5\right)}{3s^2(2s+3)}-{\psi }^{\prime }\left(s\right)-2\beta \left(s\right)\right],s>0$$

(16)

is strictly increasing with the sharp bounds $0<\sigma \left(s\right)<\frac{49}{120}$.

Proof. 

Using the relation

$$\sigma \left(s\right)=s^{5}F\left(s\right),$$

we obtain

$$\frac{d}{ds}\sigma \left(s\right)=s^4{\int }_0^{\infty }\frac{e^{-st}m\left(t\right)}{27{\left(e^{2t}-1\right)}^2}{e}^{-\frac{3t}{2}}dt,$$

where

$$\begin{array}{ccc}\hfill m\left(t\right)& =& -27e^{\frac{5t}{2}}{t}^2-54e^{\frac{7t}{2}}{t}^2-27e^{\frac{9t}{2}}{t}^2\hfill \\ & & +90e^{\frac{3t}{2}}t+135e^{\frac{5t}{2}}t-207e^{\frac{7t}{2}}t-27e^{\frac{9t}{2}}t+9e^{\frac{11t}{2}}t-6e^{2t}t+3e^{4t}t+3t\hfill \\ & & +208e^{\frac{3t}{2}}-216e^{\frac{5t}{2}}-200e^{\frac{7t}{2}}+216e^{\frac{9t}{2}}-8e^{\frac{11t}{2}}-16e^{2t}+8e^{4t}+8\hfill \\ & =& \sum _{r=27}^{\infty }V\left(r\right)\frac{2^{-2-r}{t}^r}{40425r!}\hfill \\ & & +\frac{623698635091326968990549t^{26}}{469868607935879589931253760000}+\frac{1625663466842958699257t^{25}}{278886875555484087091200000}\hfill \\ & & +\frac{122689699708378183999t^{24}}{5019963759998713567641600}+\frac{35494192493938036189t^{23}}{363765489854979244032000}\hfill \\ & & +\frac{428947480570000501t^{22}}{1159966485506949120000}+\frac{2568895942714902679t^{21}}{1937671288290017280000}\hfill \\ & & +\frac{1239688163864594033t^{20}}{276810184041431040000}+\frac{27562668794800823t^{19}}{1942527607308288000}\hfill \\ & & +\frac{10183224174059897t^{18}}{242815950913536000}+\frac{11397477274003t^{17}}{99189522432000}+\frac{7182473003897t^{16}}{24797380608000}\hfill \\ & & +\frac{16503914196013t^{15}}{24797380608000}+\frac{38836308383t^{14}}{28178841600}+\frac{719523341t^{13}}{283852800}+\frac{72073021t^{12}}{17740800}\hfill \\ & & +\frac{23771543t^{11}}{4300800}+\frac{3304163t^{10}}{537600}+\frac{23563t^9}{4480}+\frac{3473t^8}{1120}+\frac{153t^7}{160}\hfill \end{array}$$

with

$$\begin{array}{ccc}\hfill V\left(r\right)& =& -\left(698544r^2\right)5^r-\left(712800r^2\right)7^r-\left(215600r^2\right)9^r+\left(9430344r\right)5^r-\left(8850600r\right)7^r\hfill \\ & & +\left(121275r\right)8^r-\left(754600r\right)9^r+\left(264600r\right){11}^r-\left(121275r\right)4^{r+1}\hfill \\ & & +\left(1078000r\right)3^{r+2}+\left(11211200\right)3^{r+1}-\left(117600\right){11}^{r+1}-\left(1397088\right)5^{r+2}\hfill \\ & & -\left(660000\right)7^{r+2}+\left(431200\right)9^{r+2}-\left(40425\right)4^{r+3}+\left(40425\right)2^{3r+5}\hfill \\ & >& {11}^r(264600r-1293600)-4^r(485100r+2587200)\hfill \\ & & -9^r\left(215600r^2+754600r-34927200\right)-5^r\left(698544r^2-9430344r+34927200\right)\hfill \\ & & -7^r\left(712800r^2+8850600r+32340000\right),r\ge 27\hfill \\ & >& -9^r\left(215600r^2+754600r-34927200\right)-9^r\left(698544r^2-9430344r+34927200\right)\hfill \\ & & -9^r\left(712800r^2+8850600r+32340000\right)+{11}^r(264600r-1293600)\hfill \\ & & -9^r(485100r+2587200),r\ge 27\hfill \\ & >& {11}^r(264600r-1293600)-44\left(36976r^2+14999r+793800\right)9^r,r\ge 27.\hfill \end{array}$$

Using induction, we obtain

$${\left(\frac{11}{9}\right)}^r>\frac{44\left(36976r^2+14999r+793800\right)}{264600r-1293600},r\ge 27,$$

where

$$\begin{array}{ccc}& & \frac{11}{9}\frac{44\left(36976r^2+14999r+793800\right)}{264600r-1293600}-\frac{44\left(36976{(r+1)}^2+14999(r+1)+793800\right)}{264600(r+1)-1293600}\hfill \\ & & =\frac{11ϵ\left(s\right)}{33075(9r-44)(9r-35)}>0,r\ge 27\hfill \end{array}$$

with $ϵ(s+27)=5161745790+604101173s+24298807s^2+332784s^3$ and hence $V\left(r\right)>0$ for $r\ge 27$. Using the asymptotic expansions

$${\psi }^{\prime }\left(s\right)=\frac{{\pi }^2}{6}+\frac{1}{s^2}-2\zeta \left(3\right)s+\frac{{\pi }^4{s}^2}{30}+O\left(s^3\right),ass\to 0,$$

$${\psi }^{\prime }\left(s\right)=\frac{1}{s}+\frac{1}{2s^2}+\frac{1}{6s^3}-\frac{1}{30s^5}+O\left(s^{-7}\right),ass\to \infty ,$$

$$\beta \left(s\right)=-log\left(2\right)+\frac{1}{s}+\frac{{\pi }^{2}s}{12}-\frac{3}{4}\zeta \left(3\right)s^2+O\left(s^3\right),ass\to 0$$

and

$$\beta \left(s\right)=\frac{1}{2s}+\frac{1}{4s^2}-\frac{1}{8s^4}+O\left(s^{-6}\right),ass\to \infty ,$$

where the Riemann zeta function [4] is defined by $\zeta \left(s\right)={\sum }_{r=1}^{\infty }\frac{1}{r^s}$, $s>1$, then we have

$$\underset{s\to 0^{+}}{lim}\sigma \left(s\right)=0and\underset{s\to \infty }{lim}\sigma \left(s\right)=\frac{49}{120},$$

which completes the proof. □

Remark 1.

The function $\sigma \left(s\right)$ satisfies

$$\sigma \left(s\right)=\frac{s^3}{9}+O\left(s^4\right),ass\to 0^{+}and\sigma \left(s\right)=\frac{49}{120}+O\left(s^{-1}\right),ass\to \infty .$$

Furthermore, the function $\beta \left(s\right)$ satisfies

$$\beta \left(s\right)-\frac{6s^2+12s+5}{3s^2(2s+3)}+\frac{1}{2}{\psi }^{\prime }\left(s\right)=\left(\frac{{\pi }^2}{12}-\frac{2}{81}-log\left(2\right)\right)-\frac{1}{18s^2}+\frac{1}{27s}+O\left(s\right),ass\to 0$$

and

$$\beta \left(s\right)-\frac{6s^2+12s+5}{3s^2(2s+3)}+\frac{49}{240s^5}+\frac{1}{2}{\psi }^{\prime }\left(s\right)=\frac{17}{32s^6}+O\left(s^{-7}\right),ass\to \infty .$$

Remark 2.

The upper bound in (15) is better than the counterpart in (10) for $s>0$, where

$$\begin{array}{ccc}& & \left(\frac{6s^2+12s+5}{3s^2(2s+3)}-\frac{{\psi }^{\prime }\left(s\right)}{2}\right)-\frac{1}{8}\left({\psi }^{\prime }\left(\frac{s}{2}\right)+{\psi }^{\prime }\left(\frac{s+1}{2}\right)\right)\hfill \\ & & ={\int }_0^{\infty }-\frac{e^{-\frac{3t}{2}}\left(15e^{\frac{3t}{2}}t+12e^{\frac{5t}{2}}t+26e^{\frac{3t}{2}}-26e^{\frac{5t}{2}}-e^t+1\right)}{27\left(e^t-1\right)}{e}^{-st}dt\hfill \\ & & ={\int }_0^{\infty }-\frac{e^{-(3/2+s)t}}{27\left(e^t-1\right)}\sum _{r=3}^{\infty }\left[10(5r+13)3^r-\left(5\right)2^r+2(12r-65)5^r\right]\frac{2^{-r}{t}^r}{5r!}dt\hfill \\ & & <0,\mathrm{for}s>0.\hfill \end{array}$$

Remark 3.

The lower bound in (13) is better than the counterpart in (10) for $s\ge 2$. To verify that, consider the function

$$T\left(s\right)=\left(-\frac{49}{240s^5}+\frac{6s^2+12s+5}{3s^2(2s+3)}-\frac{{\psi }^{\prime }\left(x\right)}{2}\right)-\frac{1}{4}{\psi }^{\prime }\left(\frac{s}{2}+\frac{1}{4}\right),$$

then ${lim}_{s\to \infty }T(s+2)=0$ and

$$T(s+2)-T(s+4)=\frac{2q\left(s\right)}{{(s+2)}^5{(s+3)}^2{(s+4)}^5{(2s+5)}^2(2s+7)(2s+11)}>0,$$

where

$$\begin{array}{ccc}\hfill q\left(s\right)& =& 480s^{12}+19200s^{11}+348600s^{10}+3785880s^9+27307420s^8+137389520s^7\hfill \\ & & +492611791s^6+1262260470s^5+2278167311s^4+2792735404s^3\hfill \\ & & +2162321672s^2+908271720s+137545680.\hfill \end{array}$$

However, if $\pi \left(s\right)$ is a real-valued function on $s>0$ with ${lim}_{s\to \infty }\pi \left(s\right)=0$ and $\pi \left(s\right)>\pi (s+r)$ for all $s>0$, $r\in \mathbb{N}$, then $\pi \left(s\right)>0$ on $s>0$ (see [19]). Hence $T(s+2)>0$ for $s>0$ or $T\left(s\right)>0$ for $s\ge 2$.

Remark 4.

The upper (lower) bound in inequality (15) (in inequality (13)) is better than the upper (lower) bound in inequality (8) for $s>0.5$ ($s\ge 2.4$), respectively, since the two functions

$$B\left(s\right)=\frac{6s^2+12s+5}{3s^2(2s+3)}-\frac{{\psi }^{\prime }\left(s\right)}{2}-\frac{1}{4s^2}-\frac{1}{2s}$$

and

$$A\left(s\right)=\frac{6s^2+12s+5}{3s^2(2s+3)}-\frac{49}{240s^5}-\frac{{\psi }^{\prime }\left(s\right)}{2}-\frac{1}{4s^2+3}-\frac{1}{2s}$$

satisfy

$$\underset{s\to \infty }{lim}B\left(s\right)=A\left(s\right)=0$$

and

$$B\left(s\right)-B(s+1)=\frac{-24s-35}{12s^2{(s+1)}^2\left(4s^2+16s+15\right)}<0,s>0.5,$$

$$A\left(s\right)-A(s+1)=\frac{C\left(s\right)}{240s^5{(s+1)}^5(2s+3)(2s+5)\left(4s^2+3\right)\left(4s^2+8s+7\right)}>0,s>2.4,$$

where

$$\begin{array}{ccc}\hfill C(s+2.4)& =& 3840s^{11}+134016s^{10}+2082912s^9+\frac{95187392s^8}{5}+\frac{2839058008s^7}{25}\hfill \\ & & +\frac{289284768212s^6}{625}+\frac{4087044207104s^5}{3125}+\frac{7923834916229s^4}{3125}\hfill \\ & & +\frac{50631576895694s^3}{15625}+\frac{194357236141326s^2}{78125}+\frac{1771594294687077s}{1953125}\hfill \\ & & +\frac{431366838881769}{9765625}>0,s>0.\hfill \end{array}$$