Positive solutions to a semipositone superlinear elastic beam equation

Haixia Lu; Li Sun; Haixia Lu; Li Sun

doi:10.3934/math.2021250

AIMS Mathematics

2021, Volume 6, Issue 5: 4227-4237. doi: 10.3934/math.2021250

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Research article Special Issues

Positive solutions to a semipositone superlinear elastic beam equation

Haixia Lu ^{1,2
,
,},
Li Sun ¹

1.
School of Mathematics and Statistics, Jiangsu Normal University, Xuzhou, 221116, P. R. China
2.
School of arts and science, Suqian College, Suqian 223800, P. R. China

Received: 11 October 2020 Accepted: 04 January 2021 Published: 07 February 2021
MSC : 34K10, 37C25

A semipositone fourth-order two-point boundary value problem is considered. In mechanics, the problem describes the deflection of an elastic beam rigidly fastened on the left and simply supported on the right. Under some conditions concerning the first eigenvalue corresponding to the relevant linear operator, the existence of nontrivial solutions and positive solutions to this boundary value problem is obtained. The main results are obtained by using the topological method and the fixed point theory of superlinear operators.

Keywords:

topological degree,
fixed point,
nontrivial solutions and positive solutions,
elastic beam equations

Citation: Haixia Lu, Li Sun. Positive solutions to a semipositone superlinear elastic beam equation[J]. AIMS Mathematics, 2021, 6(5): 4227-4237. doi: 10.3934/math.2021250

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Abstract

1. Introduction

In ^[18] Ramanujan showed a total of 17 series for $1/\pi$ but he did not indicate how he arrived at these series. The Borwein brothers ^[5] gave rigorous proofs of Ramanujan's series for the first time and also obtained many new series for $1/\pi.$ Till now, many new Ramanujan's-type series for $1/\pi$ have been published, (see, for example, ^[4,6,8]). Chu ^[7], Liu ^[15,16] and Wei et al. ^[21,22] gave many $\pi$ -formula with free parameters by means of gamma functions and hypergeometric series. Guillera ^[10] proved a kind of bilateral semi-terminating series related to Ramanujan-like series for negative powers of $\pi$ . Moreover, Guillera and Zudilin ^[11] outlined an elementary method for proving numerical hypergeometric identities, in particular, Ramanujan-type identities for $1/\pi$ . Recently, $q$ -analogues of Ramanujan-type series for $1/\pi$ have caught the interests of many authors (see, for example, ^{[9,12,13,14,20,21]}).

Although various definitions for gamma functions are used in the literature, we adopt the following definition [23, p.76]

$\frac 1{\Gamma(z)} = ze^{\gamma z}\prod\limits_{n = 1}^\infty\left(1+\frac zn\right)e^{-\frac zn}$

where $\gamma$ is the Euler constant defined as

$\gamma = \lim\limits_{n\rightarrow \infty}\left(1+ \frac 12+\cdots+ \frac 1n-\log n\right).$

It is easy to verify that $\Gamma(1) = 1, \Gamma({ \frac 12}) = \sqrt\pi$ and $\Gamma({z+1}) = z\Gamma(z).$ It follows that for every positive integer $n$ , $\Gamma(n) = (n-1)!.$

For any complex $\alpha$ , we define the general rising shifted factorial by

$\begin{align} (z)_{{\alpha}} = \Gamma({z+{\alpha}})/\Gamma(z). \end{align}$

(1.1)

Obviously, $(z)_0 = 1.$ For every positive integer $n$ , we have

$(z)_n = \Gamma({z+n})/\Gamma(z) = z(z+1)\cdots(z+n-1)$

and

$(z)_{-n} = \Gamma({z-n})/\Gamma(z) = \frac 1{(z-1)(z-2)\ldots(z-n)}.$

For convenience, we use the following compact notations

$(a_1,a_2,\ldots,a_m)_n = (a_1)_n(a_2)_n\ldots(a_m)_n$

and

$(a)_{(n_1,n_2,\ldots,n_m)} = (a)_{n_1}(a)_{n_2}\ldots(a)_{n_m}.$

Following ^[1,3], the hypergeometric series is defined by

$_{{r+1}}F_{s}\left[\begin{array}{cccc}a_0,&a_1,&\ldots,&a_r\\&b_1,&\ldots,&b_s\end{array};z\right] = \sum\limits_{k = 0}^\infty \frac {(a_0, a_1,\ldots,a_r)_k}{(b_1,\ldots,b_s)_k} \frac {z^k}{k!},$

where $a_i, b_j(i = 0, 1, \ldots, r, j = 1, 2, \ldots, s)$ are complex numbers such that no zero factors appear in the denominators of the summand on the right hand side.

We let $F_{q:s; v}^{p:r; u}\ ({p, q, r, s, u, v\in \Bbb N_0 = \{0, 1, 2, \ldots\}})$ denote a general (Kampé de Fériet's) double hypergeometric function defined by (see ^[2,19])

$\begin{eqnarray*} &&F_{q:s;v}^{p:r;u}\left[\begin{array}{ccc}{\alpha}_1,\ldots,{\alpha}_p:&a_1,\ldots,a_r;&c_1,\ldots,c_u;\\ \beta_1,\ldots,\beta_q:&b_1,\ldots,b_s;&d_1,\ldots,d_v;\end{array}x,y\right]\\&& = \sum\limits_{m,n = 0}^\infty \frac {({\alpha}_1,\ldots,{\alpha}_p)_{m+n} (a_1,\ldots,a_r)_m(c_1,\ldots,c_u)_n}{(\beta_1,\ldots,\beta_q)_{m+n} (b_1,\ldots,b_s)_m(d_1,\ldots,d_v)_n} \frac {x^m}{m!} \frac {y^n}{n!}, \end{eqnarray*}$

where, for convergence of the double hypergeometric series,

$p+r\leq q+s+1{\quad}\text{and}{\quad} p+u\leq q+v+1,$

with equality only when $|x|$ and $|y|$ are appropriately constrained (see, for details, [19,Eq 1.3(29),p.27]).

There exist numerous identities for such series. For example, we have

Theorem 1.1 (See [,(30)] ) If $Re(e-d) > 0$ and $Re(d+e-a-b-c) > 0,$ then

$F_{1:1;1}^{0:3;3}\left[\begin{array}{ccc}-:&a,b,c;&d-a,d-b,d-c;\\d:&e;&d+e-a-b-c;\end{array}1,1\right] = \frac {\Gamma({e})\Gamma({e+d-a-b-c})\Gamma({e-d})}{\Gamma({e-a})\Gamma({e-b})\Gamma({e-c})}.$

In ^[15], Liu used the general rising shifted factorial and the Gauss summation formula to prove the following four-parameter series expansions formula, which implies infinitely many Ramanujan type series for $1/\pi$ and $\pi$ .

Theorem 1.2 For any complex ${\alpha}$ and $Re(c-a-b) > 0,$ we have

$\sum\limits_{n = 0}^\infty \frac {({\alpha})_{a+n}(1-{\alpha})_{b+n}}{n!\Gamma({c+n+1})} = \frac {({\alpha})_a(1-{\alpha})_b\Gamma({c-a-b})} {({\alpha})_{c-b}(1-{\alpha})_{c-a}}\cdot \frac {\sin \pi{\alpha}}\pi.$

Motivated by Liu's work, in this paper we derive the following result from Theorem 1.1 which enables us to give many double series expansions for $1/\pi$ and $\ \pi$ . To the best of our knowledge, most of the results in this paper have not previously appeared.

Theorem 1.3 If $d\in \Bbb N_0, Re(e-d+{\sigma}-{\delta}) > 0$ and $Re(d+e-a-b-c+{\delta}+{\sigma}-{\alpha}-\beta-{\gamma}) > 0,$ then

$\begin{align*} &\sum\limits_{m,n = 0}^\infty \frac {({\alpha})_{a+m}(\beta)_{b+m}({\gamma})_{c+m}({\delta}-{\alpha})_{d-a+n}({\delta}-\beta)_{d-b+n}({\delta}-{\gamma})_{d-c+n}} {m!n!({\delta}+d)_{m+n}({\sigma})_{e+m}({\delta}+{\sigma}-{\alpha}-\beta-{\gamma})_{d+e-a-b-c+n}}\\ & = \frac {({\alpha})_a(\beta)_b({\gamma})_c({\delta}-{\alpha})_{d-a}({\delta}-\beta)_{d-b}({\delta}-{\gamma})_{d-c}({\sigma}-{\delta})_{e-d}} {({\sigma}-{\alpha})_{e-a}({\sigma}-\beta)_{e-b}({\sigma}-{\gamma})_{e-c}} \cdot \frac {\Gamma({\sigma})\Gamma({{\sigma}-{\delta}})\Gamma({{\delta}+{\sigma}-{\alpha}-\beta-{\gamma}})}{\Gamma({{\sigma}-{\alpha}})\Gamma({{\sigma}-\beta})\Gamma({{\sigma}-{\gamma}})}. \end{align*}$

Several examples of such formulae are

$\sum\limits_{m,n = 0}^\infty \frac {( \frac 12)_m^3( \frac 12)_n^2}{m!n!(m+n)!(m+1)!(2n+1)} = \frac 4{\pi},$

$\sum\limits_{m,n = 0}^\infty \frac {(- \frac 12)_m^3( \frac 32)_n^3}{m!n!(m+n)!(n+3)!( \frac 12)_{m+1}} = \pi,$

and

$\sum\limits_{m,n = 0}^\infty \frac {(- \frac 23)_m^2( \frac 13)_n^3}{m!n!(n+1)!(2-3m)(- \frac 13)_{m+n}} = \frac {\sqrt3\pi}3.$

The remainder of the paper is organized as follows. In section 2 we give the proof of Theorem 1.3. Sections 3 and 4 are devoted to the double series expansions for $1/\pi$ and $\pi$ , respectively.

2. Proof of Theorem 1.3

First of all, by making use of (1.1), Theorem 1.3 can be restated as follows:

$\begin{equation} \begin{split} &\sum\limits_{m,n = 0}^\infty \frac {\Gamma({a+m})\Gamma({b+m})\Gamma({c+m})\Gamma({d-a+n})\Gamma({d-b+n})\Gamma({d-c+n})} {m!n!\Gamma({d+m+n})\Gamma({e+m})\Gamma({d+e-a-b-c+n})}\\ & = \frac {\Gamma(a)\Gamma(b)\Gamma(c)\Gamma({d-a})\Gamma({d-b})\Gamma({d-c})\Gamma({e-d})}{\Gamma(d)\Gamma({e-a})\Gamma({e-b})\Gamma({e-c})}. \end{split} \end{equation}$

(2.1)

From (1.1) it is easy to see that

$\begin{align*} &\Gamma({a+{\alpha}+m}) = ({\alpha})_{a+m}\Gamma({{\alpha}}), \ \Gamma({b+\beta+m}) = (\beta)_{b+m}\Gamma({\beta}),\ \Gamma({c+{\gamma}+m}) = ({\gamma})_{c+m}\Gamma({{\gamma}}),\\ &\Gamma({d-a+{\delta}-{\alpha}+n}) = ({\delta}-{\alpha})_{d-a+n}\Gamma({{\delta}-{\alpha}}),\ \Gamma({d-b+{\delta}-\beta+n}) = ({\delta}-\beta)_{d-b+n}\Gamma({{\delta}-\beta}),\\ &\Gamma({d-c+{\delta}-{\gamma}+n}) = ({\delta}-{\gamma})_{d-c+n}\Gamma({{\delta}-{\gamma}}),\ \Gamma({d+{\delta}+m+n}) = ({\delta})_{d+m+n}\Gamma({{\delta}})\\ &\Gamma({e+m+{\sigma}}) = ({\sigma})_{e+m}\Gamma({\sigma}),\ \Gamma({a+{\alpha}}) = ({\alpha})_a\Gamma({\alpha}),\ \Gamma({b+\beta}) = (\beta)_b\Gamma(\beta),\ \Gamma({c+{\gamma}}) = ({\gamma})_c\Gamma({\gamma}),\\ &\Gamma({d-a+{\delta}-{\alpha}}) = ({\delta}-{\alpha})_{d-a}\Gamma({{\delta}-{\alpha}}),\ \Gamma({d-b+{\delta}-\beta}) = ({\delta}-\beta)_{d-b}\Gamma({{\delta}-\beta}),\\ &\Gamma({d-c+{\delta}-{\gamma}}) = ({\delta}-{\gamma})_{d-c}\Gamma({{\delta}-{\gamma}}),\ \Gamma({e-d+{\sigma}-{\delta}}) = ({\sigma}-{\delta})_{e-d}\Gamma({{\sigma}-{\delta}}),\\ &\Gamma({d+{\delta}}) = ({\delta})_d\Gamma({\delta}),{\quad} \Gamma({e-a+{\sigma}-{\alpha}}) = ({\sigma}-{\alpha})_{e-a}\Gamma({{\sigma}-{\alpha}}),\\ &\Gamma({e-b+{\sigma}-\beta}) = ({\sigma}-\beta)_{e-b}\Gamma({{\sigma}-\beta}),\ \Gamma({e-c+{\sigma}-{\gamma}}) = ({\sigma}-{\gamma})_{e-c}\Gamma({{\sigma}-{\gamma}}),\\ &\Gamma({d+e-a-b-c+{\delta}+{\sigma}-{\alpha}-\beta-{\gamma}})\\& = ({\delta}+{\sigma}-{\alpha}-\beta-{\gamma})_{d+e-a-b-c}\Gamma({{\delta}+{\sigma}-{\alpha}-\beta-{\gamma}}). \end{align*}$

and we realize that $({\delta})_{d+m+n} = ({\delta})_d({\delta}+d)_{m+n}$ when $d\in \Bbb N_0.$ Replacing $(a, b, c, d, e)$ by $(a+{\alpha}, b+\beta, c+{\gamma}, d+{\delta}, e+{\sigma})$ in (2.1) and substituting above identities into the resulting equation, we get the desired result.

3. Double series expansions for $1/\pi$

In this section we will use Theorem 1.3 to prove the following double series expansion formula for $1/\pi$ .

Theorem 3.1 If $d\in \Bbb N_0, Re(e-d+1) > 0$ and $Re(d+e-a-b-c+ \frac 32) > 0,$ then

$\sum\limits_{m,n = 0}^\infty \frac {( \frac 12)_{(a+m,b+m,c+m,d-a+n,d-b+n,d-c+n)}} {m!n!(d+1)_{m+n}(2)_{e+m}( \frac 32)_{d+e-a-b-c+n}} = \frac {( \frac 12)_{(a,b,c,d-a,d-b,d-c)}(1)_{e-d}} {( \frac 32)_{(e-a,e-b,e-c)}}\cdot \frac 4{\pi}.$

Proof. Let $({\alpha}, \beta, {\gamma}, {\delta}, {\sigma}) = ( \frac 12, \frac 12, \frac 12, 1, 2)$ in Theorem 1.3. We find that

$\begin{align} \sum\limits_{m,n = 0}^\infty \frac {( \frac 12)_{(a+m,b+m,c+m,d-a+n,d-b+n,d-c+n)}} {m!n!(d+1)_{m+n}(2)_{e+m}( \frac 32)_{d+e-a-b-c+n}} = \frac {( \frac 12)_{(a,b,c,d-a,d-b,d-c)}(1)_{e-d}} {( \frac 32)_{(e-a,e-b,e-c)}}\cdot \frac {\Gamma(2)\Gamma(1)\Gamma({ \frac 32})}{\Gamma^3( \frac 32)}. \end{align}$

(3.1)

Substituting $\Gamma({ \frac 32}) = \frac {\sqrt{\pi}}2$ into (3.1) we obtain the result immediately. Putting $(a, b, c) = (0, 0, 0)$ in Theorem 3.1 we get the following general double summation formula for $1/\pi$ with two free parameters.

Corollary 3.2 If $d\in \Bbb N_0, Re(e-d+1) > 0$ and $Re(d+e+ \frac 32) > 0,$ then

$\sum\limits_{m,n = 0}^\infty \frac {( \frac 12)_{(m,d+n)}^3} {m!n!(d+1)_{m+n}(2)_{e+m}( \frac 32)_{d+e+n}} = \frac {4( \frac 12)_d^3(1)_{e-d}}{\pi( \frac 32)_e^3}.$

Setting $d = 0$ and $e = k\in\Bbb N_0$ in Corallary 3.2 we have the following result.

Proposition 3.3 Let $k$ be a nonnegative integer. Then

$\sum\limits_{m,n = 0}^\infty \frac {( \frac 12)_{(m,n)}^3}{m!n!(m+n)!(m+k+1)!( \frac 32+k)_n} = \frac {4k!}{\pi( \frac 32)_k^2}.$

Example 3.1 ( $k = 0$ in Proposition 3.3).

$\sum\limits_{m,n = 0}^\infty \frac {( \frac 12)_m^3( \frac 12)_n^2}{m!n!(m+n)!(m+1)!(2n+1)} = \frac 4{\pi}.$

If $d = e = k\in\Bbb N_0$ in Corollary 3.2 we achieve

Proposition 3.4 Let $k$ be a nonnegative integer. Then

$\sum\limits_{m,n = 0}^\infty \frac {( \frac 12)_{(m,n+k)}^3} {m!n!(k+1)_{m+n}(m+k+1)!( \frac 32)_{n+2k}} = \frac 4{\pi(2k+1)^3}.$

If we put $k = 0$ into Proposition 3.4, then we can also get Example 3.1.

4. Double series expansions for $\pi$

In this section we will prove the following theorem, which allows us to derive infinitely double series expansions for $\pi$ .

Theorem 4.1 If $d\in \Bbb N_0, Re(e-d-{\sigma}+1) > 0$ and $Re(d+e-a-b-c+2) > 0,$ then

$\begin{align*} &\sum\limits_{m,n = 0}^\infty \frac {({\sigma}-1)_{(a+m,b+m,c+m)}({\sigma})_{(d-a+n,d-b+n,d-c+n)}} {m!n!(2{\sigma}+d-1)_{m+n}({\sigma})_{e+m}(2)_{d+e-a-b-c+n}}\\ & = \frac {({\sigma}-1)_{(a,b,c)}({\sigma})_{(d-a,d-b,d-c)}(1-{\sigma})_{e-d}}{(1)_{(e-a,e-b,e-c)}} \cdot \frac {\pi}{\sin{{\sigma}\pi}}. \end{align*}$

Proof. Let $({\alpha}, \beta, {\gamma}, {\delta}) = ({\sigma}-1, {\sigma}-1, {\sigma}-1, 2{\sigma}-1)$ in Theorem 1.3. We obtain that

$\begin{align*} &\sum\limits_{m,n = 0}^\infty \frac {({\sigma}-1)_{(a+m,b+m,c+m)}({\sigma})_{(d-a+n,d-b+n,d-c+n)}} {m!n!(2{\sigma}+d-1)_{m+n}({\sigma})_{e+m}(2)_{d+e-a-b-c+n}}\\ & = \frac {({\sigma}-1)_{(a,b,c)}({\sigma})_{(d-a,d-b,d-c)}(1-{\sigma})_{e-d}} {(1)_{(e-a,e-b,e-c)}} \cdot \frac {\Gamma({\sigma})\Gamma({1-{\sigma}})\Gamma({2})}{\Gamma^3(1)}. \end{align*}$

(4.1)

Combining $\Gamma({{\sigma}})\Gamma({1-{\sigma}}) = \frac \pi{\sin{\sigma}\pi}$ with (4.1) we get the desired result immediately. Putting $a = b = c = 0$ in Theorem 4.1 we obtain the following equation.

Corollary 4.2 If $d\in \Bbb N_0, Re(e-d-{\sigma}+1) > 0$ and $Re(d+e+2) > 0,$ then

$\sum\limits_{m,n = 0}^\infty \frac {({\sigma}-1)_m^3({\sigma})_{d+n}^3} {m!n!(2{\sigma}+d-1)_{m+n}({\sigma})_{e+m}(2)_{d+e+n}} = \frac {({\sigma})_d^3(1-{\sigma})_{e-d}}{(1)_e^3} \cdot \frac {\pi}{\sin{{\sigma}\pi}}.$

4.1. $\sigma=\frac{1}{2}$ in Corollary 4.2

Letting $\sigma=\frac{1}{2}$ in Corollary 4.2, we get the following proposition.

Proposition 4.3 If $d\in \Bbb N_0, Re(e-d+ \frac 12) > 0$ and $Re(d+e+2) > 0,$ then

$\sum\limits_{m,n = 0}^\infty \frac {(- \frac 12)_m^3( \frac 12)_{d+n}^3}{m!n!(d)_{m+n}( \frac 12)_{e+m} (2)_{d+e+n}} = \frac {( \frac 12)_d^3( \frac 12)_{e-d}}{(1)_{e}^3} \pi.$

When we set $d = 1$ and $e = k\in\Bbb N = \{1, 2, 3\ldots\}$ in Proposition 4.3 we obtain

Proposition 4.4 If $k$ is a positive integer, then

$\sum\limits_{m,n = 0}^\infty \frac {(- \frac 12)_m^3( \frac 32)_n^3}{m!n!(m+n)!(n+k+2)!( \frac 12)_{m+k}} = \frac {\pi( \frac 12)_{k-1}}{(k!)^3} .$

Example 4.1 ( $k = 1$ in Proposition 4.4).

$\sum\limits_{m,n = 0}^\infty \frac {(- \frac 12)_m^3( \frac 32)_n^3}{m!n!(m+n)!(n+3)!( \frac 12)_{m+1}} = \pi.$

4.2. $\sigma=\frac{1}{3}$ in Corollary 4.2

Putting $\sigma=\frac{1}{3}$ in Corollary 4.2, we get the following proposition.

Proposition 4.5 If $d\in \Bbb N_0, Re(e-d+ \frac 23) > 0$ and $Re(d+e+2) > 0,$ then

$\sum\limits_{m,n = 0}^\infty \frac {(- \frac 23)_m^3( \frac 13)_{d+n}^3}{m!n!(d- \frac 13)_{m+n}( \frac 13)_{e+m} (2)_{d+e+n}} = \frac {2\sqrt3\pi( \frac 13)_d^3( \frac 23)_{e-d}}{3(1)_{e}^3}.$

When we set $d = 0$ and $e = k\in\Bbb N_0$ in Proposition 4.5 we obtain

Proposition 4.6 If $k$ is a nonnegative integer, then

$\sum\limits_{m,n = 0}^\infty \frac {(- \frac 23)_m^3( \frac 13)_n^3}{m!n!(- \frac 13)_{m+n}( \frac 13)_{m+k}(n+k+1)!} = \frac {2\sqrt3\pi( \frac 23)_k}{3k!^3}.$

Example 4.2 ( $k = 0$ in Proposition 4.6).

$\sum\limits_{m,n = 0}^\infty \frac {(- \frac 23)_m^2( \frac 13)_n^3}{m!n!(n+1)!(2-3m)(- \frac 13)_{m+n}} = \frac {\sqrt3\pi}3.$

Setting $d = e = k\in\Bbb N_0$ in Proposition 4.5, we get

Proposition 4.7 If $k$ is a nonnegative integer, then

$\sum\limits_{m,n = 0}^\infty \frac {(- \frac 23)_m^3( \frac 13+k)_n^3}{m!n!(n+2k+1)!(k- \frac 13)_{m+n}( \frac 13)_{m+k} } = \frac {2\sqrt3\pi}{3k!^3}.$

Therefore, Example 4.2 can also be deduced by fixing $k = 0$ in the above equation.

Example 4.3 ( $k = 1$ in Proposition 4.7).

$\sum\limits_{m,n = 0}^\infty \frac {(- \frac 23)_m^3( \frac 43)_n^3}{m!n!(n+3)!( \frac 23)_{m+n} ( \frac 43)_m} = \frac {2\sqrt3\pi}9.$

5. Conclusions

Double series expansions for $1/\pi$ and $\pi$ with several free parameters are established and many interesting formulas are obtained. A point that should be stressed is that there is an important connection between the summation formulas for double hypergeometric functions and double series expansions for the powers of $\pi$ .

Acknowledgements

The author was partially supported by the Natural Science Foundation of the Jiangsu Higher Education Institutions of China (grant 19KJB110006).

Conflicts of interest

The author declares that there is no conflict of interest in this paper.

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沈阳化工大学材料科学与工程学院沈阳 110142

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AIMS Mathematics

Positive solutions to a semipositone superlinear elastic beam equation

Related Papers:

Abstract

1. Introduction

2. Proof of Theorem 1.3

3. Double series expansions for $1/\pi$

4. Double series expansions for $\pi$

4.1. $\sigma=\frac{1}{2}$ in Corollary 4.2

4.2. $\sigma=\frac{1}{3}$ in Corollary 4.2

5. Conclusions

Acknowledgements

Conflicts of interest

References

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通讯作者: 陈斌, bchen63@163.com

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AIMS Mathematics

Positive solutions to a semipositone superlinear elastic beam equation

Related Papers:

Abstract

1. Introduction

2. Proof of Theorem 1.3

3. Double series expansions for 1/π 1/\pi

4. Double series expansions for π \pi

4.1. σ=12 \sigma=\frac{1}{2} in Corollary 4.2

4.2. σ=13 \sigma=\frac{1}{3} in Corollary 4.2

5. Conclusions

Acknowledgements

Conflicts of interest

References

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3. Double series expansions for $1/\pi$

4. Double series expansions for $\pi$

4.1. $\sigma=\frac{1}{2}$ in Corollary 4.2

4.2. $\sigma=\frac{1}{3}$ in Corollary 4.2