We use some properties of gamma functions and a summation formula for Kampé de Fériet function F0:3;31:1;1 to give many double series expansions for 1/π and π.
Citation: Long Li. Double series expansions for π[J]. AIMS Mathematics, 2021, 6(5): 5000-5007. doi: 10.3934/math.2021294
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We use some properties of gamma functions and a summation formula for Kampé de Fériet function F0:3;31:1;1 to give many double series expansions for 1/π and π.
In [18] Ramanujan showed a total of 17 series for 1/π 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/π. Till now, many new Ramanujan's-type series for 1/π have been published, (see, for example, [4,6,8]). Chu [7], Liu [15,16] and Wei et al. [21,22] gave many π-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 π. Moreover, Guillera and Zudilin [11] outlined an elementary method for proving numerical hypergeometric identities, in particular, Ramanujan-type identities for 1/π. Recently, q-analogues of Ramanujan-type series for 1/π 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]
1Γ(z)=zeγz∞∏n=1(1+zn)e−zn |
where γ is the Euler constant defined as
γ=limn→∞(1+12+⋯+1n−logn). |
It is easy to verify that Γ(1)=1,Γ(12)=√π and Γ(z+1)=zΓ(z). It follows that for every positive integer n, Γ(n)=(n−1)!.
For any complex α, we define the general rising shifted factorial by
(z)α=Γ(z+α)/Γ(z). | (1.1) |
Obviously, (z)0=1. For every positive integer n, we have
(z)n=Γ(z+n)/Γ(z)=z(z+1)⋯(z+n−1) |
and
(z)−n=Γ(z−n)/Γ(z)=1(z−1)(z−2)…(z−n). |
For convenience, we use the following compact notations
(a1,a2,…,am)n=(a1)n(a2)n…(am)n |
and
(a)(n1,n2,…,nm)=(a)n1(a)n2…(a)nm. |
Following [1,3], the hypergeometric series is defined by
r+1Fs[a0,a1,…,arb1,…,bs;z]=∞∑k=0(a0,a1,…,ar)k(b1,…,bs)kzkk!, |
where ai,bj(i=0,1,…,r,j=1,2,…,s) are complex numbers such that no zero factors appear in the denominators of the summand on the right hand side.
We let Fp:r;uq:s;v (p,q,r,s,u,v∈N0={0,1,2,…}) denote a general (Kampé de Fériet's) double hypergeometric function defined by (see [2,19])
Fp:r;uq:s;v[α1,…,αp:a1,…,ar;c1,…,cu;β1,…,βq:b1,…,bs;d1,…,dv;x,y]=∞∑m,n=0(α1,…,αp)m+n(a1,…,ar)m(c1,…,cu)n(β1,…,βq)m+n(b1,…,bs)m(d1,…,dv)nxmm!ynn!, |
where, for convergence of the double hypergeometric series,
p+r≤q+s+1andp+u≤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 [17,(30)] ) If Re(e−d)>0 and Re(d+e−a−b−c)>0, then
F0:3;31:1;1[−:a,b,c;d−a,d−b,d−c;d:e;d+e−a−b−c;1,1]=Γ(e)Γ(e+d−a−b−c)Γ(e−d)Γ(e−a)Γ(e−b)Γ(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/π and π.
Theorem 1.2 For any complex α and Re(c−a−b)>0, we have
∞∑n=0(α)a+n(1−α)b+nn!Γ(c+n+1)=(α)a(1−α)bΓ(c−a−b)(α)c−b(1−α)c−a⋅sinπαπ. |
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/π and π. To the best of our knowledge, most of the results in this paper have not previously appeared.
Theorem 1.3 If d∈N0,Re(e−d+σ−δ)>0 and Re(d+e−a−b−c+δ+σ−α−β−γ)>0, then
∞∑m,n=0(α)a+m(β)b+m(γ)c+m(δ−α)d−a+n(δ−β)d−b+n(δ−γ)d−c+nm!n!(δ+d)m+n(σ)e+m(δ+σ−α−β−γ)d+e−a−b−c+n=(α)a(β)b(γ)c(δ−α)d−a(δ−β)d−b(δ−γ)d−c(σ−δ)e−d(σ−α)e−a(σ−β)e−b(σ−γ)e−c⋅Γ(σ)Γ(σ−δ)Γ(δ+σ−α−β−γ)Γ(σ−α)Γ(σ−β)Γ(σ−γ). |
Several examples of such formulae are
∞∑m,n=0(12)3m(12)2nm!n!(m+n)!(m+1)!(2n+1)=4π, |
∞∑m,n=0(−12)3m(32)3nm!n!(m+n)!(n+3)!(12)m+1=π, |
and
∞∑m,n=0(−23)2m(13)3nm!n!(n+1)!(2−3m)(−13)m+n=√3π3. |
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/π and π, respectively.
First of all, by making use of (1.1), Theorem 1.3 can be restated as follows:
∞∑m,n=0Γ(a+m)Γ(b+m)Γ(c+m)Γ(d−a+n)Γ(d−b+n)Γ(d−c+n)m!n!Γ(d+m+n)Γ(e+m)Γ(d+e−a−b−c+n)=Γ(a)Γ(b)Γ(c)Γ(d−a)Γ(d−b)Γ(d−c)Γ(e−d)Γ(d)Γ(e−a)Γ(e−b)Γ(e−c). | (2.1) |
From (1.1) it is easy to see that
Γ(a+α+m)=(α)a+mΓ(α), Γ(b+β+m)=(β)b+mΓ(β), Γ(c+γ+m)=(γ)c+mΓ(γ),Γ(d−a+δ−α+n)=(δ−α)d−a+nΓ(δ−α), Γ(d−b+δ−β+n)=(δ−β)d−b+nΓ(δ−β),Γ(d−c+δ−γ+n)=(δ−γ)d−c+nΓ(δ−γ), Γ(d+δ+m+n)=(δ)d+m+nΓ(δ)Γ(e+m+σ)=(σ)e+mΓ(σ), Γ(a+α)=(α)aΓ(α), Γ(b+β)=(β)bΓ(β), Γ(c+γ)=(γ)cΓ(γ),Γ(d−a+δ−α)=(δ−α)d−aΓ(δ−α), Γ(d−b+δ−β)=(δ−β)d−bΓ(δ−β),Γ(d−c+δ−γ)=(δ−γ)d−cΓ(δ−γ), Γ(e−d+σ−δ)=(σ−δ)e−dΓ(σ−δ),Γ(d+δ)=(δ)dΓ(δ),Γ(e−a+σ−α)=(σ−α)e−aΓ(σ−α),Γ(e−b+σ−β)=(σ−β)e−bΓ(σ−β), Γ(e−c+σ−γ)=(σ−γ)e−cΓ(σ−γ),Γ(d+e−a−b−c+δ+σ−α−β−γ)=(δ+σ−α−β−γ)d+e−a−b−cΓ(δ+σ−α−β−γ). |
and we realize that (δ)d+m+n=(δ)d(δ+d)m+n when d∈N0. Replacing(a,b,c,d,e) by (a+α,b+β,c+γ,d+δ,e+σ) in (2.1) and substituting above identities into the resulting equation, we get the desired result.
In this section we will use Theorem 1.3 to prove the following double series expansion formula for 1/π.
Theorem 3.1 If d∈N0,Re(e−d+1)>0 and Re(d+e−a−b−c+32)>0, then
∞∑m,n=0(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(32)d+e−a−b−c+n=(12)(a,b,c,d−a,d−b,d−c)(1)e−d(32)(e−a,e−b,e−c)⋅4π. |
Proof. Let (α,β,γ,δ,σ)=(12,12,12,1,2) in Theorem 1.3. We find that
∞∑m,n=0(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(32)d+e−a−b−c+n=(12)(a,b,c,d−a,d−b,d−c)(1)e−d(32)(e−a,e−b,e−c)⋅Γ(2)Γ(1)Γ(32)Γ3(32). | (3.1) |
Substituting Γ(32)=√π2 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/π with two free parameters.
Corollary 3.2 If d∈N0,Re(e−d+1)>0 and Re(d+e+32)>0, then
∞∑m,n=0(12)3(m,d+n)m!n!(d+1)m+n(2)e+m(32)d+e+n=4(12)3d(1)e−dπ(32)3e. |
Setting d=0 and e=k∈N0 in Corallary 3.2 we have the following result.
Proposition 3.3 Let k be a nonnegative integer. Then
∞∑m,n=0(12)3(m,n)m!n!(m+n)!(m+k+1)!(32+k)n=4k!π(32)2k. |
Example 3.1 (k=0 in Proposition 3.3).
∞∑m,n=0(12)3m(12)2nm!n!(m+n)!(m+1)!(2n+1)=4π. |
If d=e=k∈N0 in Corollary 3.2 we achieve
Proposition 3.4 Let k be a nonnegative integer. Then
∞∑m,n=0(12)3(m,n+k)m!n!(k+1)m+n(m+k+1)!(32)n+2k=4π(2k+1)3. |
If we put k=0 into Proposition 3.4, then we can also get Example 3.1.
In this section we will prove the following theorem, which allows us to derive infinitely double series expansions for π.
Theorem 4.1 If d∈N0,Re(e−d−σ+1)>0 and Re(d+e−a−b−c+2)>0, then
∞∑m,n=0(σ−1)(a+m,b+m,c+m)(σ)(d−a+n,d−b+n,d−c+n)m!n!(2σ+d−1)m+n(σ)e+m(2)d+e−a−b−c+n=(σ−1)(a,b,c)(σ)(d−a,d−b,d−c)(1−σ)e−d(1)(e−a,e−b,e−c)⋅πsinσπ. |
Proof. Let (α,β,γ,δ)=(σ−1,σ−1,σ−1,2σ−1) in Theorem 1.3. We obtain that
∞∑m,n=0(σ−1)(a+m,b+m,c+m)(σ)(d−a+n,d−b+n,d−c+n)m!n!(2σ+d−1)m+n(σ)e+m(2)d+e−a−b−c+n=(σ−1)(a,b,c)(σ)(d−a,d−b,d−c)(1−σ)e−d(1)(e−a,e−b,e−c)⋅Γ(σ)Γ(1−σ)Γ(2)Γ3(1). | (4.1) |
Combining Γ(σ)Γ(1−σ)=πsinσπ 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∈N0,Re(e−d−σ+1)>0 and Re(d+e+2)>0, then
∞∑m,n=0(σ−1)3m(σ)3d+nm!n!(2σ+d−1)m+n(σ)e+m(2)d+e+n=(σ)3d(1−σ)e−d(1)3e⋅πsinσπ. |
Letting σ=12 in Corollary 4.2, we get the following proposition.
Proposition 4.3 If d∈N0,Re(e−d+12)>0 and Re(d+e+2)>0, then
∞∑m,n=0(−12)3m(12)3d+nm!n!(d)m+n(12)e+m(2)d+e+n=(12)3d(12)e−d(1)3eπ. |
When we set d=1 and e=k∈N={1,2,3…} in Proposition 4.3 we obtain
Proposition 4.4 If k is a positive integer, then
∞∑m,n=0(−12)3m(32)3nm!n!(m+n)!(n+k+2)!(12)m+k=π(12)k−1(k!)3. |
Example 4.1 (k=1 in Proposition 4.4).
∞∑m,n=0(−12)3m(32)3nm!n!(m+n)!(n+3)!(12)m+1=π. |
Putting σ=13 in Corollary 4.2, we get the following proposition.
Proposition 4.5 If d∈N0,Re(e−d+23)>0 and Re(d+e+2)>0, then
∞∑m,n=0(−23)3m(13)3d+nm!n!(d−13)m+n(13)e+m(2)d+e+n=2√3π(13)3d(23)e−d3(1)3e. |
When we set d=0 and e=k∈N0 in Proposition 4.5 we obtain
Proposition 4.6 If k is a nonnegative integer, then
∞∑m,n=0(−23)3m(13)3nm!n!(−13)m+n(13)m+k(n+k+1)!=2√3π(23)k3k!3. |
Example 4.2 (k=0 in Proposition 4.6).
∞∑m,n=0(−23)2m(13)3nm!n!(n+1)!(2−3m)(−13)m+n=√3π3. |
Setting d=e=k∈N0 in Proposition 4.5, we get
Proposition 4.7 If k is a nonnegative integer, then
∞∑m,n=0(−23)3m(13+k)3nm!n!(n+2k+1)!(k−13)m+n(13)m+k=2√3π3k!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).
∞∑m,n=0(−23)3m(43)3nm!n!(n+3)!(23)m+n(43)m=2√3π9. |
Double series expansions for 1/π and π 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 π.
The author was partially supported by the Natural Science Foundation of the Jiangsu Higher Education Institutions of China (grant 19KJB110006).
The author declares that there is no conflict of interest in this paper.
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