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Research article Special Issues

Evaluations of some Euler-type series via powers of the arcsin function

  • In this paper, by using special integrals and integer powers of the arcsin function, we obtained the recurrences or explicit expressions of some parametric Euler-type series involving multiple harmonic sums and multiple t-harmonic sums. According to our results, these Euler-type series are all expressible in terms of π, ln(2), and zeta values. In particular, by specifying the parameters, we presented as examples the evaluations of some special series, including some known ones in the literature and some new ones.

    Citation: Jiaye Lin. Evaluations of some Euler-type series via powers of the arcsin function[J]. AIMS Mathematics, 2025, 10(4): 8116-8130. doi: 10.3934/math.2025372

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  • In this paper, by using special integrals and integer powers of the arcsin function, we obtained the recurrences or explicit expressions of some parametric Euler-type series involving multiple harmonic sums and multiple t-harmonic sums. According to our results, these Euler-type series are all expressible in terms of π, ln(2), and zeta values. In particular, by specifying the parameters, we presented as examples the evaluations of some special series, including some known ones in the literature and some new ones.



    Generalized harmonic numbers and odd harmonic numbers are defined by

    H(s)n:=nk=11ksandO(s)n:=nk=11(2k1)s,

    for nN0 and sN. Multiple harmonic sums (MHSs) ζn(s) and multiple t-harmonic sums (MtSs) tn(s) are defined by

    ζn(s):=nn1>>nk11ns11nskkandtn(s):=nn1>>nk1kj=11(2nj1)sj, (1.1)

    for nN0 and s=(s1,,sk)Nk. By convention, ζn()=tn():=1 for n0, and ζn(s)=tn(s):=0 for n<k. When k=1, the MHSs and MtSs reduce respectively, to harmonic numbers and odd harmonic numbers, i.e., ζn(s)=H(s)n and tn(s)=O(s)n. Moreover, for s1>1, define the multiple zeta values (MZVs) and multiple t-values (MtVs) by

    ζ(s):=n1>>nk11ns11nskkandt(s):=n1>>nk1kj=11(2nj1)sj, (1.2)

    which can be viewed as the limits of MHSs and MtSs, respectively. The concept of MZVs was introduced by Hoffman [1] and Zagier [2] in the early 1990s, and the concept of MtVs was formally introduced by Hoffman [3] in 2019. Similarly to MZVs and MtVs, let s=(s1,,sk)Nk and σ=(σ1,,σk){±1}k with (s1,σ1)(1,1), and we define the alternating multiple zeta values (AMZVs) by

    ζ(s;σ):=n1>>nk1σn11σnkkns11nskk.

    Summation formulas on the (alternating) multiple zeta values have attracted extensive attention, and the readers are referred to the papers due to, e.g., Alegri et al. [4], Hoffman [1,5], and Shen and Cai [6].

    The linear Euler sums Sp,q:=n=1H(p)n/nq were first considered by Euler in correspondence with Goldbach in 1742 (see [7, p. 253]). Euler discovered that the linear sums Sp,q can be evaluated in terms of zeta values when p=1, p=q, p+q is odd, and (p,q){(2,4),(4,2)}. The nonlinear Euler sums were introduced by Flajolet and Salvy [8] in 1998. These sums are infinite series of the form

    Sp1p2pk,q:=n=1H(p1)nH(p2)nH(pk)nnq,

    where p1,p2,,pk,qN, with p1p2pk and q2. Flajolet and Salvy used the contour integral representations and residue computation to establish explicit formulas of several classes of Euler sums. In 2020, by permutations and compositions, Xu and Wang[9] established explicit formulas of Euler sums Sp1p2pk,q and various alternating Euler sums, and showed that these infinite series are expressible in terms of MZVs or AMZVs, respectively. More classical results on Euler sums can be found in, e.g., the works due to Bailey et al. [10], Borwein et al. [11], Chu [12,13], Wang and Lyu [14], Xu[15], Xu et al. [16], and Zheng [17].

    Recently, various Euler-type series have been investigated, including Euler sums involving negative powers of two [18], Euler-type series on odd harmonic numbers [19], and Euler-type series on hyperharmonic numbers [20,21]. Additionally, in 2017, Hoffman [22] established some general series identities by symmetric functions and MZVs. For example, he showed that

    n=1Pk(Hn,H(2)n,,H(k)n)n2=k+32ζ(k+2)12kj=2ζ(j)ζ(k+2j),n=0Pk(Hn,H(2)n,,H(k)n)(n+1)(n+2)(n+q)=1(q1)!(q1)k+1,

    for integers k0 and q2. According to the definition of the multivariate polynomials Pk, it can be verified that ζn({1}k)=Pk(Hn,H(2)n,,H(k)n), where {}k denotes k repetitions of a substring, with {}0=. Therefore, Hoffman's results can also be viewed as Euler-type series identities. Kuba and Panholzer [23], and Ma and Wang [24], further generalized Hoffman's results, and established more infinite series identities on Euler-type series involving ζn({1}k) and ζn({1}k), where the star version ζn(s) was obtained from the definition of ζn(s) by replacing n>n1>>nk1 by nn1nk1. More recent works on various Euler-type series can be found in the papers of Campbell and Chen [25], Chen and Wang [26], Liu and Wang [27], and Liu et al. [28]. Note that infinite series involving (odd) harmonic numbers and central binomial coefficients have also attracted extensive attention recently. For example, Campbell et al.[29] used the theory of colored multiple zeta values and various special integrals to evaluate this kind of infinite series, and Nimbran et al.[30] used the series expansions of integer powers of the arcsin function to determine the values of such series. More recent results of this type can be found in, e.g., the papers of Wang and Xu[31], Wang and Yuan[32], and Wei and Xu[33].

    In this paper, we will use the series expansions of arcsinn(x) as well as some integrals of trigonometric functions and inverse trigonometric functions to obtain the explicit evaluations of more Euler-type series. For n=1,2,3, the series expansions of arcsinn(x) were given in Edwards' textbook [34, pp. 77-82], and the expansion of arcsin4(x) was recorded in Ramanujan's notebooks [7, Proposition 15]. In the classical book [35, Eq (46)] of Schwatt, a general formula for the series expansion of arcsinn(x), for nN, was presented. However, two elegant formulas of arcsinn(x) were presented by Borwein and Chamberland [36, Eqs (1.1) and (1.3)], which can be reformulated as

    arcsin2p+1(x)(2p+1)!=n=p(2nn)4ntn({2}p)(2n+1)x2n+1,forp0, (1.3)
    arcsin2p(x)(2p)!=n=p4np(2nn)ζn1({2}p1)n2x2n,forp1. (1.4)

    By means of the expansions (1.3) and (1.4), the following parametric series involving MHSs ζn({2}p) and MtSs tn({2}p) will be studied in this paper:

    n=pζn({2}p)(n+1)(n+m+1),n=ptn({2}p)(2n+1)(2n+2m+1), (1.5)
    n=pζn({2}p)(2n+k)(2n+k+1),n=ptn({2}p)(2n+k1)(2n+k), (1.6)

    for mN, pN0, and k=0,1,2. Note that according to our previous results [37, Lemma 2.1], ζn({r}k) and tn({r}k) can be expressed in terms of harmonic numbers and odd harmonic numbers by the complete Bell polynomials:

    ζn({r}k)=1k!Yk(H(r)n,1!H(2r)n,,(1)k1(k1)!H(kr)n),tn({r}k)=1k!Yk(O(r)n,1!O(2r)n,,(1)k1(k1)!O(kr)n).

    For example,

    ζn(r)=H(r)n,ζn(r,r)=12{(H(r)n)2H(2r)n},tn(r)=O(r)n,tn(r,r)=12{(O(r)n)2O(2r)n}.

    Therefore, the parametric series in (1.5) and (1.6) are indeed Euler-type. We establish in Section 2 the recurrences of the series in (1.5) by the integral π20θnsin2m(θ)dθ, and derive in Sections 3–5 the explicit expressions of the series in (1.6) by the integrals θ0arcsinn(x)/x3dx, π20θn/sin2(θ)dθ, and θ0tncos(t)dt. The results show that the above general Euler-type series are all expressible in terms of powers of π, ln(2), and zeta values. By specifying the parameters, we present as examples the evaluations of some special Euler-type series, including some known ones appearing in recent papers due to Hoffman [22], Campbell and Chen [25], and Chen and Wang [26].

    In this section, we consider two kinds of infinite series Z(p)m and T(p)m, defined by

    Z(p)m:=n=pζn({2}p)(n+1)(n+m+1)andT(p)m:=n=ptn({2}p)(2n+1)(2n+2m+1),

    for integers m1 and p0. These series are closely related to the integral

    I(n,m):=1n!π20θnsin2m(θ)dθ,forn,m0.

    According to Orr's result [38, Eq (14)], the following expression for I(n,m) holds:

    I(n,m)=πn+1(2mm)22m+n+1(n+1)!mk=1n12j=1(1)j(2mm+k)πn2j+122m+nk2j(n+12j)!+δn+12,n+12mk=1(1)k+n+12(2mm+k)22m+nkn+1.

    In the next two theorems, we will show that by means of the integral I(n,m), the recurrences of the series Z(p)m and T(p)m can be established.

    Theorem 2.1. For integers m1 and p0, the series Z(p)m satisfy the following recurrence:

    mk=1a(m,k)kZ(p)k=(2mm)π2p+222m1(2p+3)!4p+2πI(2p+2,m), (2.1)

    where

    a(m,k)=2k4m(2m2kmk)(2kk).

    Therefore, the series Z(p)m is expressible in terms of powers of π.

    Proof. Applying a change of variable xsin(θ) to Eq (1.4) gives

    θ2p(2p)!=n=p4np(2nn)ζn1({2}p1)n2sin2n(θ). (2.2)

    Multiply (2.2) with sin2m(θ), integrate from 0 to π2, and use Wallis' formula. Then we obtain

    I(2p,m)=n=p4np(2nn)ζn1({2}p1)n2π20sin2n+2m(θ)dθ=π2n=p(2n+2mn+m)(2nn)ζn1({2}p1)4p+mn2=π2n=pζn1({2}p1)4pn2mk=1(112n+2k).

    Define f2n(m)=mk=1(112n+2k). Based on one of our previous results [37, Lemma 4.1], fl(m) can be expanded as fl(m)=1+mk=1a(m,k)l+2k, with a(m,k)=2k4m(2m2kmk)(2kk). Substituting this expansion back, and using the definition of MZVs as well as the method of partial fraction decomposition, we have

    I(2p,m)=π2n=pζn1({2}p1)4pn2(1+mk=1a(m,k)2n+2k)=π24pn=pζn1({2}p1)n2+π4p+1n=pmk=1a(m,k)kζn1({2}p1)(1n21n(n+k))=πζ({2}p)24p+πζ({2}p)4p+1mk=1a(m,k)kπ4p+1mk=1a(m,k)kZ(p1)k.

    The summation in the second term can be computed by Gould [39, Eq (3.90)]:

    mk=1a(m,k)k=24mmk=1(2m2kmk)(2kk)=2(2mm)4m2, (2.3)

    which, together with the fact that ζ({2}p)=π2p(2p+1)! (see [40, Proposition 2.3]), gives the desired recurrence.

    Example 2.1. When m=1, Eq (2.1) gives the following explicit expression:

    n=pζn({2}p)(n+1)(n+2)=p+1j=1(1)j+1π2p+22j(2p+32j)!. (2.4)

    Setting p=1,2 in the above expression yields

    n=0H(2)n(n+1)(n+2)=π261,n=0ζn(2,2)(n+1)(n+2)=π4120π26+1. (2.5)

    Note that Eq (2.5) was presented by Hoffman [22, Corollary 1]. Similarly, when m=2,3, setting p=1,2 gives

    n=0H(2)n(n+1)(n+3)=π281116, (2.6)
    n=0ζn(2,2)(n+1)(n+3)=π416011π296+4364,n=0ζn(2,2)(n+1)(n+4)=11π42160341π23888+1181323328, (2.7)

    where Eqs (2.6) and (2.7) can be found in Sofo [41, Lemma 1.1].

    In the next theorem, we derive the recurrence of the series T(p)m.

    Theorem 2.2. For integers m1 and p0, the series T(p)m satisfies the following recurrence:

    mk=1a(m,k)kT(p)k=(2mm)(π2)2p+222m1(2p+2)!2I(2p+1,m), (2.8)

    where a(m,k)=2k4m(2m2kmk)(2kk). Therefore, the series T(p)m is expressible in terms of powers of π.

    Proof. Applying a change of variable xsin(θ) to Eq (1.3), the series expansion turns into

    θ2p+1(2p+1)!=n=p(2nn)4ntn({2}p)(2n+1)sin2n+1(θ). (2.9)

    Multiply (2.9) with sin2m(θ), integrate from 0 to π2, and use Wallis' formula as well as the expansion of fl(m). Then we obtain

    I(2p+1,m)=n=p(2nn)tn({2}p)4n(2n+1)π20sin2n+2m+1(θ)dθ=4mn=p(2nn)tn({2}p)(2n+2mn+m)(2n+1)(2n+2m+1)=n=ptn({2}p)(2n+1)2f2n+1(m)=n=ptn({2}p)(2n+1)2+mk=1a(m,k)n=ptn({2}p)(2n+1)2(2n+2k+1). (2.10)

    By partial fraction decomposition and the definition of MtVs, we have

    n=ptn({2})p(2n+1)2=t({2}p+1),n=ptn({2}p)(2n+1)2(2n+2k+1)=12k(t({2}p+1)T(p)k).

    Setting m=0 in (2.10) gives t({2}p)=1(2p)!(π2)2p (see also [42, Lemma 1]). Combining the above results with (2.3) yields Eq (2.8).

    Example 2.2. When m=1, Eq (2.8) can be reformulated as

    n=0tn({2}p)(2n+1)(2n+3)=pj=1(1)jπ2p+22j4p+1(2p+22j)!+(1)p4p+1.

    In particular, setting p=1,2 yields

    n=0O(2)n(2n+1)(2n+3)=π23218,n=0tn(2,2)(2n+1)(2n+3)=π41536π2128+132. (2.11)

    Equation (2.11) was presented by Chen and Wang [26, Example 4.6]. Similarly, setting m=2, p=1,2 and m=3, p=1,2 in Eq (2.8) gives

    n=0O(2)n(2n+1)(2n+5)=3π2128112,n=0tn(2,2)(2n+1)(2n+5)=π4204811π22048+148,n=0O(2)n(2n+1)(2n+7)=11π25762033240,n=0tn(2,2)(2n+1)(2n+7)=11π427648341π282944+1823116640.

    Based on Eqs (1.3) and (1.4) as well as the integral sin(θ)0arcsinn(x)x3dx, we can establish the expressions of another two kinds of infinite series.

    Theorem 3.1. For an integer p0, the following series are expressible in terms of powers of π, ln(2), and zeta values:

    n=pζn({2}p)2n(2n+1)=pk=1(1)k(2k+1)(4k1)π2pζ(2k+1)(2p2k)!+(1)pζ(2p+1)πln(2)(2p+1)! (3.1)

    and

    n=ptn({2}p)2n(2n1)=pk=1(1)k(2k+1)(14k)(π2)2p2kζ(2k+1)(2p2k)!16k+(1)p1(2p+1)ζ(2p+1)4p(π2)2pln(2)(2p)!. (3.2)

    Proof. Consider the integral 1(2p)!sin(θ)0arcsin2p(x)x3dx. On the one hand, by integration by parts twice, we have

    1(2p)!sin(θ)0arcsin2p(x)x3dx=θ2p2(2p)!sin2(θ)θ2p1cot(θ)2(2p1)!+12(2p2)!sin(θ)0arcsin2p2(x)xdx.

    On the other hand, expanding arcsin2p(x) by (1.4), using partial fraction decomposition, and applying Eq (2.2), the integral on the left can be rewritten as

    1(2p)!sin(θ)0arcsin2p(x)x3dx=n=p4np(2nn)ζn1({2}p1)sin2n2(θ)n2(2n2)=n=p4np(2nn)(ζn1({2}p1)sin2n2(θ)n(2n2)ζn1({2}p1)sin2n2(θ)2n2)=n=p4np(2nn)ζn1({2}p1)sin2n2(θ)n(2n2)θ2p2(2p)!sin2(θ).

    Combining the above two expressions gives

    n=p4np(2nn)ζn1({2}p1)sin2n2(θ)n(2n2)=12(2p2)!sin(θ)0arcsin2p2(x)xdxθ2p1cot(θ)2(2p1)!. (3.3)

    Now, integrate both sides of Eq (3.3) from 0 to π2. Then by Wallis' formula, the left-hand side of (3.3) turns into

    n=p4np(2nn)ζn1({2}p1)n(2n2)π20sin2n2(θ)dθ=π4pn=pζn1({2}p1)(2n2)(2n1).

    The integral of the first term on the right-hand side of (3.3) can be computed by the following formula (see [37, Eq (5.2)]):

    ζ(3,{2}p)=4p+2(2p+2)!ππ20dθsin(θ)0arcsin2p+2(x)xdx=p+1n=1(1)nπ2p+22n(122n)(2p+32n)!22n2nζ(2n+1)+(1)p+12ζ(2p+3),

    and the integral of the second term can be evaluated by Orr's result [43, Eq (2.2)]:

    π20θncot(θ)dθ=(π2)nln(2)+n2k=1(1)kn!(14k)πn2kζ(2k+1)(n2k)!2n+δn2,n2(1)n2n!ζ(n+1)2n. (3.4)

    Substituting the above formulas into (3.3) gives (3.1).

    Similarly, expanding the integral 1(2p+1)!sin(θ)0arcsin2p+1(x)x3dx in two different ways yields

    n=p(2nn)4ntn({2}p)sin2n1(θ)2n1=1(2p1)!sin(θ)0arcsin(x)2p1xdxθ2pcot(θ)(2p)! (3.5)

    (see also [37, Theorem 3.7]). Integrate both sides of (3.5) from 0 to π2. Using Wallis' formula for the left, and the integral expression of t(3,{2}p) [37, Eq (5.1)] as well as the integral (3.4) for the right, we obtain Eq (3.2).

    Example 3.1. Setting p=1,2,3 in Eq (3.1) gives

    n=1H(2)n2n(2n+1)=π2ln(2)6+5ζ(3)4,n=1ζn(2,2)2n(2n+1)=π4ln(2)120+3π2ζ(3)859ζ(5)16,n=1ζn(2,2,2)2n(2n+1)=π6ln(2)5040+3π4ζ(3)16025π2ζ(5)32+377ζ(7)64.

    Setting p=1,2,3 in Eq (3.2) yields

    n=1O(2)n2n(2n1)=π2ln(2)8+21ζ(3)16,n=1tn(2,2)2n(2n1)=π4ln(2)384+9π2ζ(3)128155ζ(5)256,n=1tn(2,2,2)2n(2n1)=π6ln(2)46080+3π4ζ(3)204875π2ζ(5)2048+889ζ(7)4096.

    In this section, we establish the expressions of two Euler-type series by the integral of θn/sin2(θ).

    Theorem 4.1. For an integer p0, the following series are expressible in terms of powers of π, ln(2), and zeta values:

    n=pζn({2}p)(2n+1)(2n+2)=π2pln(2)(2p+1)!+pk=1(1)k(14k)π2p2kζ(2k+1)(2p+12k)! (4.1)

    and

    n=ptn({2}p)2n(2n+1)=pk=1(1)k(4k1)(π2)2p2kζ(2k+1)(2p2k)!16k+(π2)2pln(2)(2p)!+(1)pζ(2p+1)4p. (4.2)

    Proof. Consider the integral π20θnsin2(θ)dθ. On the one hand, by integration by parts, we have

    π20θnsin2(θ)dθ=π20θnd(cot(θ))=nπ20θn1cot(θ)dθ, (4.3)

    which can be computed by Eq (3.4). On the other hand, expanding θ2p by Eq (2.2) and using Wallis' formula, we obtain

    π20θ2psin2(θ)dθ=π2(2p)!4p1n=pζn1({2}p1)2n(2n1),

    which, together with (4.3), gives Eq (4.1). Similarly, expanding θ2p+1 by Eq (2.9), using Wallis' formula, and applying Eq (4.3), we obtain Eq (4.2).

    Example 4.1. Setting p=1,2,3 in Eq (4.1) gives

    n=0H(2)n(2n+1)(2n+2)=π2ln(2)63ζ(3)4,n=0ζn(2,2)(2n+1)(2n+2)=π4ln(2)120π2ζ(3)8+15ζ(5)16,n=0ζn(2,2,2)(2n+1)(2n+2)=π6ln(2)5040π4ζ(3)160+5π2ζ(5)3263ζ(7)64. (4.4)

    Here, Eq (4.4) can be found in Campbell and Chen [25, p. 15], and Chen and Wang [26, Example 4.1]. Setting p=1,2,3 in Eq (4.2) yields

    n=1O(2)n2n(2n+1)=π2ln(2)87ζ(3)16,n=1tn(2,2)2n(2n+1)=π4ln(2)3843π2ζ(3)128+31ζ(5)256,n=1tn(2,2,2)2n(2n+1)=π6ln(2)46090π4ζ(3)2048+15π2ζ(5)2048127ζ(7)4097.

    In this section, we establish two series identities based on Eqs (2.2) and (2.9), which give the evaluations of the following two Euler-type series.

    Theorem 5.1. For an integer p0, the following series are expressible in terms of powers of π, ln(2), and zeta values:

    n=pζn({2}p)(2n+2)(2n+3)=pk=0pkj=1(1)k+j(4kj4k)π2p2k2jζ(2j+1)(2p+12k2j)!+(1ln(2))pk=0(4)kπ2p2k(2p+12k)! (5.1)

    and

    n=ptn({2}p)(2n+1)(2n+2)=(1)pln(2)+(1ln(2))p1k=0(1)k(π2)2p2k(2p2k)!(1)ppk=1ζ(2k+1)4k+p1k=0pkj=1(1)k+j(14j)(π2)2p2k2jζ(2j+1)(2p2k2j)!16j. (5.2)

    Proof. On the one hand, by integration by parts, we have

    1(2p)!θ0t2pcos(t)dt=pk=0(1)kθ2p2ksin(θ)(2p2k)!+p1k=0(1)kθ2p2j1cos(θ)(2p2k1)!.

    On the other hand, replace θ by t in (2.2), multiply both sides by cos(t), and integrate from 0 to θ. Then we have

    1(2p)!θ0t2pcos(t)dt=n=p4np(2nn)ζn1({2}p1)n2(2n+1)sin2n+1(θ)=n=p4np(2nn)ζn1({2}p1)n2sin2n+1(θ)2n=p4np(2nn)ζn1({2}p1)n(2n+1)sin2n+1(θ)=θ2psin(θ)(2p)!2n=p4np(2nn)ζn1({2}p1)n(2n+1)sin2n+1(θ),

    where the last two steps can be obtained by partial fraction decomposition and Eq (2.9). Combining the above two results gives us an identity. Divide both sides of this identity by sin(θ), and integrate from 0 to π2. Then we have

    2n=p4np(2nn)ζn1({2}p1)n(2n+1)π20sin2n(θ)dθ=pk=1(1)k+1(2p2k)!π20θ2p2kdθ+p1k=0(1)k+1(2p2k1)!π20θ2p2k1cot(θ)dθ.

    Using Wallis' formula for the left, and applying the integral (3.4) for the second term on the right, we obtain Eq (5.1). Similarly, by means of the integral π20t2p+1cos(t)dt, we have

    n=p(2nn)4ntn({2}p)sin2n+2(θ)2n+2=pk=1(1)k+1θ2p2k+1sin(θ)(2p2k+1)!+pk=0(1)k+1θ2p2kcos(θ)(2p2k)!+(1)p (5.3)

    (see also [37, Theorem 3.8]). Divide (5.3) by sin(θ) and integrate from 0 to π2. The integral of the left-hand side is easy to calculate by Wallis' formula. The right-hand side turns into

    pk=1(1)k+1(2p2k+1)!π20θ2p2k+1dθ+p1k=0(1)k+1(2p2k)!π20θ2p2kcot(θ)dθ+(1)p+1π20cos(θ)1sin(θ)dθ.

    The first and third terms can be computed directly. For the second term, use the integral (3.4). Finally, we obtain Eq (5.2).

    Example 5.1. Setting p=1,2 in Eq (5.1) gives

    n=0H(2)n(2n+2)(2n+3)=π2ln(2)6+π26+3ζ(3)4+4ln(2)4,n=0ζn(2,2)(2n+2)(2n+3)=π4ln(2)120+π412015ζ(5)16+π2ζ(3)82π233ζ(3)+2π2ln(2)316ln(2)+16, (5.4)

    where Eq (5.4) can be found in Chen and Wang [26, Example 4.2]. Setting p=1,2 in Eq (5.2) yields

    n=0O(2)n(2n+1)(2n+2)=π2ln(2)8+π28+7ζ(3)16ln(2),n=0tn(2,2)(2n+1)(2n+2)=π4ln(2)384+π43847ζ(3)16+3π2ζ(3)128+π2ln(2)8π2831ζ(5)256+ln(2).

    In this paper, by using integer powers of the arcsin function and special integrals, we obtained the explicit expressions or recurrences of some parametric Euler-type series:

    n=pζn({2}p)(n+1)(n+m+1),n=ptn({2}p)(2n+1)(2n+2m+1),n=pζn({2}p)(2n+k)(2n+k+1),n=ptn({2}p)(2n+k1)(2n+k),

    for mN, pN0, and k=0,1,2. These Euler-type series are all expressible in terms of π, ln(2), and zeta values.

    The author declares he (she) have not used Artificial Intelligence (AI) tools in the creation of this article.

    The author would like to thank the anonymous referees for their valuable comments and suggestions and Professor Weiping Wang for his valuable discussions and kind help.

    The author declares there is no conflict of interest.



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